### 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
v
FT
ms

Ls

v f 

T


k
2
  2 f 
T
1
T
f
k
2

2
TRANSVERSE WAVES ON STRINGS
Wave speed v (speed of propagation)
The string (linear density ) must be under
tension FT for wave to propagate
v
FT

linear density
ms

Ls
• increases with increasing tension FT
• decreases with increasing mass per unit
length 
• independent of amplitude or frequency
3
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.
use the ISEE method
[Ans: 100 m.s-1, 3.14x102 rad.s-1, 2.00x10-2 s, 2.00 m]
4
Solution 1
F = 25.0 N  = 2.50 g.m-1 = 2.50×10-3 kg.m-1 f = 50.0 Hz
A = 5.00 mm = 5×10-3 m v = ? m.s-1  = ? rad.s-1 T = ? s  = ? m
Speed of a transverse wave on a string
v
v
T


T

speed of a wave
v f 
25
-1
-1
m.s

100
m.s
2.5 103
  2  f   2   50  rad.s-1  3.14 102 rad.s -1
1 1
T 
s  2.00  102 s
f 50
v 100
v f   
m  2.00 m
f
50
5
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?
6
Reflection of waves at a fixed end
Reflected wave is inverted
 PHASE CHANGE
Reflection of waves at a free end
Reflected wave is not inverted
ZERO PHASE CHANGE
7
Refection of a pulse - string with boundary condition at
junction like a fixed end
Incident pulse
Reflected pulse
(180°) out of phase
with incident wave
Transmitted pulse
Heavy string exerts
a downward force on
light string when pulse
arrives
8
CP 510
Refection of a pulse - string with boundary condition at
junction like a free end
Incident pulse
Reflected pulse
Reflected wave: in phase
or 0°C phase difference
Transmitted pulse
Heavy string pulls light
string up when pulse arrives,
string stretches then recovers
producing reflected pulse
9
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
10
11
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)
12
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)
y( x, t )  A sin(kx  t )  A sin(kx  t )
 2 A sin(kx) cos(t )
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 fd
at 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
fundamental
3rd harmonic
fd = 200 Hz
13
14
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 
LN
Speed transverse wave along string
Natural frequencies of vibration

2
2L
v
v

 f  N
N

2L
FT
v

 1  FT
fN 
 N 
N
 2L  
v
CP 511
15
Why do musicians have to tune their string instruments before a concert?
different string - 
Fingerboard
bridges - change L
tuning knobs (pegs)
Body of instrument (belly)
resonant chamber - amplifier
L
1
f1 
2L
v
FT

FT

f N  N f1
f 
1
L
FT

N  1, 2, 3,...
CP 518
16
Modes of vibrations of a vibrating string fixed at both ends
Natural frequencies of vibration
2L

N
v
v
f  N

2L

LN
node

antinode

2

Fundamental f1  v
2L
 1  FT
fN 
 N 
N
 2L  

v
f N  Nf1
CP 518
17
Harmonic series
120
Nth harmonic or (N-1)th overtone
N = 2L / N = 1 / N
100
fN = N f1
80
N=3
N=2
3nd harmonic (2 nd overtone)
3 = L = 3 / 2
f3 = 3 f1
2nd
60
40
harmonic (1st overtone)
2 = L = 1 / 2
f2 = 2 f1
20
N=1
fundamental or first harmonic
1 = 2L f1 = (1/2L).(FT / )
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
position along string
Resonance (“large” amplitude oscillations) occurs when the string is excited or
driven at one of its natural frequencies.
22 23
CP 518
1
violin – spectrum
18
10
50
9
40
8
30
7
20
6
10
5
0
4
-10
-20
3
-30
2
-40
1
-50
0
1
2
3
4
5
6
7
8
9
0
10 11 12 13 14 15
0.002
0.004
0.006
0.008
0.006
0.008
time t (s)
harmonics (fundamental f 1 = 440 Hz)
viola – spectrum
14
50
40
12
30
10
20
10
8
0
-10
6
-20
4
-30
-40
2
-50
0
1
2
3
4
5
6
7
8
9
10 11 12 13 14 15
harmonics (fundamental f 1 = 440 Hz)
0
0.002
0.004
time t (s)
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
PRACTICE ONLY
MAKES PERMANENT
Rearrange equations then substitute numbers
Evaluate:
sensible ?
units ?
significant figures ?
19
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.
20
Solution 2
L1 = 900 mm = 0.900 m m = 3.6 g = 3.610-3 kg
L = 600 mm = 0.600 m FT = 520 N
 = m / L1 = (3.610-3 / 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 f 1
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
22
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
v=f
LA = LB
NA = NB
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
24
Why does a tree howl?
Problem 4
The branches of trees vibrate because of the wind.
The vibrations produce the howling sound.
N
A
Length of limb L = 2.0 m
Transverse wave speed in wood
v = 4.0103 m.s-1
Fundamental L =  / 4  = 4 L
v=f
f = v /  = (4.0 103) / {(4)(2)} Hz
f = 500 Hz
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