PHY2053
Summer 2013
Exam 3 Solutions
1. By Pascal’s principle, the pressure needed to lift the mass is distributed throughout the
hydraulic fluid. The pressure is
Pout 
Fout
mg
(1250kg )(9.8 m/s2 )

 1.68 105 Pa
2
2
2
Aout  (rout )  [( 6 in)(2.54 10 m/in)]
By Pascal’s principle, Pout = Pin.
Pin 
Fin
Ain
Fin  Pin Ain  (1.68 105 Pa)( [(1in)(2.54 102 m/in)]2  340 N
2. The pressure at the bottom of the tube is due to the weight of the water, the weight of the
oil, and the atmosphere.
P2  P1  O ghO  W ghW
 1.01105 Pa  (800kg/m 3 )(9.8 m/s2 )(15 m)  (1000kg/m 3 )(9.8 m/s2 )(12.5 m)
 3.4 105 Pa
3. There two ways to solve the problem. One is to use the density of steel to find the
volume of the bolt. First find the mass of the bolt
W  mg
W 0.770 N
m

 0.0786kg
g 9.8 m/s2
Use the density to find the volume,

V
m
V
m


0.0786kg
 1.0  105 m3
3
7860kg/m
The other way to solve this problem is to use Archimedes’ principle. The buoyant force
is the difference in the actual weight and the apparent weight,
FB  W  Wapp  0.770 N  0.672 N  0.098N
1
Now use Archimedes’ principle
FB  gV
V 
F
0.098 N

 1.0  105 m 3
g (1000kg/m 3 )(9.8 m/s 2 )
4. The volume flow rate is related to the speed of the water,
V
 Av
t
1 V
1 V
1
0.080 m3
v
 2

 13.2 m/s
A t r t  ((0.0254m) / 2) 2 12 s
This velocity is created as the water flows from down from the tower. Using Bernoulli’s
equation
P1  gy1  12 v1  P2  gy2  12 v2
2
2
Associate the 1 subscript with the tower and the 2 subscript with the faucet. Since the
tower and the faucet are open to the atmosphere, P1 = P2 = Patm. The tower has such a
large volume and its level falls very slowly, v1 = 0.
Patm  gy1  12  (0) 2  Patm  gy2  12 v2
2
g ( y1  y2 )  12 v2 2
2
y1  y2 
v2
(13.2 m/s)2

 8.8 m
2 g 2(9.8 m/s 2 )
5. Use Poiseuille’s law
V  P / L 4

r
t
8 
V 8 L
8 (1900m)(0.20 Pa  s)
P 
 (4.5 104 m3 /s)
 860 Pa
4
t  r

(0.15 cm) 4
6. The stress in the cable is
stress 
F mg (250 kg )(9.8 m/s2 )
 2 
 5.42 106 Pa
2
A r
 (0.012 m)
Young’s modulus is
2
stress
strain
stress 5.42 106 Pa
strain 

 2.7 105
11
Y
2.0 10 Pa
Y
Since strain = L/L this is the fractional increase in length.
7. The volume compresses 0.10% = 0.001. The relationship defining the bulk modulus,
P   B
V
 (90 109 Pa)(0.001)  9.0 107 Pa
V
8. The spring constant is
k
mg (3 kg )(9.8 m/s2 )

 118 N/m
x
0.25 m
The period of a simple harmonic oscillator is
m
(3 kg )
 2
 1.0 s
k
(118 N/m)
T  2
9. The period of a pendulum is
T  2
L
g
Forming a ratio
TM

TE
L
gM
2
L
gE
2

gE
gM
gE
9.8 m/s 2
 (2.0 s)
 3.2 s
gM
3.8 m/s 2
TM  TE
10. Use the conservation of energy
U1  K1  U 2  K 2
1
2
kA2  0  12 kx2  K 2
2
3
The amplitude is A = 0.25 cm. Solving for K2,
K 2  12 kA2  12 kx2  12 k ( A2  x2 )  12 (50 N/m)((0.25 m) 2  (0.10 m) 2 )  1.3 J
2
2
11. Since vm = A, the amplitude can be found,
vm
A


4 m/s
 0.667 m
6 rad/s
The position function for the given velocity function is
y  Asint  (0.667m) sin[(6 rad/s)(3s)]  0.50 m
Your calculator must be in radian mode to get this result.
12. The intensity is a function of distance,
I
P
4r 2
Forming a ratio
P
2
2
 r1 
I 2 4r2

  
P
I1
 r2 
2
4r1
Solving for r2
I 2  r1 
 
I1  r2 
r1

r2
r2  r1
2
I2
I1
I1
0.1 W/m 2
 (2 m)
 6. 3 m
I2
0.01W/m 2
13. The speed of a wave in string is
4
F
v
v2 

F

F
v2
m F

L v2
F
75 N
m 2L
(5 m)  0.019 kg
v
(140 m/s) 2

14. The velocity of the wave is related to the wavenumber and angular frequency
v

k
  vk  (60 m/s)(3.0 rad/m)  180 rad/s
The general equation of a wave traveling in the –x direction is
y  A cos(t  kx)
Substituting the values above,
y  A cos((180rad/s)t  (3.0 rad/m) x)
15. Since the end of the rope is a free end, the reflected pulse will have the same orientation
as the incident pulse. Both pulses will point upward when they meet and
A  A1  A2  (3 cm)  (4 cm)  7 cm
16. The speed of the wave can be found
v
F


mg

(2.10 kg )(9.8 m/s2 )
 761m/s
3.55 105 kg/m

For the lowest frequency, the wavelength is twice the length of the string,  = 2L = 4 m.
The frequency is
f 
v


761m/s
 190 Hz
4m
5
17. The decibel scale is defined as
I 

 I0 
  10 log
Solving for I,
I 

I
 0
  10 log
I 

 I0 
 / 10  log
I
 10 / 10
I0
I  I 010 / 10  (1.0 1012 W/m2 )1070 / 10  1.0 105 W/m2
18. The frequencies for a pipe open at both ends is given by
fn  n
v
2L
where n = 1, 2, 3, etc.
f1  1
v
340 m/s

 170 Hz
2L
2(1m)
The next frequency corresponds to n = 2
f1  2
v
340 m/s

 340 Hz
2L
1m
The frequencies are 170 Hz and 340 Hz.
19. Two waves with similar frequencies create beats.
20. The vehicles are traveling in different directions
vs
vo
6
The sound travels in the same direction as the sound. Therefore vs > 0. The car moves
against the sound waves and vo < 0. Using the equation for the Doppler effect
 vo
1
v
f0  f s 
v
 1 s

v

 (25 m/s) 

1


340 m/s 
  (550 Hz)
 660 Hz
35 m/s 


1



340 m/s 


7