Standing sound waves
Standing sound waves
Sound in fluids is a wave composed of longitudinal vibrations
of molecules. The speed of sound in a gas depends on the
temperature. For air at room temperature, the speed of
sound is about 340 m/s.
At a solid boundary, the vibration amplitude must be zero
(a standing wave node).
Physical picture of
particle motions (sound
wave in a closed tube)
node
antinode node
antinode node
graphical picture
Standing sound waves in tubes – Boundary Conditions
-there is a node at a closed end
-less obviously, there is an antinode at an open end
(this is only approximately true)
antinode
node
antinode
graphical picture
Exercise: Sketch the first three standing-wave patterns for a
pipe of length L, and find the wavelengths and frequencies if:
a) both ends are closed
b) both ends are open
c) one end open
a) Pipe with both ends closed
L
b) Pipe with both ends open
L
c) Pipe with one closed end, one open end
L  14 1
L
L  34 3
L  54 5
Beats Chap 18
Given two nearly identical harmonic waves with slightly
different frequencies and wave numbers but moving in
the same direction.
y1  A sin(k1 x  1t )
y2  A sin(k2 x  2t )
v

1
k1


2
k2
 wavevelocity
y  y1  y2  A sin(k1 x  1t )  A sin(k2 x  2t )

 2 A sin (
k k
1
2
2
)x 

 
1
2
2
    
t  cos
k k
1
2
x
 
2
This is the product of two traveling waves.
1
2
2

t
Define : k 
k k
2
1
2
and  
 
1
2
2
so regular wave velocity

is nearly the
k
same as the individual waves.
k1  k2
  2
and   1
2
2

so group velocity, vgroup =
is much different
k
from the individual waves.
Define : k 
in general ,    (k )
and
vgroup
d

dk
For waves on a string,
T
   (k ) 
so v
group


T

k
v
In this case the group and wave velocities are the same.
y  y1  y2  2 A sin  kx  t   cos  k x   t 
Nearly same as
initial wave
Much different than initial
wave. This is the GROUP
wave. It has much longer
wavelength and much slower
frequency
time
in phase
180o out of phase
in phase
t
Temporal Beats
Two waves of different frequencies traveling in the
same direction produce a fluctuation in amplitude.
Since the frequencies are different, the two vibrations
drift in and out of phase with each other, causing the
total amplitude to vary with time.
y
time
1 beat
Note: maximum intensity when “amplitude” part is 2A
# beats/second = beat frequency
= twice the group frequency
 f1  f 2 
2

 2 
=f  f  f
b
1
2
The beat frequency (number of beats per second)
is equal to the difference between the frequencies.
The frequency of the combined waves is:
f f
2
1
2
Quiz:
One tuning fork has a frequency 440Hz and another
has a frequency 450Hz.
a) What is the beat frequency of the sound heard with
both tuning forks vibrating?
A) 890 Hz
B) 445 Hz
C) 10 Hz
D) 5 Hz
What is the actual sound frequency heard in this case?
Spatial beats