CSULA
Summer 2006
I)
(SOLUTION) TAKE-HOME EXAM (Midterm II)
Professor: Rafael Obregon
PHYSICS 202
CONCEPTUAL QUESTIONS
1) g) 2 2v .
By increasing the mass of the hanging object by a factor of four, you have increased the
tension by a factor of four. Since the rubber band has doubled its length, its mass per
unit length or linear mass density has gone down by a factor of two. Therefore,
v1 
T

and v2 
4Te
1
2


8Te

2 2
Te

 (2 2)v1.
2) c) The same as the rate for the original string.
Equation 16.21 relates the energy rate transfer to the linear mass density, frequency,
amplitude and speed. The speed is in turn related to the tension and the linear mass
density so that the equation may be rewritten,
P  12  2 A2 v  12  4 2 f 2 A2
Te

 2 2 f 2 A2  Te .
Therefore, if one doubles the linear mass density and halves the tension, the rate of
energy transfer must remain the same to keep the same amplitude and frequency.
3) d) The power from the sound is spread over a much bigger area by the time it
reaches Station B.
Equation 17.7 describes the intensity that is associated with the power from the source,
equally distributed over a sphere of radius r. Station B is 3 times as far from the blast as
the first station. Since: I  1/r2
Then the intensity should be 1/32 or 1/9 less at Station B than at Station A. The fact that
the decrease in intensity does roughly scale by this amount indicates the decrease is
due mainly to the power distributed over a larger sphere.
4) c) The observer will not hear Doppler Effect.
Let’s analyze the case of the wind blowing from the source towards the listener. The
moving air (wind) has the same effect as if the speed of sound had been increased by
an amount equal to the wind speed. The wavelength of the sound waves (distance that
a wave travels during one period of time) will be increased by the same percentage that
the wind speed is relative to the still-air speed of sound. Since the frequency is the
speed divided by the wavelength f = v/, the frequency does not change, and so there is
no Doppler effect to hear. Alternatively, the wind has the same effect as if the air were
not moving but the source and listener were moving at the same speed in the same
direction.
II) Straightforward problems (10 points ea):
1) Calculate the pressure amplitude of a 2.00-kHz sound wave in air if the displacement
amplitude is equal to 2.00 × 10–8 m.
Pmax  vsmax  air  2 f  vsound smax  1.29kg / m3  6.28 x 2 x103 s 1   343m / s   2 x108 m  
Pmax  0.111N / m2  0.111Pa
2) The speed of a longitudinal wave in a solid is given by: v  B  .
Call:
Less dense material 1.
More dense material 2.
The less dense material will have the higher speed, since the speed is inversely
proportional to the square root of the density.
v1
v2

B 1
B 2

2
 2  1.41
1
III) Intermediate problems (15 points ea):
1) The total distance traveled by each sound will be the width of the inlet (W).
W  vsound  air tair  vsound  saltwater tsaltwater 
W  va ta  vswtsw
(1)
The sound through the air has a longer time of travel than the sound in saltwater,
by 4.5 s:
ta  tsw  4.5s
(2)
va  343m / s
From Table 17.1:
vsw  1533m / s
(3)
By substituting (2) and (3) in (1):
va  tsw  4.5   vswtsw   343m / s  tsw   343m / s  4.5s   1533m / s  t sw 
1533m / s  343m / s  tsw  1543.5m  1190m / s  tsw  1543.5m 
1543.5
s  1.30s 
1190
W  vswtsw  (1533m / s )(1.30s )  1,990m  1.99km
tsw 
2) The bug moves in SHM as the wave passes. The maximum K of a particle in SHM is
the total energy, which is given by: Etotal  12 kA2 . Compare the two K maxima.
2
 A   2.0 cm  4
 1 2  2  
 
K1 2 kA1  A1   3.0 cm  9
K2
1
2
kA22
2
IV) Challenging problems (15 points ea):
1) The speed of the pulses is found from the tension and mass per unit length of the wire.
v
Te


Te

m/ L
225 N
 150m / s
 0.100kg  / 10.0m 
The total distance traveled by the two pulses will be the length of the wire. The second
pulse has a shorter time of travel than the first pulse, by 1/30 s.
1 
1
1

  2vt1  v  2vt1  L  v 
30
30
 30 
1
10.0 m    s  150 m s 
 30 
L  d1  d 2  vt1  vt2  vt1  v  t1 
L
t1 
1
v
30 
2v
2 150 m s 
 0.05s  d1  vt1  150 m s  0.05s   7.5m
The two pulses meet 7.5 m from the end where the first pulse originated.
2) Using equation from page 516:
s
with s  smax sin  kx  t  & B  v2 , then:
x
  smax sin  kx  t  
P    v 2
   v 2 smax k cos  kx  t 
x
 2 v 2 smax
2
2
But: k 
 P    v 2 smax
cos  kx  t    




P   B

 cos  kx  t 
