Transcript lecture32

Standing waves on a string (review)
2L
1


 2L
L  2 1
1
1
L  22 2  2
L  32 3
L  n2 1
2L
2 
L
2
2L
3 
3
n 
v
2L
n
v
fn 
n
 nf1
n
2L
Different boundary conditions:
•Both ends fixed (see above)
•Both ends free (similar to both ends fixed )
•One end fixed and on end free (next slide)
n=1,2,3...
f n  nf1
Standing waves on a string (review)
One end fixed and on end free
n=1
L  14 1
n=3
L  34 3
L  n4 n
v
n
Tn
 n f n
v
vn
fn 

 f1n
n 4 L
4L
n 
n
n  1,3,5...
Standing waves in tubes (longitudinal)
Waves in tubes (pipes) can be described in terms of:
•displacement vibrations of the fluid
•pressure variations in the fluid
A pressure node is a displacement antinode and vice versa
Open and both ends closed pipes
Closed (both ends): displacement
Open:
pressure
pressure
displacement
n=1
L  12 1
n=2
L  22 2
n=3
L  32 3
v
n
Tn
L  n2 n
 n f n
fn 
v
n

vn
 nf1
2L
2L
n 
n
n  1,2,3...
One end open pipes (stopped pipes)
displacement
pressure
n=1
L  14 1
n=3
L  34 3
L  n4 n
v
n
Tn
 n f n
fn 
v
n

vn
 f1n
4L
4L
n 
n
n  1,3,5...
Example: Standing sound waves are produced in a pipe that is 0.6 m long.
For the first overtone, determine the locations along the pipe (measured
from the left end) of the displacement nodes. The pipe is closed at the left
end and open at the right end.
L  0.6m
Nodes at 0.0 m and 0.4 m
4 L 4  0.6m

 0.8m
3
3
3 / 2  0.4m
3 
Sound
Acoustic waves in the range of frequencies: 20Hz -20,000Hz
Sound waves:
• can travel in any solid, liquid or gas
• travel faster in a medium that is more dense
• in liquids and gases sound waves are longitudinal ONLY!
• longitudinal and transversal sound waves could propagate in in solids
Sound in air is a longitudinal wave that contains regions of low and high pressure
Pressure sensor
Vibrating tuning fork
These pressure variations are usually small – a “loud” sound changes
the pressure by 2.0x10-5 atm
Speed of Sound
Speed of waves in strings (review): v 
Speed of sound waves:
B is bulk modulus, defined as
 is density
v
F

B

dV
1
 p
V
B
Speed of sound waves in a gas:
B  p0
 is ratio of heat capacities,
p0 is the equilibrium pressure of the gas
v
RT
M
Example: Speed of Sound in Air
The speed of sound in an ideal gas is
v
 RT
, where
M
  "ratio of heat capacities"
R  molar gas constant
= 8.314 J/mol  K
T  temperature in K
M  molar mass in kg
Air is a diatomic gas, so
  1.40 .
The molar mass of dry air, containing
about 20% O 2 and 80% N 2 , is
M  0.0290 kg/mol .
Let the ambient temperature be
T  22 °C = 72 °F = 295 K .
Then
v
(1.40)(8.31 J/mol  K)(295 K)
.
0.0290 kg
 344 m/s
This agrees well with experiments.
Speed of Sound in Some Common Substances
Substance
1. Air (20 oC)
Speed (m/s)
344
2. Helium
1,006
3. Water
1,140
4. Lead
1,200
5. Human tissue
1,540
6. Aluminum
5,100
7. Iron and steel
5,200
Pressure variations
dV y

V
x
dV
1
 p
V
B
y
p  B
x
yx, t   A coskx  t 
px, t   BkAsin kx  t 
Example: Displacement Wave Amplitude of Sound in Air
Let the pressure wave amplitude of a sound wave in air at pressure p0  1.00 atm
be pmax  0.030 Pa (a typical value). The frequency of the wave is f  1000 Hz
and the speed of sound in the air is v  344 m/s .
What is the displacement wave amplitude of the sound wave?
Solution :
pmax
(1)
Bk
1 atm = 1.013 kPa and  = 1.40 for air.
pmax  BAk, so A 
B   p0  (1.40)(1.013  10 5 Pa)  1.42  10 5 Pa.
2 f
v
v
Putting (3) and (2) into (1) gives
k

A
and   2 f , so k 
(2)
(3)
pmax v
(0.030 Pa)(344 m/s)

 1.2  10 8 m  12 nm .
5
2 Bf 2 (1.42  10 Pa)(1000 Hz)
Properties of Sound
We classify sounds according to their
waveforms:
- A “pure tone” is a sound with a
sinusoidal waveform.
- This is a sound with a single frequency;
produced by tuning fork, etc.
- A “complex tone” is a sound that
repeats itself but is not sinusoidal
Most sounds are like this!
A note from a musical instrument will
be mostly sinusoidal, but have a
character all its own that is specific
to the instrument.
“Noise” is sound with a complex
waveform that does not repeat
No definite wavelength or
frequency
piano note
Perception of Sound
We use three qualities to characterize how we perceive sound:
1. Pitch. 2. Loudness. 3. Tone quality.
1. The pitch of a sound is how “high” or “low” we perceive a sound to be
(directly related to how we perceive frequency); combinations of notes that
are “pleasing” to the ear have frequencies that are related by a simple wholenumber ratio (Pythagoras)
2. The loudness of a sound is how we perceive the amplitude of the sound
wave
The unit of sound loudness is the decibel (dB)
Quiet: 30 dB; Moderate: 50 dB; Noisy: 70 dB; Very loud: 90 dB; Problems: 120 dB
3. The tone quality of a sound is how we distinguish sounds of the same pitch
and loudness (how we perceive the qualities of the waveform)
Musical Instruments
Anything that causes pressure vibrations creates sound!
Stringed musical instruments produce sound by vibrating a wire or string
A plucked string on a guitar produces a
different sound than a violin
The tension in the string is used to adjust
the wave speed and the frequency
Wind instruments produce sound by pressure
waves in a tube
The sound reflects partly at the open end
Valves change the effective length of the tube
Complex Waves
To produce a pure fundamental
tone on a guitar string:
Pull the template away fast.
Then the string will vibrate in its
fundamental mode.
Pluck a guitar string in the middle:
Fourier Theorem: The initial shape is a
superposition of the sinusoidal shapes
for the fundamental mode and for higher
harmonics.
The sound produced is the fundamental
frequency plus higher harmonics.
Different musical instruments sound
different even when playing the same
tone partly because the harmonic
contents are different.
Fourier Analysis
Fourier Analysis
"Fourier analysis" allows the amplitudes An
to be determined from y(x, t  0) (see text
supplement "Waves"). For our y(x, t  0),
one obtains:
Fourier Theorem for an Even Function

y(x,t  0)   An cos(kn x)
n=0
Two fixed ends: kn 
2
n
 n
y(x,t  0)   An cos 
 L
n=0

and n 

x .

2L
, so
n
8A
if n is odd
2 2
.
 n
An  0 if n is even
An 