PHY2053 Summer 2012
Exam 3
Solutions
1.
The cross-sectional area of the cable is
r2
A
(1.2 10 2 m) 2
4.52 10 4 m 2
The force exerted by the mass is due to its weight
F W
mg
(250 kg )(9.80 m/s 2 )
2450 N
Young’s modulus is
F
A
L
L
2.
L
L
F
AY
2450 N
2.7 10
(4.52 10 m 2 )(2.0 1011 Pa)
5
4
The definition of bulk modulus
P
V
3.
Y
V
V
PV
B
B
(300 1.01 10 5 Pa)(2.00 m3 )
50 10 9 Pa
1.2 10 3 m3
The angular frequency can be found from the period
2
T
2
0.3 s
20.9 rad/s
The angular frequency depends on the spring constant and mass
k
4.
k
m
2
m (20.9 rad/s)2 (3 kg ) 1300 N/m
The period of a simple pendulum is given by
T
L
g
2
Since mass does not appear in the equation, the change of mass does not affect the
period. Forming a ratio
T2
T1
2
L2
g
2
L1
g
L2
L1
Solve for T2,
T2
5.
L2
L1
T1
2L
L
(3.6 s)
5.1 s
Conservation of energy
K1 U1
1
2
2
1
1
2
mv
2
mv1
K2 U 2
2
1
1
2
kx
2
kx1
mv2
mv2
2
2
1
2
kx2
kx2
2
2
When the mass is at its maximum amplitude, its velocity is zero.
2
kx1
2
kx1
mv1
mv1
2
mv2
2
m(0) 2
A
6.
2
kx2
2
kA2
m 2
v1
k
0.75 kg
(0.80 m/s )2 (0.15 m) 2
30 N/m
2
x1
The angular frequency can be found from the period
2
T
The equation for the velocity is
2
1.5 s
4.19 rad/s
0.20 m
x
7.
A sin t
(0.50 m)(4.19 rad/s) sin[( 4.19 rad/s)(5 s)]
The intensity is defined as
P
4 r2
I
Forming a ratio
I2
I1
P
2
4 r2
P
2
4 r1
r1
r2
2
Solving for I2,
r1
r2
I2
8.
2
2
I1
3m
(10 W/m2 ) 5.6 W/m2
4m
The speed of a wave in a string is given by
v
F
The tension (F) in the string is due to the suspended mass,
F
and
Mg
(50 kg )(9.8 m/s 2 ) 490 N
is the mass per unit length of the string
m
L
0.100 kg
8m
0.0125 kg/m
The speed of the wave is
v
F
490 N
0.0125 kg/m
198 m/s
1.8 m/s
The time needed for the wave depends on the distance and the speed of the wave,
x
v
t
8m
198 m/s
0.040 s
9.
In a transverse wave, the individual particles of the medium move perpendicularly
to the direction of the wave’s travel.
10.
The speed of the wave is related to its frequency and wavelength,
v
f
v
f
340 m/s
440 Hz
0.77 m
None of the choices are correct.
11.
The general equation for a wave traveling along the +x-direction is
y( x, t )
A cos( t kx)
The angular frequency can be found from the linear frequency
2 f
2 (25 Hz) 157 rad/s
The wavenumber can be found from the speed of the wave and ,
v
k
k
v
157 rad/s
50 m/s
3.14 rad/m
Substituting into the general equation
y( x, t )
12.
A cos((157 rad/s)t (3.14 rad/m) x)
When the pulse on the right reflects back from the wall, it will be inverted
because the rope is a fixed point. The upright and inverted pulses cancel when
they overlap. The combined amplitude is zero.
13.
The general equation for a standing wave is
y
2 A sin t cos kx
The wavelength is twice the distance between two nodes,
2 (0.30 m)
0.60 m
which can be used to find the wavenumber
k
2
2
0.60 m
10.5 rad/m
The angular frequency is related to the linear frequency
2 f
2 (60 Hz)
377 rad/s
Substituting into the general equation
y
2 Asin(( 377 rad/s)t ) cos((10.5 rad/m) x)
14.
A wave can spread around an obstacle when the wavelength is similar to the size
of the obstacle. This is called diffraction.
15.
The speed of the wave in the rope is related to the tension in the rope and its linear
density,
v
F
The linear density is
m
L
The speed of the wave is
0.160 kg
4m
0.040 kg/m
F
v
400 N
0.040 kg/m
100 m/s
The equation for the frequencies in a rope is
nv
2L
fn
For the second harmonic, n = 2
nv
2L
fn
16.
2(100 m/s )
2(4 m)
25 Hz
The decibel scale is defined as
(10 dB) log
I
I0
Taking the difference of two dB readings
2
(10 dB) log
1
I2
I0
(10 dB) log
(10 dB) log
I2
I0
(10 dB) log
I2 I0
I1 I 0
(10 dB) log
I2
I1
log
I2
I0
I1
I0
Solving for I2/I1,
2
1
I2
I1
17.
(10 dB) log
I2
I1
10 (
10 (90 74) 10
2
1)
10
40
The possible wavelengths in an open pipe are related to the length of the pipe
2L
n
n
For the fundamental frequency, = 2L = 2(0.077 m) = 0.154 m. The speed of the
wave is related to its frequency and wavelength
v
f
18.
f
v
340 m/s
0.154 m
2200 Hz
For a pipe with one end open, the wavelength of the fundamental frequency is
related to the length of the pipe, = 4L. The corresponding frequency is
f
v
v
4L
Forming a ratio
f2
f1
v 4 L2
v 4 L1
L1
L2
L1
2 L1 / 3
3
2
We have f2 = 3f1/2.
19.
20.
The lighter tuning fork has frequency f2. Since it has lighter mass, it vibrates
quicker and it will have the higher frequency.
fbeat
f2
f1
f2
f1
fbeat
440 Hz 6 Hz
446 Hz
Since the source moves in the same direction as the sound, vS > 0.The Doppler
shift for a moving source is
fO
1
fS
1 vS / v
1
(1000 Hz) 1100 Hz
1 (31 m/s ) /(340 m/s )