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Simple harmonic motion

Vibration / Oscillation to-and-fro repeating

movement

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Simple harmonic motion S.H.M. A special kind of oscillation

X YO

X, Y : extreme points O: centre of oscillation / equilibrium position A: amplitude

A

A special relation between A special relation between the displacement and the displacement and acceleration of the particleacceleration of the particle

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Exploring the acceleration and displacement of S.H.M.

- 1 .5

- 1

- 0 .5

0

0 .5

1

1 .5

0 4 5 9 0 1 3 5 1 8 0 2 2 5 2 7 0 3 1 5 3 6 0 4 0 5

time

displacement

acceleration

aa– – xx graph graph

- 0 .3

- 0 .2

- 0 .1

0

0 .1

0 .2

0 .3

- 0 .5 - 0 .4 - 0 .3 - 0 .2 - 0 .1 0 0 .1 0 .2 0 .3 0 .4 0 .5

displacement

acceleration

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Exploring the acceleration and displacement of S.H.M.

a – x graph a

x

a a ∝∝ x x a a ∝∝ -x -x

Definition of Simple harmonic motion (S.H.M.)Definition of Simple harmonic motion (S.H.M.) An oscillation is said to be an S.H.M if An oscillation is said to be an S.H.M if (1)(1) the the magnitude magnitude of acceleration is directly proportional to of acceleration is directly proportional to

distance from a fixed point (centre of oscillation), and distance from a fixed point (centre of oscillation), and

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-3 -2 -1 0 1 2 3-3 -2 -1 0 1 2 3x

aa(-) (-) aa(-) (-) aa(-) (-) aa(+) (+) aa(+) (+) aa(+) (+) aa = 3 = 3kk aa = 2 = 2kk aa = = kk aa = 0 = 0 aa = = kk aa = 2 = 2kk aa = 3 = 3kk

Definition of Simple harmonic motion (S.H.M.)Definition of Simple harmonic motion (S.H.M.) An oscillation is said to be an S.H.M if An oscillation is said to be an S.H.M if (1)(1) the magnitude of acceleration is directly proportional to the magnitude of acceleration is directly proportional to

distance from a fixed point, and distance from a fixed point, and (2)(2) the acceleration is always the acceleration is always directed towards that pointdirected towards that point..

Note: For S.H.M., direction of accelerationacceleration and displacementdisplacement is always opposite to each other.

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Equations of S.H.M.

P

O

r

aa = = rr22 r2 sin

r2 cos

Y’ O’ P’ P’ X’

x

x

For the projection P’ • Moves from X’ to O’ to Y’ and returns through O’ to X’ as P completes each revolution.Displacement• Displacement from O • x = r cos = r cos tAccelerationAcceleration of P’ = component of acceleration of P along the x-axis • a = -r2 cos (-ve means directed towards O)∴ aa = - = -22 xxThe motion of P’ is simple harmonic.

= t

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Equations of S.H.M.

P

O

r

Y’ O’ P’ P’ X’

x

x

PeriodThe period of oscillation of P’ = time for P to make one revolution T = 2 / angular speed ∴ T = 2T = 2//

VelocityVelocity of P’ = component of velocity of P along the x-axisvv = - = -rr sin sin = - = -rr sin sin tt

r

rsin

rcos

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Equations of S.H.M.

P

O

r

Y’ O’ P’ P’ X’

x

x

Motion of P’ Amplitude of oscillation

= Radius of circle ⇒⇒ AA = = rr

Displacement x:xx = A cos = A cos

Velocity v:vv = - = -AA sin sin

Acceleration a:aa = - = -22AA cos cos

r

rsin

rcos

A

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Relation between the amplitude of oscillation A and x, , and v:

2222

22222

22

1

cos,sin

xAv

Axv

A

x

A

v

A

x

A

v

Note: Maximum speed = A at x = 0 (at centre of oscillation / equilibrium positio

n).

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Example 1Example 1A particle moving with S.H.M. has velocities of 4 cm sA particle moving with S.H.M. has velocities of 4 cm s-1-1 and and 3 cm s3 cm s-1-1 at distances of 3 cm and 4 cm respectively from its at distances of 3 cm and 4 cm respectively from its equilibrium. Findequilibrium. Find(a)(a) the amplitude of the oscillationthe amplitude of the oscillation Solution:

By v2 = 2(A2 – x2)when x = 3 cm, v = 4 cm s-1,

x = 4 cm, v = 3 cm s-1. 42 = 2(A2 – 32) --- (1) 32 = 2(A2 – 42) --- (2)(1)/(2):16/9 = (A2 – 9) / (A2 – 16)9A2 – 81 = 16 A2 - 256 A2 = 25A = 5 cm∴ amplitude = 5 cm

4 ms-1

3 ms-1O

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(b)(b) the period,the period,(c)(c) the velocity of the particle as it passes through the velocity of the particle as it passes through the equilibrium position.the equilibrium position.

(b) Put A = 5 cm into (1) 42 = 2(52 – 32) 2 = 1⇒ = 1 rad s-1

T = 2/ = 2 s(c) at equilibrium position, x = 0

By v2 = 2(A2 – x2)v2 = 12(52 – 02)v = 5 cm s-1

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Isochronous oscillations

Definition: period of oscillation is independent of its amplitude.

Examples: Masses on springs and simple pendulums

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Phase difference of x-t, v-t and a-t graphs

0 T/4 T/2 3T/4 T 0 T/4 T/2 3T/4 T 0 T/4 T/2 3T/4 T

A A 2A

x v a

t t t

tAx cos tAa cos2tAv sin

2A

x

y

v

aA A

Vectors x, v and a rotate with the same angular velocity . Their projections on the y-axis give the above x-t, v-t and a-t graphs.

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Phase difference of x-t, v-t and a-t graphs

2A

x

y

v

a

A A

Note:1 a leads v by 90o or T/4.

(v lags a by 90o or T/4)2 v leads x by 90o or T/4.

(x lags v by 90o or T/4)3 a leads x by 180oor T/2.

(a and x are out of phase or antiphase)

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Energy of S.H.M.(Energy and displacement)

From equation of S.H.M. v2 = 2(A2 – x2), ∴ K.E. = ½ mv2 = ½ m2(A2 –

x2)

x

K.E.

0 A-ANote:1. K.E. is maximum when x = 0 (equilibrium position) 2. K.E. is minimum at extreme points (speed = 0)

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Potential energy P.E. = ½ kx2 ∵ 2 = k/m ∴ P.E. = ½ m2x2

P.E. is maximum at extreme points.

(Spring is most stretched.) P.E. is minimum when x = 0.

(Spring is not stretched)

x

P.E.

0 A-A

x = Ax = -A

Centre of oscillation

ix

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Total energy = K.E. + P.E.

x

Total energy

0 A-A

Energy

K.E.

P.E.

½ m2A2

2222222

2

1

2

1

2

1AmxmxAm (constant)

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Energy and time

tAvandtAx sincos

tAmtAmmv 22222 sin2

1sin

2

1

2

1

tAmtAmxm 2222222 cos2

1cos

2

1

2

1

22

2

1Am

From equation of S.H.M.

K.E. =

P.E. =

Total energy = K.E. + P.E.

(constant)

ttAmtAmtAm 2222222222 cossin2

1cos

2

1sin

2

1

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Total energy

Energy

P.E. = ½ m2A2cos2t

0 T/4 T/2 3T/4 T Time

½ m2A2

K.E. = ½ m2A2sin2t

t = 0t = T/2

Centre of oscillation

ix

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Examples of S.H.M.Mass on spring – horizontal oscillationHooke’s law: F = kx where k is the force constant and x is the extension By Newton’s second law

T = -makx = -maa = -(k/m)x

which is in the form of a = -2x Hence, the motion of the mass i

s simple harmonic, and 2 = k/m Period of oscillation

Natural length (l)

Extension (x)

T

Centre of oscillation

ix

k

m

22

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Mass on spring – vertical oscillation At equilibrium,

T’ = mgke = mg

Displaced from equilibrium,T – mg = -mak(e + x) – mg = -ma

Natural length (l)

TT

Extension at equilibrium

Displacement from

equilibrium

x

e

mgmg

T’T’

mgmg

Centre of oscillation

In equilibrium

Spring unstretched

Displaced from

equilibrium which is in the form of a = -2x Hence, the motion of the mass i

s simple harmonic

and 2 = k/m. Period of oscillation

mamgxk

mgk )(

xm

ka

k

m

22

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Effective mass of spring

Not only the mass oscillates when it is released, but also the spring itself.

The period of oscillation is affected by the mass of the spring.

Hence, the equation k

mT 2

k

mmT s

2should be rewritten as

where ms is the effective mass of the spring.

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Measurement of effective mass of spring To find the effective mass, we can do an experiment by

using different masses m and measure the corresponding periods T.

Use the results to plot a graph of T2 against m which is a straight line but it does not pass through the origin.

x

xx

x

x

T2

m

Line of best fit

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k

mmT s

22 4sm

km

kT

222 44

∴ or

k

mmT s

2

∴

x

xx

x

x

T2

m

k

24slope =

smk

24y-intercept =

∴ effective mass ms

24

ky-intercept

In theory, effective mass of a spring is about ⅓ of the mass of

string.

Usually, we would neglect the effective mass for simplicity.

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Combined Springs OscillationCase 1: Springs in parallel

Let x be the common extension of the spring.

the springs are in parallel, ∴ upward force FF = FF11 + FF22

F = k1x + k2x = (k1 + k2)x Note: k1 + k2 is the equivalent force

constant of the system. When the mass is set into vibration,

the oscillation is simple harmonic. Period of oscillation

FF

F2 = k2x

where k = k1 + k2

k

mT 2

F1 = k1x

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Case 2: Springs in series Let x1 and x2 be the extensions of the first

and the second spring respectively. The total extension x = x1 + x2 ∵ the springs are in series, ∴ upward force F = F1 = F2

F = k1x1 = k2x2

FF

FF11 = = k k11xx11

F2 = k2 x2

22

11 k

Fxand

k

Fx

∵ x = x1 + x2

21 k

F

k

Fx

21

11

kk

xF

kxF

21

111

kkkor where

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Case 2: Springs in series

FF

FF11 = = k k11xx11

F2 = k2 x2

Note: the equivalent force constant of the

system is k where 21

111

kkk

k

mT 2

When the mass is set into vibration, the oscillation is simple harmonic.

Period of oscillation

21

111

kkk where

.

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Case 3: The mass is connected by two springs on both sides

T1

Equilibrium position

T2

Under compression

Under extension

Force constant k1

Force constant k2

x

Suppose the springs are initially unstretched.

When the mass is displaced to the right by x, 1st spring is extended but 2nd spring is compressed.Resultant force on the bob F = T1 + T2

∴ F = k1x + k2x = (k1 + k2)xNote: k1 + k2 is the equivalent force constant of the system.

The oscillation is simple harmonic. Period of oscillation

k

mT 2 where k = k1 + k2.

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Simple pendulum Resolve tangentially (perpendicular to the string)

mg sin = -ma where a is the acceleration along the arcIf is small (i.e. <10o), sin ≈ and x ≈ l ,mg sin = -ma becomesmg = -maa = -g(x/l) = -(g/l)xwhich is in the form of a = -2x

Hence, the motion of the bob is simple harmonic and 2 = g/l

Period of oscillation T

OP

A

mgmg cos

mg sin

l

x

T

g

l

22

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A simple pendulum has a period of 2 s and an A simple pendulum has a period of 2 s and an amplitude of swing 5 cm. amplitude of swing 5 cm. Calculate the maximum magnitudes of Calculate the maximum magnitudes of (a)(a) velocity, and velocity, and (b)(b) acceleration of the bob.acceleration of the bob.

Solution:

(a) maximum magnitude of velocity

= A = (5) = 5 cm s-1 (b) maximum magnitude of velocity

= 2A = 5cm s-2

1srad2

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By

T