p4 oscillations all

7/21/2019 P4 Oscillations All.

http://slidepdf.com/reader/full/p4-oscillations-all 1/69

14 (a) (i) Define simple harmonic motion.

...................................................................................................................................

...................................................................................................................................

...................................................................................................................................

(ii) On the axes of Fig. 4.1, sketch the variation with displacement x of the accelerationa of a particle undergoing simple harmonic motion.

[4]Fig. 4.1

(b) A strip of metal is clamped to the edge of a bench and a mass is hung from its free end

as shown in Fig.4.2.

Fig. 4.2

clamp

metal strip

mass

0

a

x 0

9702/4 M/J/02

ai s f m a s o o @ gm ai l . c om

C el l N o. 0

3 4 6 7 0 3 0 0 3 9



2

The end of the strip is pulled downwards and then released. Fig. 4.3 shows the variationwith time t of the displacement y of the end of the strip.

Fig. 4.4

On Fig. 4.4, show the corresponding variation with time t of the potential energy E p ofthe vibrating system. [3]

(c) The string supporting the mass breaks when the end of the strip is at its lowest point inan oscillation. Suggest what change, if any, will occur in the period and amplitude of thesubsequent motion of the end of the strip.

period: ..............................................................................................................................

amplitude: .....................................................................................................................[2]

E p

0.05

0

0.10 0.15 0.20 t / s

y

0

0.05

0

0.10 0.15 0.20 t / s0

Fig. 4.3 ai s f m a s o o @ gm ai l . c om

C el l N o. 0

3 4 6 7 0 3 0 0 3 9



33 A student sets out to investigate the oscillation of a mass suspended from the free end of a

spring, as illustrated in Fig. 3.1.

Fig. 3.1

The mass is pulled downwards and then released. The variation with time t of thedisplacement y of the mass is shown in Fig. 3.2.

Fig. 3.2

(a) Use information from Fig.3.2

(i) to explain why the graph suggests that the oscillations are undamped,

...................................................................................................................................

2

1

-1

-2

00

0.5 1.0 1.5 2.0 2.5 t /s

y/ cm

spring

mass oscillationof mass

9702/4 O/N/02


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



4

(ii) to calculate the angular frequency of the oscillations,

angular frequency = ........................................ rads–1

(iii) to determine the maximum speed of the oscillating mass.

speed = ........................................ ms–1

[6]

(b) (i) Determine the resonant frequency f 0 of the mass-spring system.

f 0 = ........................................ Hz

(ii) The student finds that if short impulsive forces of frequency f 0

are impressed onthe mass-spring system, a large amplitude of oscillation is obtained. Explain this

observation.

...................................................................................................................................

...................................................................................................................................

...................................................................................................................................[3]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



53 An aluminium sheet is suspended from an oscillator by means of a spring, as illustrated in

Fig. 3.1.

Fig. 3.1

An electromagnet is placed a short distance from the centre of the aluminium sheet.

The electromagnet is switched off and the frequency f of oscillation of the oscillator isgradually increased from a low value. The variation with frequency f of the amplitude a ofvibration of the sheet is shown in Fig. 3.2.

Fig. 3.2

a

00.9f 0 f 0 f

oscillator

spring

aluminiumsheet

electromagnet

9702/4/M/J03


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



6

A peak on the graph appears at frequency f 0.

(a) Explain why there is a peak at frequency f 0.

..........................................................................................................................................

..........................................................................................................................................

..................................................................................................................................... [2]

(b) The electromagnet is now switched on and the frequency of the oscillator is againgradually increased from a low value. On Fig. 3.2, draw a line to show the variation withfrequency f of the amplitude a of vibration of the sheet. [3]

(c) The frequency of the oscillator is now maintained at a constant value. The amplitude ofvibration is found to decrease when the current in the electromagnet is switched on.

Use the laws of electromagnetic induction to explain this observation.

..........................................................................................................................................

..........................................................................................................................................

..........................................................................................................................................

..........................................................................................................................................

..........................................................................................................................................

..................................................................................................................................... [4]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



7


C el l N o. 0

3 4 6 7 0 3 0 0 3 9




C el l N o. 0

3 4 6 7 0 3 0 0 3 9

8




C el l N o. 0

3 4 6 7 0 3 0 0 3 9

94 A vertical spring supports a mass, as shown in Fig. 4.1.

Fig. 4.1

The mass is displaced vertically then released. The variation with time t of the displacementy from its mean position is shown in Fig. 4.2.

Fig. 4.2

y / cm+2

0

–1

–2

+1

1.41.21.00.80.60.40.20

t / s

A student claims that the motion of the mass may be represented by the equation

y = y 0sinω t .

(a) Give two reasons why the use of this equation is inappropriate.

1. .....................................................................................................................................

..........................................................................................................................................

2. .....................................................................................................................................

.................................................................................................................................... [2]

9702/04/M/J/04




C el l N o. 0

3 4 6 7 0 3 0 0 3 9

10

(b) Determine the angular frequency ω of the oscillations.

angular frequency = .................................. rad s–1 [2]

(c) The mass is a lump of plasticine. The plasticine is now flattened so that its surface areais increased. The mass of the lump remains constant and the large surface area ishorizontal.

The plasticine is displaced downwards by 1.5 cm and then released.On Fig. 4.2, sketch a graph to show the subsequent oscillations of the plasticine. [3]

3 The vibrations of a mass of 150 g are simple harmonic. Fig. 3.1 shows the variation withdisplacement x of the kinetic energy E k of the mass.

Fig. 3.1

(a) On Fig. 3.1, draw lines to represent the variation with displacement x of

(i) the potential energy of the vibrating mass (label this line P),

(ii) the total energy of the vibrations (label this line T).

[2]

(b) Calculate the angular frequency of the vibrations of the mass.

0-2-4-6 2 4 60

4

8

12

16

E k /mJ

x / cm

angular frequency = ......................................... rads–1 [3]

9702/04/O/N/04




C el l N o. 0

3 4 6 7 0 3 0 0 3 9

11

(c) The oscillations are now subject to damping.

(i) Explain what is meant by damping .

...................................................................................................................................

...................................................................................................................................

...............................................................................................................................[2]

(ii) The mass loses 20% of its vibrational energy. Use Fig. 3.1 to determine the newamplitude of oscillation. Explain your working.

amplitude = ............................................... cm [2]

4 A tube, closed at one end, has a constant area of cross-section A. Some lead shot is placedin the tube so that the tube floats vertically in a liquid of density ρ , as shown in Fig. 4.1.

Fig. 4.1

The total mass of the tube and its contents is M .When the tube is given a small vertical displacement and then released, the verticalacceleration a of the tube is related to its vertical displacement y by the expression

a = – y ,

where g is the acceleration of free fall.

Aρ g

M

tu e, area o

cross-section A

lead shot

liquid,density

9702/04/M/J/05




C el l N o. 0

3 4 6 7 0 3 0 0 3 9

12

(a) Define simple harmonic motion .

..........................................................................................................................................

..........................................................................................................................................

......................................................................................................................................[2]

(b) Show that the tube is performing simple harmonic motion with a frequency f given by

f = .Aρ g

M

1

2π

(c) Fig. 4.2 shows the variation with time t of the vertical displacement y of the tube inanother liquid.

Fig. 4.2

(i) The tube has an external diameter of 2.4 cm and is floating in a liquid of density950kgm –3. Assuming the equation in (b), calculate the mass of the tube and its

contents.

mass = ..................................... kg [3]

(ii) State what feature of Fig. 4.2 indicates that the oscillations are damped.

...................................................................................................................................

...............................................................................................................................[1]

–3

–2

–1

0

1

2

3

0 0.40.2 0.6 0.8 1.0 1.2 1.4

y / cm

t / s

[3]




C el l N o. 0

3 4 6 7 0 3 0 0 3 9

134 The centre of the cone of a loudspeaker is oscillating with simple harmonic motion of

frequency 1400Hz and amplitude 0.080 mm.

(a) Calculate, to two significant figures,

(i) the angular frequency ω of the oscillations,

ω = ………………………………. rad s–1 [2]

(ii) the maximum acceleration, in m s–2, of the centre of the cone.

acceleration = ……………………………….. m s–2 [2]

(b) On the axes of Fig. 4.1, sketch a graph to show the variation with displacement x of theacceleration a of the centre of the cone.

[2]

00 x

a

9702/04/O/N/05




C el l N o. 0

3 4 6 7 0 3 0 0 3 9

14

(c) (i) State the value of the displacement x at which the speed of the centre of the cone

is a maximum.

x = ……………………………… mm [1]

(ii) Calculate, in m s–1, this maximum speed.

speed = ……………………………. m s–1

[2]

4 A piston moves vertically up and down in a cylinder, as illustrated in Fig. 4.1.

Fig. 4.1

The piston is connected to a wheel by means of a rod that is pivoted at the piston and at thewheel. As the piston moves up and down, the wheel is made to rotate.

(a) (i) State the number of oscillations made by the piston during one complete rotation ofthe wheel.

number = ………………………. [1]

cylinder

wheel

pivot

piston

pivot P

9702/04/M/J/06



15

(ii) The wheel makes 2400 revolutions per minute. Determine the frequency ofoscillation of the piston.

frequency = ………………………. Hz [1]

(b) The amplitude of the oscillations of the piston is 42 mm.

Assuming that these oscillations are simple harmonic, calculate the maximum valuesfor the piston of

(i) the linear speed,

speed = …………………………. ms–1 [2]

(ii) the acceleration.

acceleration = …………………………. m s–2 [2]

(c) On Fig. 4.1, mark a position of the pivot P for the piston to have

(i) maximum speed (mark this position S), [1]

(ii) maximum acceleration (mark this position A). [1]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



17

State

(i) the name of the phenomenon illustrated in Fig. 3.2,

.............................................................................................................................. [1]

(ii) the frequency f 0 at which maximum amplitude occurs.

frequency = ………………………… Hz [1]

(c) Suggest and explain how the apparatus in Fig. 3.1 could be modified to make the peakon Fig. 3.2 flatter, without significantly changing the frequency f 0 at which the peakoccurs.

..........................................................................................................................................

..........................................................................................................................................

..........................................................................................................................................

..................................................................................................................................... [3]

7 A magnet is suspended vertically from a fixed point by means of a spring, as shown inFig. 7.1.

spring

coil

magnet

R

Fig. 7.1

9702/04/M/J/07


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



18One end of the magnet hangs inside a coil of wire. The coil is connected in series with aresistor R.

(a) The magnet is displaced vertically a small distance D and then released. Fig. 7.2 shows the variation with time t of the vertical displacement d of the magnet from

its equilibrium position.

0.2 0.4 0.6 0.8 1.0 1.2 1.400

+D

–D

d

t /s

Fig. 7.2 (i) State and explain, by reference to electromagnetic induction, the nature of the

oscillations of the magnet.

..................................................................................................................................

..................................................................................................................................

..................................................................................................................................

..................................................................................................................................

..................................................................................................................................

............................................................................................................................. [5]

(ii) Calculate the angular frequency ω 0 of the oscillations.

ω 0 = …………….……………… rad s–1 [2]

(b) The resistance of the resistor R is increased. The magnet is again displaced a vertical distance D and released. On Fig. 7.2, sketch the variation with time t of the displacement d of the magnet. [2]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



19

(c) The resistor R in Fig. 7.1 is replaced by a variable-frequency signal generator of constantr.m.s. output voltage.

The angular frequency ω of the generator is gradually increased from about 0.7ω 0 toabout 1.3ω 0, where ω 0 is the angular frequency calculated in (a)(ii).

(i) On the axes of Fig. 7.3, sketch a graph to show the variation with ω of the amplitude

A of the oscillations of the magnet. [2]

A

00.7

0 1.0

0 1.3

0ω ω ω

ω

Fig. 7.3

(ii) State the name of the phenomenon illustrated in the graph of Fig. 7.3.

............................................................................................................................. [1]

(iii) Briefly describe one situation where the phenomenon named in (ii) is useful andone situation where it should be avoided.

useful: .......................................................................................................................

..................................................................................................................................

avoid: ........................................................................................................................

............................................................................................................................. [2]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



203 A spring is hung from a fixed point. A mass of 130 g is hung from the free end of the spring,

as shown in Fig. 3.1.

mass 130g

spring

Fig. 3.1

The mass is pulled downwards from its equilibrium position through a small distance d and is

released. The mass undergoes simple harmonic motion.Fig. 3.2 shows the variation with displacement x from the equilibrium position of the kineticenergy of the mass.

–1.0 –0.5 0 +0.5x / cm

+1.0

3.0

2.0

1.0

0

kinetic energy/ mJ

Fig. 3.2

9702/04/O/N/07


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



21

(a) Use Fig. 3.2 to

(i) determine the distance d through which the mass was displaced initially,

d = ............................................ cm [1]

(ii) show that the frequency of oscillation of the mass is approximately 4.0 Hz.

[6]

(b) (i) On Fig. 3.2, draw a line to represent the total energy of the oscillating mass. [1]

(ii) After many oscillations, damping reduces the total energy of the mass to 1.0 mJ.For the oscillations with reduced energy,

1. state the frequency,

frequency = .............................................Hz

2. using the graph, or otherwise, state the amplitude.

amplitude = ............................................ cm [2]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



22

3 A tube, closed at one end, has a uniform area of cross-section. The tube contains somesand so that the tube floats upright in a liquid, as shown in Fig. 3.1.

tube

d

sand

liquid

Fig. 3.1

When the tube is at rest, the depth d of immersion of the base of the tube is 16 cm. The tube is displaced vertically and then released. The variation with time t of the depth d of the base of the tube is shown in Fig. 3.2.

15

16

17

0 1.0 2.0 3.0

d / cm

t / s

Fig. 3.2

(a) Use Fig. 3.2 to determine, for the oscillations of the tube,

(i) the amplitude,

amplitude = ........................................... cm [1]

(ii) the period.

period = ........................................... s [1]

9702/04/M/J/08


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



23(b) (i) Calculate the vertical speed of the tube at a point where the depth d is 16.2 cm.

speed = ........................................... cm s–1 [3]

(ii) State one other depth d where the speed will be equal to that calculated in (i).

d = ........................................... cm [1]

(c) (i) Explain what is meant by damping .

..................................................................................................................................

..................................................................................................................................

..................................................................................................................................

..............................................................................................................................[2]

(ii) The liquid in (b) is now cooled so that, although the density is unchanged, there is

friction between the liquid and the tube as it oscillates. Having been displaced, thetube completes approximately 10 oscillations before coming to rest.

On Fig. 3.2, draw a line to show the variation with time t of depth d for the first 2.5 sof the motion. [3]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



24

3 The needle of a sewing machine is made to oscillate vertically through a total distance of22 mm, as shown in Fig. 3.1.

needle at itsmaximum height

22mm

8.0mmcloth

Fig. 3.1

The oscillations are simple harmonic with a frequency of 4.5 Hz.The cloth that is being sewn is positioned 8.0 mm below the point of the needle when theneedle is at its maximum height.

(a) State what is meant by simple harmonic motion .

..........................................................................................................................................

..........................................................................................................................................

......................................................................................................................................[2]

(b) The displacement y of the point of the needle may be represented by the equation

y = a cosω t .

(i) Suggest the position of the point of the needle at time t = 0.

..............................................................................................................................[1]

(ii) Determine the values of

1. a , a = .......................................... mm [1]

2. ω .

ω = ..................................... rad s –1 [2]

9702/04/O/N/08


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



26

At time zero, the peg is at P, producing a shadow on the screen at S.At time t , the disc has rotated through angle θ . The peg is now at R, producing a shadowat Q.

(a) Determine,

(i) in terms of ω and t , the angle θ ,

............................................................................................................................ [1]

(ii) in terms of ω , t and r , the distance SQ.

............................................................................................................................ [1]

(b) Use your answer to (a)(ii) to show that the shadow on the screen performs simple

harmonic motion.

..........................................................................................................................................

..........................................................................................................................................

.................................................................................................................................... [2]

(c) The disc has radius r of 12 cm and is rotating with angular speed ω of 4.7 rad s –1.

Determine, for the shadow on the screen,

(i) the frequency of oscillation,

frequency = ......................................... Hz [2]

(ii) its maximum speed.

speed = .................................... cm s –1 [2]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



274 The variation with time t of the displacement x of the cone of a loudspeaker is shown inFig. 4.1.

0

0.2

0.3

x / mm

0.1

– 0.2

– 0.1

– 0.3

0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.60

t / ms

Fig. 4.1

(a) Use Fig. 4.1 to determine, for these oscillations,

(i) the amplitude,

amplitude = ........................................ mm [1]

(ii) the frequency.

frequency = .......................................... Hz [2]

(b) State two times at which

(i) the speed of the cone is maximum,

time ............................... ms and time ............................... ms [1]

(ii) the acceleration of the cone is maximum.

time ............................... ms and time ............................... ms [1]

(c) The effective mass of the cone is 2.5 g.

Use your answers in (a) to determine the maximum kinetic energy of the cone.

kinetic energy = ............................................ J [3]

9702/41/O/N/09


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



28

(d) The loudspeaker must be designed so that resonance of the cone is avoided.

(i) State what is meant by resonance .

..................................................................................................................................

..................................................................................................................................

............................................................................................................................ [2]

(ii) State and briefly explain one other situation in which resonance should beavoided.

..................................................................................................................................

..................................................................................................................................

..................................................................................................................................

................................................................................................................. [2]

3 The variation with displacement x of the acceleration a of the centre of the cone of aloudspeaker is shown in Fig. 3.1.

–0.3 –0.2 –0.1 0 0.1 0.2 0.3

–750

–500

–250

0

250

500

750

a /ms –2

x

/mm

Fig. 3.1

9702/42/O/N/09


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



29

(a) State the two features of Fig. 3.1 that show that the motion of the cone is simpleharmonic.

1. ......................................................................................................................................

2. ......................................................................................................................................

[2]

(b) Use data from Fig. 3.1 to determine the frequency, in hertz, of vibration of the cone.

frequency = ........................................... Hz [3]

(c) The frequency of vibration of the cone is now reduced to one half of that calculatedin (b).

The amplitude of vibration remains unchanged.

On the axes of Fig. 3.1, draw a line to represent the variation with displacement x of theacceleration a of the centre of the loudspeaker cone.

[2]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



303 (a) State what is meant by

(i) oscillations ,

..................................................................................................................................

..............................................................................................................................[1]

(ii) free oscillations,

..................................................................................................................................

..............................................................................................................................[1]

(iii) simple harmonic motion .

..................................................................................................................................

..................................................................................................................................

..............................................................................................................................[2]

(b) Two inclined planes RA and LA each have the same constant gradient. They meet attheir lower edges, as shown in Fig. 3.1.

L R

ball

A

Fig. 3.1

A small ball moves from rest down plane RA and then rises up plane LA. It then movesdown plane LA and rises up plane RA to its original height. The motion repeats itself.

State and explain whether the motion of the ball is simple harmonic.

..........................................................................................................................................

..........................................................................................................................................

......................................................................................................................................[2]

9702/41/M/J/10


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



31

2 A long strip of springy steel is clamped at one end so that the strip is vertical. A mass of 65 gis attached to the free end of the strip, as shown in Fig. 2.1.

mass65g

clamp

springy

steel

Fig. 2.1

The mass is pulled to one side and then released. The variation with time t of the horizontaldisplacement of the mass is shown in Fig. 2.2.

2

1

displacement

/ cm

0

–1

–2

0.10.10 0.2 0.3 0.4 0.5 0.6 0.7

t / s

Fig. 2.2

The mass undergoes damped simple harmonic motion.

(a) (i) Explain what is meant by damping .

..................................................................................................................................

..................................................................................................................................

............................................................................................................................ [2]

9702/42/M/J/10


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



32 (ii) Suggest, with a reason, whether the damping is light, critical or heavy.

..................................................................................................................................

..................................................................................................................................

............................................................................................................................ [2]

(b) (i) Use Fig. 2.2 to determine the frequency of vibration of the mass.

frequency = ......................................... Hz [1]

(ii) Hence show that the initial energy stored in the steel strip before the mass isreleased is approximately 3.2 mJ.

[2]

(c) After eight complete oscillations of the mass, the amplitude of vibration is reduced from1.5 cm to 1.1 cm. State and explain whether, after a further eight complete oscillations,the amplitude will be 0.7 cm.

..........................................................................................................................................

..........................................................................................................................................

.................................................................................................................................... [2]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



33

3 A student sets up the apparatus illustrated in Fig. 3.1 in order to investigate the oscillations ofa metal cube suspended on a spring.

variable-frequencyoscillator

thread

pulley

spring

metalcube

Fig. 3.1

The amplitude of the vibrations produced by the oscillator is constant.The variation with frequency of the amplitude of the oscillations of the metal cube is shownin Fig. 3.2.

0

42 6 8 10

5

10

15

amplitude

/mm

frequency/Hz

20

Fig. 3.2

(a) (i) State the phenomenon illustrated in Fig. 3.2.

..............................................................................................................................[1]

(ii) For the maximum amplitude of vibration, state the magnitudes of the amplitude andthe frequency.

amplitude = ............................................. mm

frequency = ............................................... Hz[1]

9702/41/O/N/10


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



34 (b) The oscillations of the metal cube of mass 150 g may be assumed to be simple

harmonic. Use your answers in (a)(ii) to determine, for the metal cube,

(i) its maximum acceleration,

acceleration = ...................................... m s –2 [3]

(ii) the maximum resultant force on the cube.

force = .......................................... N [2]

(c) Some very light feathers are attached to the top surface of the cube so that the feathersextend outwards, beyond the vertical sides of the cube.

The investigation is now repeated. On Fig. 3.2, draw a line to show the new variation with frequency of the amplitude of

vibration for frequencies between 2 Hz and 10 Hz. [2]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



353 A cylinder and piston, used in a car engine, are illustrated in Fig. 3.1.

C

A

D

cylinder

piston

B

Fig. 3.1

The vertical motion of the piston in the cylinder is assumed to be simple harmonic.The top surface of the piston is at AB when it is at its lowest position; it is at CD when at itshighest position, as marked in Fig. 3.1.

(a) The displacement d of the piston may be represented by the equation

d = – 4.0 cos(220t )

where d is measured in centimetres.

(i) State the distance between the lowest position AB and the highest position CD ofthe top surface of the piston.

distance = .......................................... cm [1]

(ii) Determine the number of oscillations made per second by the piston.

number = ................................................ [2]

9702/43/O/N/10


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



9702/41/M/J/11

37

(ii) In cylinder Z, the oscillations of the piston lead those of the piston in cylinder X by aphase angle of 240° ( 4

3p rad).

Complete the diagram of cylinder Z, for this instant, by drawing

1. a line to show the top surface of the piston, [1]

2. an arrow to show the direction of movement of the piston. [1]

(iii) For the piston in cylinder Y, calculate its speed for this instant.

speed = ..................................... cm s –1 [2]

3 (a) Define simple harmonic motion .

..........................................................................................................................................

..........................................................................................................................................

......................................................................................................................................[2]

(b) A tube, sealed at one end, has a total mass m and a uniform area of cross-section A.The tube floats upright in a liquid of density ρ with length L submerged, as shown inFig. 3.1a.

tube

x

liquid

density ρL

+L + x

Fig. 3.1a Fig. 3.1b

The tube is displaced vertically and then released. The tube oscillates vertically in theliquid.

At one time, the displacement is x , as shown in Fig. 3.1b.

Theory shows that the acceleration a of the tube is given by the expression

a = –A ρg

m x .


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



38

(i) Explain how it can be deduced from the expression that the tube is moving withsimple harmonic motion.

..................................................................................................................................

..................................................................................................................................

..............................................................................................................................[2]

(ii) The tube, of area of cross-section 4.5 cm2, is floating in water of density1.0 × 103 kg m–3.

Calculate the mass of the tube that would give rise to oscillations of frequency 1.5 Hz.

mass = ............................................. g [4]

5 A bar magnet is suspended vertically from the free end of a helical spring, as shown in

Fig. 5.1.

V

helical

spring

magnet

coil

Fig. 5.1

One pole of the magnet is situated in a coil. The coil is connected in series with a

high-resistance voltmeter. The magnet is displaced vertically and then released. The variation with time t of the reading V of the voltmeter is shown in Fig. 5.2.

9702/42/M/J/11


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



39

0.500

1.0 1.5 2.0 2.5 t / s

V

Fig. 5.2

(a) (i) State Faraday’s law of electromagnetic induction.

..................................................................................................................................

..................................................................................................................................

............................................................................................................................ [2]

(ii) Use Faraday’s law to explain why

1. there is a reading on the voltmeter,

..................................................................................................................................

............................................................................................................................ [1]

2. this reading varies in magnitude,

..................................................................................................................................

............................................................................................................................ [1]

3. the reading has both positive and negative values.

..................................................................................................................................

............................................................................................................................ [1]

(b) Use Fig. 5.2 to determine the frequency f 0 of the oscillations of the magnet.

= .......................................... Hz [2]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



40

(c) The magnet is now brought to rest and the voltmeter is replaced by a variable frequencyalternating current supply that produces a constant r.m.s. current in the coil.

The frequency of the supply is gradually increased from 0.7 f 0 to 1.3 f 0, where f 0 is thefrequency calculated in (b).

On the axes of Fig. 5.3, sketch a graph to show the variation with frequency f of theamplitude A of the new oscillations of the bar magnet.

0.7 f 0

0

A

f 0 1.3 f 0

f

[2]Fig. 5.3

(d) (i) Name the phenomenon illustrated on your completed graph of Fig. 5.3.

............................................................................................................................ [1]

(ii) State one situation where the phenomenon named in (i) is useful.

..................................................................................................................................

............................................................................................................................ [1]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



41

3 A bar magnet is suspended from the free end of a helical spring, as illustrated in Fig. 3.1.

coil

magnet

helical

spring

Fig. 3.1

One pole of the magnet is situated in a coil of wire. The coil is connected in series with aswitch and a resistor. The switch is open.

The magnet is displaced vertically and then released. As the magnet passes through its restposition, a timer is started. The variation with time t of the vertical displacement y of themagnet from its rest position is shown in Fig. 3.2.

0

1.0

–1.0

–2.0

2.0

y / cm

4.03.02.01.00 5.0 6.0 7.0 8.0 9.0 10.0

t / s

Fig. 3.2

At time t = 4.0 s, the switch is closed.

(a) Use Fig. 3.2 to

(i) state the evidence for the magnet to be undergoing free oscillations during theperiod t = 0 to t = 4.0 s,

..................................................................................................................................

..............................................................................................................................[1]

9702/41/O/N/11


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



42

(ii) state, with a reason, whether the damping after time t = 4.0 s is light, critical orheavy,

..................................................................................................................................

..................................................................................................................................

..............................................................................................................................[2]

(iii) determine the natural frequency of vibration of the magnet on the spring.

frequency = ........................................... Hz [2]

(b) (i) State Faraday’s law of electromagnetic induction.

..................................................................................................................................

..................................................................................................................................

..............................................................................................................................[2]

(ii) Explain why, after time t = 4.0 s, the amplitude of vibration of the magnet is seen todecrease.

..................................................................................................................................

..................................................................................................................................

..................................................................................................................................

..................................................................................................................................

..................................................................................................................................

..............................................................................................................................[4]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



43

3 (a) Define simple harmonic motion .

..........................................................................................................................................

..........................................................................................................................................

..................................................................................................................................... [2]

(b) A horizontal plate is vibrating vertically, as shown in Fig. 3.1.

vertical oscillations

frequency 4.5Hz

cube,

mass 5.8g

plate

Fig. 3.1

The plate undergoes simple harmonic motion with a frequency of 4.5 Hz andamplitude 3.0 mm.

A metal cube of mass 5.8 g rests on the plate.

Calculate, for the cube, the energy of oscillation.

energy = ............................................. J [3]

(c) The amplitude of oscillation of the plate in (b) is gradually increased. The frequencyremains constant.

At one particular amplitude, the cube just loses contact momentarily with the plate.

(i) State the position of the plate in its oscillation at the point when the cube losescontact.

..................................................................................................................................

..................................................................................................................................

............................................................................................................................. [2]

9702/43/O/N/11


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



44

(ii) Calculate this amplitude of oscillation.

amplitude = ............................................ m [2]

4 A small metal ball is suspended from a fixed point by means of a string, as shown in Fig. 4.1.

string

ball

x

Fig. 4.1

The ball is pulled a small distance to one side and then released. The variation with time t ofthe horizontal displacement x of the ball is shown in Fig. 4.2.

0

2

4

x / cm

6

–6

–4

–2

0 0.2 0.4 0.6 0.8 1.0 t / s

Fig. 4.2The motion of the ball is simple harmonic.

(a) Use data from Fig. 4.2 to determine the horizontal acceleration of the ball for adisplacement x of 2.0 cm.

acceleration = ....................................... m s–2 [3]

9702/41/M/J/12


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



45

(b) The maximum kinetic energy of the ball is E K. On the axes of Fig. 4.3, sketch a graph to show the variation with time t of the kinetic

energy of the ball for the first 1.0 s of its motion.

E K

00

kinetic

energy

t / s0.2 0.4 0.6 0.8 1.0

Fig. 4.3[3]

2 A ball of mass 37 g is held between two fixed points A and B by two stretched helical springs,as shown in Fig. 2.1.

ball

mass 37g

A B

Fig. 2.1

The ball oscillates along the line AB with simple harmonic motion of frequency 3.5 Hz andamplitude 2.8 cm.

(a) Show that the total energy of the oscillations is 7.0 mJ.

[2]

(b) At two points in the oscillation of the ball, its kinetic energy is equal to the potentialenergy stored in the springs.

Calculate the magnitude of the displacement at which this occurs.

displacement = ............................................ cm [3]

9702/42/M/J/12


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



46

(c) On the axes of Fig. 2.2 and using your answers in (a) and (b), sketch a graph to showthe variation with displacement x of

(i) the total energy of the system (label this line T), [1]

(ii) the kinetic energy of the ball (label this line K), [2]

(iii) the potential energy stored in the springs (label this line P). [2]

0 1 2 3–3 –2 –1

2

0

4

6

8

energy

/ mJ

x / cm

Fig. 2.2

(d) The arrangement in Fig. 2.1 is now rotated through 90° so that the line AB is verticaland the ball oscillates in a vertical plane.

Suggest one form of energy, other than those in (c), that must be taken into considerationwhen plotting new graphs to show energy changes with displacement.

......................................................................................................................................[1]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



47

4 A ball is held between two fixed points A and B by means of two stretched springs, as shownin Fig. 4.1.

ball

A B

Fig. 4.1

The ball is free to oscillate horizontally along the line AB. During the oscillations, the springsremain stretched and do not exceed their limits of proportionality.

The variation of the acceleration a of the ball with its displacement x from its equilibrium

position is shown in Fig. 4.2.

–3

–5

–10

–15

–2 –1 00

1 2 3

5

10

15

x / cm

a / ms–2

Fig. 4.2

9702/41/O/N/12


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



48

(a) State and explain the features of Fig. 4.2 that indicate that the motion of the ball issimple harmonic.

..........................................................................................................................................

..........................................................................................................................................

..........................................................................................................................................

..........................................................................................................................................

......................................................................................................................................[4]

(b) Use Fig. 4.2 to determine, for the oscillations of the ball,

(i) the amplitude,

amplitude = .......................................... cm [1]

(ii) the frequency.

frequency = ........................................... Hz [3]

(c) The arrangement in Fig. 4.1 is now rotated through 90° so that the line AB is vertical.The ball now oscillates in a vertical plane.

Suggest one reason why the oscillations may no longer be simple harmonic.

..........................................................................................................................................

......................................................................................................................................[1]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



49

2 A small frictionless trolley is attached to a fixed point A by means of a spring. A secondspring is used to attach the trolley to a variable frequency oscillator, as shown in Fig. 2.1.

trolley

A

variable frequency

oscillator

Fig. 2.1

Both springs remain extended within the limit of proportionality.Initially, the oscillator is switched off. The trolley is displaced horizontally along the line joiningthe two springs and is then released.The variation with time t of the velocity v of the trolley is shown in Fig. 2.2.

0.3

v /ms–1

0.1

0

0.2

−0.2

−0.3

−0.1

0t / s

0.2 0.4 0.6 0.8 1.0 1.2

Fig. 2.2

(a) (i) Using Fig. 2.2, state two different times at which

1. the displacement of the trolley is zero,

time = ........................... s and time = ........................... s [1]

2. the acceleration in one direction is maximum.

time = ........................... s and time = ........................... s [1]

(ii) Determine the frequency of oscillation of the trolley.

frequency = ........................................... Hz [2]

9702/43/O/N/12


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



50

(ii) Determine the frequency of oscillation of the trolley.

frequency = ........................................... Hz [2]

(iii) The variation with time of the displacement of the trolley is sinusoidal. The variationwith time of the velocity of the trolley is also sinusoidal.

State the phase difference between the displacement and the velocity.

phase difference = ................................................. [1]

(b) The oscillator is now switched on. The amplitude of vibration of the oscillator is constant.The frequency f of vibration of the oscillator is varied.

The trolley is forced to oscillate by means of vibrations of the oscillator. The variation with f of the amplitude a 0 of the oscillations of the trolley is shown in

Fig. 2.3.

f

a 0

Fig. 2.3

By reference to your answer in (a), state the approximate frequency at which theamplitude is maximum.

frequency = ........................................... Hz [1]

(c) The amplitude of the oscillations in (b) may be reduced without changing significantlythe frequency at which the amplitude is a maximum. State how this may be done andgive a reason for your answer.

You may draw on Fig. 2.1 if you wish.

..........................................................................................................................................

..........................................................................................................................................

..........................................................................................................................................

..........................................................................................................................................

......................................................................................................................................[2]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



513 A ball is held between two fixed points A and B by means of two stretched springs, as shown

in Fig. 3.1.

ballA B

Fig. 3.1

The ball is free to oscillate along the straight line AB. The springs remain stretched and themotion of the ball is simple harmonic.

The variation with time t of the displacement x of the ball from its equilibrium position isshown in Fig. 3.2.

Fig. 3.2

(a) (i) Use Fig. 3.2 to determine, for the oscillations of the ball,

1. the amplitude,

amplitude = ........................................... cm [1]

2. the frequency.

frequency = ............................................ Hz [2]

2.0

1.0

0

–1.0

–2.0

0.20 0.4 0.6 1.2t / s

x / cm

0.8 1.0

9702/41/M/J/13


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



52

(ii) Show that the maximum acceleration of the ball is 5.2 m s–2.

[2]

(b) Use your answers in (a) to plot, on the axes of Fig. 3.3, the variation with displacement x

of the acceleration a of the ball.

00

a / ms–2

x / 10–2 m

Fig. 3.3 [2]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



53

(c) Calculate the displacement of the ball at which its kinetic energy is equal to one half ofthe maximum kinetic energy.

displacement = ........................................... cm [3]

3 A mass of 78 g is suspended from a fixed point by means of a spring, as illustrated in Fig. 3.1.

spring

mass78g

Fig. 3.1

The stationary mass is pulled vertically downwards through a distance of 2.1 cm and thenreleased.The mass is observed to perform simple harmonic motion with a period of 0.69 s.

(a) The mass is released at time t = 0.

For the oscillations of the mass,

(i) calculate the angular frequency ω ,

ω = ...................................... rad s–1 [2]

(ii) determine numerical equations for the variation with time t of

1. the displacement x in cm,

..................................................................................................................................

[2] ..............................................................................................................................

9702/42/M/J/13


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



54

2. the speed v in m s–1.

..................................................................................................................................

............................................................................................................................. [2]

(b) Calculate the total energy of oscillation of the mass.

energy = ............................................... J [2]

3 A metal ball is suspended from a fixed point by means of a string, as illustrated in Fig. 3.1.

x

ball

string

Fig. 3.1

The ball is given a small displacement and then released. The variation with time t of thedisplacement x of the ball is shown in Fig. 3.2.

0

–1.0

1.0

x / cm

2.0

–2.0

0.40 0.8 1.2 1.6 2.0t / s

Fig. 3.2

9702/41/O/N/13


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



55

(a) (i) State two times at which the speed of the ball is a maximum.

time = ............................ s and time = ............................ s [1]

(ii) Show that the maximum speed of the ball is approximately 0.08 m s–1.

(b) The variation with displacement x of the potential energy E P of the oscillations of theball is shown in Fig. 3.3.

–1.0–1.5 –0.5 0 0.5 1.0 1.5

x / cm

10

5

15

20

25

0

energy/mJ

E P

Fig. 3.3

(i) On the axes of Fig. 3.3, sketch a graph to show the variation with displacement x ofthe kinetic energy of the ball. [2]

(ii) The amplitude of the oscillations reduces over a long period of time. After many oscillations, the amplitude of the oscillations is 0.60 cm.

Use Fig. 3.3 to determine the total energy of the oscillations of the ball for oscillationsof amplitude 0.60 cm. Explain your working.

energy = .............................................. J [2]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



56

4 A student investigates the energy changes of a mass oscillating on a vertical spring, as shown inFig. 4.1.

spring

mass

Fig. 4.1

The student draws a graph of the variation with displacement x of energy E of the oscillation, asshown in Fig. 4.2.

–1.5 –1.0 –0.5x / cm

0 0.5 1.0 1.5

2.0

2.5

1.5

1.0

0.5

0

E / mJ

Fig. 4.2

(a) State whether the energy E represents the total energy, the potential energy or the kineticenergy of the oscillations.

.............................................................................................................................................. [1]

9702/41/M/J/14


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



57

(b) The student repeats the investigation but with a smaller amplitude. The maximum value of Eis now found to be 1.8 mJ.

Use Fig. 4.2 to determine the change in the amplitude. Explain your working.

change in amplitude = .................................................. cm [3]

4 (a) State what is meant by simple harmonic motion .

...................................................................................................................................................

...................................................................................................................................................

.............................................................................................................................................. [2]

(b) A small ball rests at point P on a curved track of radius r , as shown in Fig. 4.1.

x

P

curved track,

radius r

Fig. 4.1

The ball is moved a small distance to one side and is then released. The horizontaldisplacement x of the ball is related to its acceleration a towards P by the expression

a = −gx

r

where g is the acceleration of free fall.

(i) Show that the ball undergoes simple harmonic motion.

...........................................................................................................................................

...........................................................................................................................................

...........................................................................................................................................

...................................................................................................................................... [2]

(ii) The radius r of curvature of the track is 28 cm.

Determine the time interval τ between the ball passing point P and then returning to

point P.

τ = ..................................................... s [3]

9702/41/O/N/14


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



58

(c) The variation with time t of the displacement x of the ball in (b) is shown in Fig. 4.2.

2 t

00

3 t 4 t t

x

Fig. 4.2

Some moisture now forms on the track, causing the ball to come to rest after approximately15 oscillations.

On the axes of Fig. 4.2, sketch the variation with time t of the displacement x of the ball forthe first two periods after the moisture has formed. Assume the moisture forms at time t = 0.

[3]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



59

1 A light spring is suspended from a fixed point. A bar magnet is attached to the end of the spring,as shown in Fig. 1.1.

spring

bar magnet

cardboard cup

Fig. 1.1

In order to shield the magnet from draughts, a cardboard cup is placed around the magnet butdoes not touch it.The magnet is displaced vertically and then released. The variation with time t of the verticaldisplacement y of the magnet is shown in Fig. 1.2.

2.0

1.0

y / cm

0

–2.0

–1.0

0 0.1 0.2 0.3 0.4 0.5 0.6t / s

Fig. 1.2

9702/43/O/N/14


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



60

The mass of the magnet is 130 g.

(a) For the oscillations of the magnet, use Fig. 1.2 to

(i) determine the angular frequency ω ,

ω = ............................................. rad s−1 [2]

(ii) show that the maximum kinetic energy of the oscillating magnet is 6.4 mJ.

[2]

(b) The cardboard cup is now replaced with a cup made of aluminium foil.

During 10 complete oscillations of the magnet, the amplitude of vibration is seen to decreaseto 0.75 cm from that shown in Fig. 1.2. The change in angular frequency is negligible.

(i) Use Faraday’s law of electromagnetic induction to explain why the amplitude of theoscillations decreases.

...........................................................................................................................................

...........................................................................................................................................

...........................................................................................................................................

...........................................................................................................................................

...................................................................................................................................... [3]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



61

3 (a) f 0 is at natural frequency of spring (system) ................................. B1this is at the driver frequency ....................................................... B1(allow 1 mark for recognition that this is resonance)

[2]

(b) line: amplitude less at all frequencies ......................................... B1

peak flatter .......................................................................... B1peak at f 0 or slightly below f 0 ................................................ B1 [3]

(c) (aluminium) sheet cuts the magnetic flux/field.............................. B1(so) currents/e.m.f. induced in the (metal) sheet .......................... B1these currents dissipate energy ...................................................M1less energy available for the oscillations ...................................... A1so amplitude smaller .................................................................... A0(‘current opposes motion of sheet’ scores one of the last twomarks)

[4]

A/AS LEVEL EXAMINATIONS - JUNE 2003 9702 04

2 (a) (i) a,ω and x identified ………(-1 each error or omission) ................. B2

(ii) (-)ve because a and x in opposite directionsOR a directed towards mean position/centre................................ B1 [3]

(b) (i) forces in springs are k (e + x ) and k (e – x ) ....................................C1

resultant = k (e + x ) – k(e – x ) ......................................................M1 = 2kx ............................................................................ A0 [2]

(ii) F = ma ....................................................................................... B1a = -2kx /m .................................................................................... A0(-)ve sign explained...................................................................... B1 [2]

(iii) ω 2 = 2k /m .....................................................................................C1

(2πf )2 = (2 x 120)/0.90 ..................................................................C1f = 2.6 Hz ..................................................................................... A1 [3]

(c) atom held in position by attractive forcesatom oscillates,not just two forces OR 3D not 1Dforce not proportional to xany two relevant points, 1 each, max 2 ........................................ B2 [2]

A/AS LEVEL EXAMINATIONS - NOVEMBER 2003 9702 04

4 (a) e.g. amplitude is not constant or wave is damped B1

do not allow 'displacement constant'should be (-)cos, (not sin) B1 [2]

(b) T = 0.60 s C1

ω = 2π/T = 10.5 rad s-1 (allow 10.4 → 10.6) A1 [2]

(c) same period B1displacement always less M1amplitude reducing appropriately A1 [3]for 2 nd and 3rd marks, ignore the first quarter period

Total [7]

A/AS LEVEL EXAMINATIONS - JUNE 2004 9702 04


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



62

3 (a) (i) reasonable shape as ‘inverse’ of k.e. line 1

(ii) straight line, parallel to x-axis at 15 mJ 1 [2]

(b) either (max) kinetic energy (= ½ mv 2 ) = ½ mω 2a0

2 1

15 x 10-3 = ½ x 0.15 x ω 2 x (5.0 x 10-2)2 1

ω = 8.9(4) rad s-1 1

or (k.e. = ½ mv2

), v = 0.44(7) m s-1

1ω = v/a = (0.447)/(5.0 x 10-2) 1

ω = 8.9(4) rad s-1 1 [3]

(c) (i) either loss of energy (from the system) or amplitude decreasesor additional force acting (on the mass) 1either continuous/gradual loss or force always opposing motion 1 [2]

(ii) either (now has 80% of its) p.e./k.e. = 12 mJ or loss in k.e. = 3 mJ 1new amplitude = 4.5 cm (allow ± 0.1 cm) 1 [2]

A LEVEL – NOVEMBER 2004 9702 4

A LEVEL - JUNE 2005 9702 4

4 (a) acceleration proportional to displacement (from a fixed point) M1

or a = - ω 2 x with a, ω and x explained

and directed towards a fixed point A1 [2]

or negative sign explained

(b) for s.h.m., a = (-)ω 2 x B1

identifies ω 2 as A ρ g/M and therefore s.h.m. (may be implied) B1

2πf = ω B1

hence f =π 2

1√

M

Apg A0 [3]

(c) (i) T = 0.60 s or f = 1.7 Hz C1

0.60 = (2π√M )/√(π × {1.2 × 10-2}2 × 950 × 9.81) C1M = 0.0384 kg A1 [3]

(ii) decreasing peak height/amplitude B1 [1]

5 (a) field strength = potential gradient [- sign not required] B1 [1][allow E = ∆V /∆ x but not E = V /d ]

(b) No field for x < r B1

for x > r , curve in correct direction, not going to zero B1discontinuity at x = r (vertical line required) B1 [3]

A LEVEL – NOVEMBER 2005 9702 4

4 (a) (i) ω = 2πf ………………………………………………………....………….. C1

= 2π × 1400= 8800 rad s–1 ………………………………………………………….. A1 [2]

(ii) a0 = (–)ω 2 x 0 ……………………………..………………………………… C1

= (8800)2 × 0.080 × 10–3

= 6200 m s–2 …………………….……………………………………. A1 [2]

(b) straight line through origin with negative gradient …….…………….... M1end points of line correctly labelled …………………………………….. A1 [2]

(c) (i) zero displacement ………………………………………………………… B1 [1]

(ii) v = ω x 0 ……………………………………………………………………. C1

= 8800 × 0.080 × 10–3

= 0.70 m s–1 ……………………………………………………………. A1 [2]

GCE A/AS Level – May/June 2006 9702 04

4 (a) (i) 1.0 B1 [1]

(ii) 40 Hz B1 [1]

(b) (i) speed = 2πfa C1

= 2π × 40 × 42 × 10-3

= 10.6 m s-1 A1 [2]

(ii) acceleration = 4π2f 2

a C1

= (80π)2 × 42 × 10-3

= 2650 m s-2

A1 [2]

(c) (i) S marked correctly (on ‘horizontal line through centre of wheel) B1

(ii) A marked correctly (on ‘vertical line’ through centre of wheel) B1 [2]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9



GCE A/AS LEVEL - OCT/NOV 2006 9702 04

3 (a) use of a = –ω 2 x clear C1

either ω = √(2k /m) or ω 2 = (2k /m) B1

ω = 2 πf C1

f = (1/2 π)√(2 x 300)/0.240) B1= 7.96 ≈ 8 Hz A0 [4]

(b) (i) resonance B1 [1]

(ii) 8 Hz B1 [1]

(c) (increase amount of) damping B1without altering (k or) m …(some indirect reference is acceptable) B1sensible suggestion B1 [3]

May/June 2007 9702 04

7 (a) (i) oscillations are damped/amplitude decreases B1as magnet moves, flux is cut by coil B1e.m.f./current is induced in the coil B1causing energy loss in load OR force on magnet B1energy is derived from oscillations of magnetOR force opposes motion of magnet B1 [5]

(ii) T = 0.60 s C1

ω 0 (= 2π/T ) = 10.5 rad s–1 A1 [2]

(b) sketch: sinusoidal wave with period unchanged or slightly smaller M1same initial displacement, less damping A1 [2]

(c) (i) sketch: general shape – peaked curve M1

peak at ω 0 and amplitude never zero A1 [2]

(ii) resonance B1 [1]

(iii) useful: e.g. child on swing, microwave oven heating B1avoid: e.g. vibrating panels, vibrating bridges B1 [2]

(for credit, stated example must be put in context )

October/November 2007 9702 04

1 (a) (i) angle subtended at centre of circle .......................................................................B1

arc equal in length to the radius ...........................................................................B1 [2]

(ii) arc = r θ and for one revolution, arc = 2πr .............................................................M1

so, θ = 2πr /r = 2π ..................................................................................................A0 [1]

(b) (i) either weight provides/equals the centripetal forceor acceleration of free fall is centripetal acceleration ....................................B1

9.8 = 0.13 × ω 2 ......................................................................................................M1

ω = 8.7 rad s-1 .......................................................................................................A0 [2]

(ii) force in cord = weight + centripetal force (can be an equation) ..........................C1

force in cord = (L – 13) × 5/1.8 or force constant = 5.0/1.8 ................................C1

(L – 13) × 5/1.8 = 5.0 + 5/9.8 × L × 10-2× 8.72 ..................................................C1

L = 17.2 cm ...........................................................................................................A1 [4](constant centripetal force of 5.0 N gives L = 16.6 cm allow 2/4)

May/June 2008 9702 04

3 (a) (i) amplitude = 0.5 cm A1 [1]

(ii) period = 0.8 s A1 [1]

(b) (i) ω = 2π / T C1= 7.85 rad s–1

correct use of v = ω √( x 02 – x 2) B1

= 7.85 × √({0.5 × 10–2}2 – {0.2 × 10–2}2)= 3.6 cm s–1 A1 [3]

(if tangent drawn or clearly implied (B1)

3.6 ± 0.3 cm s–1 (A2)

but allow 1 mark for > ± 0.3 but Ğ ± 0.6 cm s–1 )

(ii) d = 15.8 cm A1 [1]

(c) (i) (continuous) loss of energy / reduction in

amplitude (from the oscillating system) B1caused by force acting in opposite direction to the motion / friction /viscous forces B1 [2]

(ii) same period / small increase in period B1line displacement always less than that on Fig.3.2 (ignore first T/4) M1peak progressively smaller A1 [3]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9

63



GCE A/AS LEVEL – October/November 2008 9702 04

3 (a) acceleration / force (directly) proportional to displacement M1and either directed towards fixed point

or acceleration & displacement in opposite directions A1 [2]

(b) (i) maximum / minimum height / 8mm above cloth / 14mm below cloth B1 [1]

(ii) 1. a = 11mm A1 [1]

2. ω = 2πf C1= 2π × 4.5= 28.3 rad s–1 (do not allow 1 s.f.) A1 [2]

(c) (i) v = ω a C1

= 28.3 × 11 × 10–3

= 0.31 m s–1 (do not allow 1 s.f.) A1 [2]

(ii) v = ω √(a2 – y 2)y = 3mm C1

= 28.3 × 10–3 √(112 – 32) C1 = 0.30 m s–1 (allow 1 s.f.) A1 [3]

May/June 2009 9702 04

4 (a) (i) (θ =) ω t (allow any subject if all terms given) B1 [1]

(ii) (SQ =) r sinω t (allow any subject if all terms given) B1 [1]

(b) this is the solution of the equation a = –ω 2 x M1

a = –ω 2 x is the (defining) equation of s.h.m. A1 [2]

(c) (i) f = ω / 2π C1

= 4.7 / 2π = 0.75 Hz A1 [2]

(ii) v = r ω (r must be identified) C1= 4.7 × 12= 56cms–1 A1 [2]


4 (a) (i) amplitude = 0.2 mm .......................................................................................... A1 [1]

(ii) period = 1.2 ms .................................................................................................C1frequency = 830 Hz .......................................................................................... A1 [2]

(b) (i) any two of zero, 0.6 ms and 1.2 ms ....................................................................A1 [1]

(ii) any two of 0.3 ms, 0.9 ms, 1.5 ms ......................................................................A1 [1]

(c) either v = ω x 0 = 2πfx 0

= 2π × 830 × 0.2 × 10-3 = 1.05 m s-1 or slope of graph = 1.0 m s-1 ……(allow ± 0.1 m s-1) .......................................C1E K = ½mv 2

= ½ × 2.5 × 10-3× 1.052 .....................................................................................C1

= 1.4 × 10-3 J ...................................................................................................... A1 [3]

(d) (i) large / maximum amplitude of vibration ..............................................................B1

when impressed frequency equals natural frequency of vibration ......................B1 [2]

(ii) e.g. metal panels on machinery vibrate / oscillate ........................................... (M1)motor in machine impresses frequency on panel ......................................(A1)

e.g. car suspension system vibrates / oscillates................................................. (M1)going over bumps would give large amplitude vibrations .............................(A1)

any feasible example, M1 + A1 ............................................................................... [2]


3 (a) straight line through origin ......................................................................................... B1negative gradient .......................................................................................................B1 [2]

(b) a = -ω 2 x and ω = 2πf ..........................................................................................C1

750 = (2πf )2 × 0.3 × 10-3 ...........................................................................................C1f = 250 Hz ................................................................................................................A1 [3]

(c) straight line between(-0.3,+190) and (+0.3,-190) ...................................................... A2 [2](allow 1 mark for end of line incorrect by one grid square or line does not extend to+/- 0.3 mm)

[Total: 7]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9

64



GCE AS/A LEVEL – May/June 2010 9702 41


2 (a) (i) reduction in energy (of the oscillations) (B1)reduction in amplitude / energy of oscillations (B1)due to force (always) opposing motion / resistive forces (B1) [2]any two of the above, max 2

(ii) amplitude is decreasing (very) gradually / oscillations would

continue (for a long time) /many oscillations M1light damping A1 [2]

(b) (i) frequency = 1 / 0.3= 3.3 Hz A1 [1]

allow points taken from time axis giving f = 3.45 Hz

(ii) energy = ½ mv 2 and v = ω a C1

= ½ × 0.065 × (2π/0.3)2× (1.5 × 10

–2)2 M1

= 3.2 mJ A0 [2]

(c) amplitude reduces exponentially / does not decrease linearly M1so will be not be 0.7 cm A1 [2]

GCE AS/A LEVEL – October/November 2010 9702 41

3 (a) (i) resonance B1 [1]

(ii) amplitude 16mm and frequency 4.6Hz A1 [1]

(b) (i) a = (–)ω 2 x and ω = 2πf C1

a = 4π2 × 4.62× 16 × 10–3 C1

= 13.4ms–2 A1 [3]

(ii) F = ma C1= 50 × 10–3 × 13.4= 2.0N A1 [2]

(c) line always ‘below’ given line and never zero M1peak is at 4.6Hz (or slightly less) and flatter A1 [2]

GCE A LEVEL – October/November 2010 9702 43

3 (a) (i) 8.0 cm A1 [1]

(ii) 2πf = 220 C1f = 35 (condone unit ) A1 [2]

(iii) line drawn mid-way between AB and CD (allow ±2 mm) B1 [1]

(iv) v = ω a C1= 220 × 4.0= 880 cm s–1 A1 [2]

(b) (i) 1. line drawn 2 cm above AB (allow ±2 mm) B1 [1]2. arrow pointing upwards B1 [1]

(ii) 1. line drawn 2 cm above AB (allow ±2 mm) B1 [1]2. arrow pointing downwards B1 [1]

(iii) v = ω √(a2 – x 2)

= 220 × √(4.02 – 2.02) C1= 760 cm s–1 A1 [2]

(incorrect value for x, 0/2 marks)


3 (a) acceleration / force proportional to displacement from a fixed point M1acceleration / force (always) directed towards that fixed point / in oppositedirection to displacement A1 [2]

(b) (i) A ρ g / m is a constant and so acceleration proportional to x B1negative sign shows acceleration towards a fixed point / in oppositedirection to displacement B1 [2]

(ii) ω 2

= ( A ρ g / m) C1ω = 2πf C1

(2 × π × 1.5)2 = ({4.5 × 10–4 × 1.0 × 103 × 9.81} / m) C1m = 50g A1 [4]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9

6566




5 (a) (i) (induced) e.m.f. proportional to M1rate of change of (magnetic) flux (linkage) / rate of flux cutting A1 [2]

(ii) 1. moving magnet causes change of flux linkage B1 [1]2. speed of magnet varies so varying rate of change of flux B1 [1]

3. magnet changes direction of motion (so current changes direction) B1 [1]

(b) period = 0.75s C1frequency = 1.33Hz A1 [2]

(c) graph: smooth correctly shaped curve with peak at f 0 M1 A never zero A1 [2]

(d) (i) resonance B1 [1]

(ii) e.g. quartz crystal for timing / production of ultrasound A1 [1]


3 (a) (i) amplitude remains constant B1 [1]

(ii) amplitude decreases gradually M1light damping A1 [2]

(iii) period = 0.80 s C1frequency = 1.25 Hz (period not 0.8 s, then 0/2) A1 [2]

(b) (i) (induced) e.m.f. is proportional to M1rate of change/cutting of (magnetic) flux (linkage) A1 [2]

(ii) a current is induced in the coil M1as magnet moves in coil A1current in resistor gives rise to a heating effect M1thermal energy is derived from energy of oscillation of the magnet A1 [4]


3 (a) acceleration proportional to displacement/distance from fixed point M1and in opposite directions/directed towards fixed point A1 [2]

(b) energy = ½mω 2 x 0

2 and ω = 2πf C1

= ½ × 5.8 × 10–3 × (2π × 4.5)2 × (3.0 × 10–3)2 C1= 2.1 × 10–5 J A1 [3]

(c) (i) at maximum displacement M1above rest position A1 [2]

(ii) acceleration = (–)ω 2 x 0 and acceleration = 9.81 or g C1

9.81 = (2π × 4.5)2 × x 0

x 0 = 1.2 × 10–2

m A1 [2]


4 (a) a = (–)ω 2 x and ω = 2π /T C1

T = 0.60s C1

a = (4π2 × 2.0 × 10–2) / (0.6)2

= 2.2 ms–2 A1 [3]

(b) sinusoidal wave with all values positive B1all values positive, all peaks at E K and energy = 0 at t = 0 B1period = 0.30s B1 [3]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9

667




2 (a) energy = ½mω 2a2 and ω = 2πf C1

= ½ × 37 × 10–3 × (2π × 3.5)2 × (2.8 × 10–2)2 M1= 7.0 × 10–3J A0 [2]

(allow 2π × 3.5 shown as 7π)

Energy = ½ mv 2 and v = r ω (C1)

Correct substitution (M1)Energy = 7.0 × 10–3J (A0)

(b) E K = E P

½mω 2 (a2 – x 2) = ½mω

2 x

2 or E K or E P = 3.5mJ C1 x = a/√2 = 2.8 /√2 or E K = ½mω

2 (a2 – x 2) or E P = ½mω 2 x

2 C1= 2.0cm A1 [3]

(E K or E P = 7.0mJ scores 0/3)

Allow: k = 17.9 (C1)E = ½ kx

2 (C1) x = 2.0cm (A1)


4 (a) straight line through origin M1shows acceleration proportional to displacement A1negative gradient M1

shows acceleration and displacement in opposite directions A1 [4]

(b) (i) 2.8cm A1 [1]

(ii) either gradient = ω 2 and ω = 2πf or a = –ω

2 x and ω = 2πf C1gradient = 13.5 / (2.8 × 10–2) = 482

ω = 22rad s–1 C1

frequency = (22/2π =) 3.5Hz A1 [3]

(c) e.g. lower spring may not be extendede.g. upper spring may exceed limit of proportionality/ elastic limit(any sensible suggestion) B1 [1]


2 (a) (i) 1. 0.1s, 0.3s, 0.5s, etc (any two) A1 [1]

2. either 0, 0.4s, 0.8s, 1.2sor0.2s, 0.6s, 1.0s (any two) A1 [1]

(ii) period = 0.4s C1frequency = (1/0.4 =) 2.5Hz A1 [2]

(iii) phase difference = 90° or ½ π rad B1 [1]

(b) frequency = 2.4 – 2.5Hz B1 [1]

(c) e.g. attach sheet of card to trolley M1increases damping / frictional force A1e.g. reduce oscillator amplitude (M1)reduces power/energy input to system (A1) [2]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9

68



3 (a) (i) 1. amplitude = 1.7cm A1 [1]

2. period = 0.36cm C1frequency = 1/0.36frequency = 2.8Hz A1 [2]

(ii) a = (–)ω 2

x andω

= 2π

/T C1acceleration = (2π/0.36)2 × 1.7 × 10–2 M1= 5.2m s–2 A0 [2]

(b) graph: straight line, through origin, with negative gradient M1from (–1.7 × 10–2, 5.2) to (1.7 × 10–2, –5.2) A1 [2]

(if scale not reasonable, do not allow second mark)

(c) either kinetic energy = ½mω 2( x 0

2 – x 2)or potential energy = ½mω

2 x 2 and potential energy = kinetic energy B1½mω

2( x 0 – x 2) = ½ × ½mω 2 x 0

2 or ½mω 2 x 2 = ½ × ½mω

2 x 02 C1

x 02 = 2 x 2

x = x 0 / √2 = 1.7 / √2= 1.2cm A1 [3]



3 (a) (i) ω = 2π / T

= 2π / 0.69 C1= 9.1 rad s–1 A1 [2]

(allow use of f = 1.5 Hz to give ω = 9.4 rad s–1)

(ii) 1. x = 2.1 cos 9.1t 2.1 and 9.1 numerical values B1use of cos B1 [2]

2. v 0 = 2.1 × 10–2× 9.1 (allow ecf of value of x 0 from (ii)1.)

v 0 = 0.19 m s–1 B1v = v 0 sin 9.1t (allow cos 9.1t if sin used in (ii)1.) B1 [2]

(b) energy = either ½ mv 02 or ½ mω

2 x 0

2

= either ½ × 0.078 × 0.192 or ½ × 0.078 × 9.12× (2.1 × 10–2)2 C1

= 1.4 × 10–3 J A1 [2]

3 (a) (i) any two from 0.3(0) s, 0.9(0) s, 1.50 s (allow 2.1 s etc.) B1 [1]

(ii) either v = ω x and ω = 2π/T C1

either v = (2π/1.2) × 1.5 × 10–2 M1either v = 0.079 ms–1 A0 [2]or gradient drawn clearly at a correct position (C1)

working clear (M1)to give (0.08 ± 0.01) ms–1 (A0)

GCE A LEVEL – October/November 2013 9702 42

(b) (i) sketch: curve from (±1.5, 0) passing through (0, 25) M1sketch: reasonable shape (curved with both intersections between

y = 12.0→13.0 ) A1 [2]

(ii) at max. amplitude potential energy is total energy B1


4 (a) kinetic (energy)/KE/E K B1 [1]

(b) either change in energy = 0.60 mJ or max E proportional to (amplitude)2/equivalent numerical working B1

new amplitude is 1.3 cm B1change in amplitude = 0.2 cm B1 [3]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9

69



Cambridge International AS/A Level – October/November 2014 9702 41

4 (a) acceleration/ force proportional to displacement (from a fixed point) M1either acceleration and displacement in opposite directionsor acceleration always directed towards a fixed point A1 [2]

(b) (i) g and r are constant so a is proportional to x B1negative sign shows a and x are in opposite directions B1 [2]

(ii) ω 2 = g / r and ω = 2π /T C1

ω 2 = 9.8 / 0.28= 35 C1

T = 2π / √35 = 1.06stime interval τ = 0.53s A1 [3]

(c) sketch: time period constant (or increases very slightly) M1drawn line always ‘inside’ given loops A1successive decrease in peak height A1 [3]

Cambridge International AS/A Level – October/November 2014 9702 43

1 (a) (i) either ω = 2π /T or ω = 2πf and f = 1 / T C1

ω = 2π / 0.30= 20.9rad s–1 (accept 2 s.f.) A1 [2]

(ii) kinetic energy = ½mω 2 x 0

2 or v = ω x 0 and ½mv 2 C1

= ½ × 0.130 × 20.92× (1.5 × 10–2)2 = 6.4 × 10–3J A1 [2]

(b) (i) as magnet moves, flux is cut by cup/aluminium giving rise to induced e.m.f.(in cup) B1

induced e.m.f. gives rise to currents and heating of the cup B1thermal energy derived from oscillations of magnet so amplitude decreases B1or

induced e.m.f. gives rise to currents which generate a magnetic field (B1)the magnetic field opposes the motion of the magnet so amplitude decreases (B1) [3]

(ii) either use of ½mω 2 x 0

2 and x 0 = 0.75 cm or x 0 is halved so ¼ energy C1to give new energy = 1.6mJ

either loss in energy = 6.4 – 1.6 or loss = ¾ × 6.4 giving loss = 4.8mJ A1 [2]

(c) q = mc ∆θ

4.8 × 10–3 = 6.2 × 10–3× 910 × ∆θ C1

∆θ = 8.5 × 10–4K A1 [2]


C el l N o. 0

3 4 6 7 0 3 0 0 3 9

p4 oscillations all

Documents