waves_03

TRANSVERSE WAVES ON STRINGS

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Animations courtesy of Dr. Dan Russell, Kettering University

Travelling transverse waves, speed of propagation, wave function, string tension, linear density, reflection (fixed and free ends), interference, boundary conditions, standing waves, stationary waves, SHM, string musical instruments, amplitude, nodes, antinodes, period, frequency, wavelength, propagation constant (wave number), angular frequency, normal modes of vibrations, natural frequencies of vibration, fundamental, harmonics, overtones, harmonic series, frequency spectrum, radian, phase, sinusoidal functions

waves_03: MINDMAP SUMMARY - TRANSVERSE WAVES ON STRINGS

v FT

s

s

m

L

2

2 1 22v f f T k

T k T f

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The string (linear density ) must be under tension FT for wave to propagate

• increases with increasing tension FT

• decreases with increasing mass per unit length

• independent of amplitude or frequency

v FT

s

s

m

L

TRANSVERSE WAVES ON STRINGS

Wave speed v (speed of propagation)

linear density

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Problem 1

A string has a mass per unit length of 2.50 g.m-1 and is put under a tension of 25.0 N as it is stretched taut along the x-axis. The free end is attached to a tuning fork that vibrates at 50.0 Hz, setting up a transverse wave on the string having an amplitude of 5.00 mm. Determine the speed, angular frequency, period, and wavelength of the disturbance.

[Ans: 100 m.s-1, 3.14x102 rad.s-1, 2.00x10-2 s, 2.00 m]

use the ISEE method

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Solution 1

F = 25.0 N = 2.50 g.m-1 = 2.50×10-3 kg.m-1 f = 50.0 HzA = 5.00 mm = 5×10-3 m v = ? m.s-1 = ? rad.s-1 T = ? s = ? m

Tv

Speed of a transverse wave on a string speed of a wave

v f

-1 -13

-1 2 -1

2

25m.s 100 m.s

2.5 10

2 2 50 rad.s 3.14 10 rad.s

1 1s 2.00 10 s

50

100m 2.00 m

50

Tv

f

Tf

vv f

f

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Pulse on a rope

• When pulse reaches the attachment point at the wall the pulse is reflected

• If attachment is fixed the pulse inverts on reflection

• If attachment point can slide freely of a rod, the pulse reflects without inversion

• If wave encounters a discontinuity, there will be some reflection and some transmission

• Example: two joined strings, different . What changes across the discontinuity - frequency, wavelength, wave speed?

Reflection of waves at a fixed end

Reflected wave is inverted / 2 PHASE CHANGE

Reflection of waves at a free end

Reflected wave is not invertedZERO PHASE CHANGE

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Refection of a pulse - string with boundary condition at junction like a fixed end

Incident pulse

Transmitted pulseReflected pulse

Reflected wave rad(180°) out of phasewith incident wave

Heavy string exertsa downward force onlight string when pulsearrives

CP 510

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Refection of a pulse - string with boundary condition at junction like a free end

Incident pulse

Transmitted pulseReflected pulse

Reflected wave: in phasewith incident wave, 0 rad or 0°C phase difference

Heavy string pulls lightstring up when pulse arrives, string stretches then recovers producing reflected pulse

CP 510

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Refection of a pulse - string with boundary condition at junction like a fixed end

Refection of a pulse - string with boundary condition at junction like a free end

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• If we try to produce a traveling harmonic wave on a rope, repeated reflections from the end produces a wave traveling in the opposite direction - with subsequent reflections we have waves travelling in both directions

• The result is the superposition (sum) of two waves traveling in opposite directions

• The superposition of two waves of the same amplitude travelling in opposite directions is called a standing wave

• Examples: transverse standing waves on a string with both ends fixed (e.g. stringed musical instruments); longitudinal standing waves in an air column (e.g. organ pipes and wind instruments)

STANDING WAVES

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( , ) sin( ) sin( )

2 sin( ) cos( )

y x t A kx t A kx t

A kx t

oscillationamplitude

Standing waves on strings

CP 511

each point oscillates with SHM, period T = 2 /

Two waves travelling in opposite directions with equal displacement amplitudes and with identical periods and wavelengths interfere with each other to give a standing (stationary) wave (not a travelling wave - positions of nodes and antinodes are fixed with time)

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A string is fixed at one end and driven by a small amplitude sinusoidal driving force fd

at the other end.

The natural frequencies of vibration of the string (nodes at each end) arefo = 150 Hz, 300 Hz, 450 Hz, 600 Hz, ...

Click the next image after an animation stops

The string vibrates at the frequency of the driving force. When the string is excited at one of its natural frequencies, large amplitude standing waves are set up on the string (resonance).

fd = 150 Hz fd = 200 Hzfd = 450 Hz

fundamental 3rd harmonic

string_150.avi string_450.avi string_200.avi

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• String fixed at both ends

• A steady pattern of vibration will result if the length corresponds to an integer number of half wavelengths

• In this case the wave reflected at an end will be exactly in phase with the incoming wave

• This situations occurs for a discrete set of frequencies

2L N

2

2

L v vf N

N L

CP 511

T1

2NN

Fvf N

L

TFv

Speed transverse wave along string

Natural frequencies of vibration

Boundary conditions

Standing waves on strings

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Why do musicians have to tune their string instruments before a concert?

t u n i n g k n o b s ( p e g s )- a d j u s t F TB o d y o f i n s t r u m e n t ( b e l l y )

r e s o n a n t c h a m b e r - a m p l i f i e r

d i f f e r e n t s t r i n g -

b r i d g e s - c h a n g e L

TFv

T1 F

fL

L

T1

1

2

Ff

L 1 1 , 2 , 3 , . . .Nf N f N

F i n g e r -b o a r d

CP 518

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L N2

2L

N

f v

N

v

2L

node antinode

Fundamental

f1 v

2L

fN Nf1

CP 518

Modes of vibrations of a vibrating string fixed at both ends

Natural frequencies of vibration

T1

2NN

Fvf N

L

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0

20

40

60

80

100

120

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

position along string

N = 1 fundamental or first harmonic 1 = 2L f1 = (1/2L).(FT / )

Resonance (“large” amplitude oscillations) occurs when the string is excited or driven at one of its natural frequencies.

Harmonic series

Nth harmonic or (N-1)th overtone

N = 2L / N = 1 / N fN = N f1

N = 2 2nd harmonic (1st overtone) 2 = L = 1 / 2 f2 = 2 f1

N = 3 3nd harmonic (1st overtone) 3 = L = 3 / 2 f3 = 2 f1

CP 51822 23

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0

1

2

3

4

5

6

7

8

9

10

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15harmonics (fundamental f 1 = 440 Hz)

-50

-40

-30

-20

-10

0

10

20

30

40

50

0 0.002 0.004 0.006 0.008

time t (s)

0

2

4

6

8

10

12

14

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15harmonics (fundamental f 1 = 440 Hz)

-50

-40

-30

-20

-10

0

10

20

30

40

50

0 0.002 0.004 0.006 0.008

time t (s)

violin – spectrum

viola – spectrum

CP 518

Problem solving strategy: I S E E

Identity: What is the question asking (target variables) ? What type of problem, relevant concepts, approach ?

Set up: Diagrams Equations Data (units) Physical principals

Execute: Answer question Rearrange equations then substitute numbers

Evaluate: Check your answer – look at limiting cases sensible ? units ? significant figures ?

PRACTICE ONLY MAKES PERMANENT

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20Problem 2

A guitar string is 900 mm long and has a mass of 3.6 g. The distance from the bridge to the support post is 600 mm and the string is under a tension of 520 N.

1 Sketch the shape of the wave for the fundamental mode of vibration

2 Calculate the frequency of the fundamental.

3 Sketch the shape of the string for the sixth harmonic and calculate its frequency.

4 Sketch the shape of the string for the third overtone and calculate its frequency.

Solution 2 L1 = 900 mm = 0.900 m m = 3.6 g = 3.610-3 kgL = 600 mm = 0.600 m FT = 520 N = m / L1 = (3.610-3 / 0.9) kg.m-1 = 0.004 kg.m-1

v = (FT / ) = (520 / 0.004) m.s-1= 360.6 m.s-1

1 = 2L = (2)(0.600) m = 1.200 m

Fundamental frequency f1 = v / 1 = (360.6 / 1.2) Hz = 300 Hz

fN = N f1

sixth harmonic N = 6 f6 = (6)(300) Hz = 1800 Hz = 1.8 kHz

third overtone = 4th harmonic N = 4 f4 = (4)(300) Hz = 1200 Hz = 1.2 kHz

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Problem 3

A particular violin string plays at a frequency of 440 Hz.

If the tension is increased by 8.0%, what is the new frequency?

Solution 3

fA = 440 Hz fB = ? Hz

FTB = 1.08 FTA

A = B A = B LA = LB NA = NB

v = f v = (FT / )

string fixed at both ends L = N /2 = 2L / N

natural frequencies fN = N v / 2L = (N / 2L).(FT / )

fB / fA = (FTB / FTA)

fB = (440)(1.08) Hz = 457 Hz

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Why does a tree howl?The branches of trees vibrate because of the wind.The vibrations produce the howling sound.

N A

Fundamental mode of vibration

Problem 4

Length of limb L = 2.0 mTransverse wave speed in wood v = 4.0103 m.s-1

Fundamental L = / 4 = 4 Lv = f f = v / = (4.0 103) / {(4)(2)} Hzf = 500 Hz

standing_1.avi

http://www.isvr.soton.ac.uk/spcg/tutorial/tutorial/Tutorial_files/Web-standing-membrane.htm

Standing waves in membranes

NB the positions of the nodes and antinodes

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CHLADNI PLATES

Some of the animations are from the web site

http://paws.kettering.edu/~drussell/demos.html