General Physics (RE) Final Exam, Fall 2010
1.
(20%)
A uniform rod of mass M and length L is pivoted about
a horizontal axis at one end and attached to a vertical
spring whose constant is k. For small angle
displacements from the equilibrium position
(indicated by the dashed line), what is the period of
the oscillation?
Ans:
For small angle displacements,
L   y = change of spring length.
Also the force exerted by the spring is nearly perpendicular to the rod so that
d 2
I 2   L k y   L2 k 
dt
With I 
1
M L2 , we have
3
d 2
3k
 
2
dt
M
The angular frequency of the oscillation is therefore
which corresponds to a period
2.
T  2

3k
,
M
M
.
3k
(20%)
A bat is flitting about in a cave, navigating via ultrasonic bleeps. Assume that the
sound emission frequency of the bat is 39000 Hz. During one fast swoop directly
toward a flat wall surface, the bat is moving at 0.025 times the speed of sound in
air. What frequency does the bat hear reflected off the wall?
Ans.
Consider a wavefront a distance  away from the wall. It’ll hit the wall and be
reflected in time T equal to the period of the sound wave. During which time, the
bat will have moved a distance v T towards the wall and then release the next wave
front. Let the 1st reflected wavefront take time t to reach the bat. t is then the
period of the wave that the bat recieves.
Since the bat will have moved a distance
vt, we have
ct  vt  cT  vT

v
1
cv
cT
t
T
v
cv
1
c
Hence
1 1  0.025
f 
39000  41 kHz
t 1  0.025
To check the result, let the 2nd front take time t’ after the 1st to reach the bat.
When the bat recieves the 1st front, it is c t away from the wall while the 2nd front
is v t away. When the bat recieves the 2nd front, it’ll have travelled a distance v
t’.
Hence,
ct  vt   ct  vt 
so that t  t
as expected.
3. (20%)
Ice has formed on a shallow pond, and a steady state has been reached, with the air
above the ice at 5.0C and the bottom of the pond at 4.0C. If the total depth of
ice and water is 2.0 m, how thick is the ice?
Assume the thermal conductivity of ice and water to be 0.40 and 0.12 cal/mCs,
respectively.
Ans.
Let the ice be t thick. The total heat flow per unit area is
5.0 C
4.0 C
  0.12 cal / m C  s 
0
 0.40 cal / m C  s 
tm
 2.0  t  m
Or
 0.40 
5.0
  0.12
t

4.0
0
 2.0  t 
which gives
t  1.6 m
4. (20%)
The thermal expansion coefficient of air at 0C and 1 atm is 0.00366 /C. Use it to
define the Kelvin scale by means of the ideal gas law.
Ans.
From PV  nRT we get

1  V 
1 n R 1
 



V   T  P V  P  T
Hence, 0C corresponds to
5.
or
T
1

1
 273.2 K
0.00366
(20%)
In the figure, water stands at depth D = 35.0 m
behind the vertical upstream face of a dam of
width W = 300 m. Find
(a) the net horizontal force on the dam from the
gauge pressure of the water and
(b) the net torque due to that force about a line
through O parallel to the width of the dam.
Ans.
D
(a)
F   P dA    g h W dh 
0

1
 g W D2
2
1
2
1000 kg / m 3   9.8 m / s 2   300 m   35.0 m   1.8  109 N

2
(b) Let r be the vector from O to a point on the dam at depth h. Then
D
    g h r sin  W dh
0
where  is the angle between r & the horizontal.
D
1  1
1
    g h  D  h  W dh   g W  D 3  D 3    g W D 3
3  6
2
0

1
3
1000 kg / m 3   9.8 m / s 2   300 m   35.0 m   2.1  1010 N  m

6