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waves_03

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waves_03

TRANSVERSE WAVES

ON STRINGS

Two sine waves travelling in opposite directions standing wave

Some animations courtesy of Dr. Dan Russell, Kettering University

waves_03: MINDMAP SUMMARY - TRANSVERSE WAVES ON STRINGS

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

TRANSVERSE WAVES ON STRINGS

Wave speed v (speed of propagation)

- The string (linear density ) must be under tension FTfor wave to propagate
- increases with increasing tension FT
- decreases with increasing mass per unit length
- independent of amplitude or frequency

linear density

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

Solution 1

F = 25.0 N = 2.50 g.m-1 = 2.50×10-3 kg.m-1f = 50.0 Hz

A = 5.00 mm = 5×10-3 m v = ? m.s-1 = ? rad.s-1T = ? s = ? m

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

- 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

Reflection of waves at a free end

Reflected wave is inverted

PHASE CHANGE

Reflected wave is not inverted

ZERO PHASE CHANGE

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

CP 510

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

CP 510

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

STANDING WAVES

- 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 on strings

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)

amplitude

oscillation

each point oscillates

with SHM, period T = 2 /

CP 511

A string is fixed at one end and driven by a small amplitude sinusoidal driving force fdat the other end.

The natural frequencies of vibration of the string (nodes at each end) are

fo = 150 Hz, 300 Hz, 450 Hz, 600 Hz, ...

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= 450 Hz

fd= 200 Hz

fundamental

3rd harmonic

Standing waves on strings

- 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

Boundary conditions

Speed transverse wave along string

Natural frequencies of vibration

CP 511

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

CP 518

Modes of vibrations of a vibrating string fixed at both ends

Natural frequencies of vibration

Fundamental

node

antinode

CP 518

Harmonic series

Nth harmonic or (N-1)th overtone

N = 2L / N = 1 / NfN = N f1

N = 3 3nd harmonic (2 ndovertone)

3 = L = 3 / 2 f3 = 3f1

N = 2 2nd harmonic (1st overtone)

2 = L = 1 / 2 f2 = 2 f1

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.

CP 518

22 23

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

Problem 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 kg

L = 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

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 = BA = BLA = LBNA = NB

v = fv = (FT / )

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

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

fB / fA = (FTB / FTA)

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

Problem 4

Why does a tree howl?

The branches of trees vibrate because of the wind.

The vibrations produce the howling sound.

Length of limb L = 2.0 m

Transverse wave speed in wood

v = 4.0103 m.s-1

Fundamental L = / 4 = 4 L

v = f

f = v / = (4.0 103) / {(4)(2)} Hz

f = 500 Hz

N

A

Fundamental

mode of vibration

Standing waves in membranes

NB the positions of the nodes and antinodes

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

CHLADNI PLATES

Some of the animations are from the web site

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