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Physics II
Homework III
CJ
Chapter 14; 6, 16, 20, 23, 34, 44, 51, 76
14.6. Model: The oscillation is the result of simple harmonic motion.
Visualize: Please refer to Figure Ex14.6.
Solve: (a) The amplitude A  20 cm.
(b) The period T  4.0 s, thus
f 
1
1

 0.25 Hz
T 4.0 s
(c) The position of an object undergoing simple harmonic motion is x  t   A cos t  0  . At
t  0 s, x0  10 cm. Thus,
10 cm 
 20 cm cos0  0  cos1   20 cm   cos1   2   
10 cm

1



2
rad  120
3
Because the object is moving to the right at t  0 s, it is in the lower half of the circular motion diagram and thus must have a
phase constant between  and 2 radians. Therefore, 0   23  rad  120 .
14.16. Model: The mass attached to the spring is in simple harmonic motion.
Solve: (a) The period is
T  2
m
 2
k
 0.507 kg
 20 N m 
 1.00 s
(b) The angular frequency is   2 T  2 1.0 s  2 rad s.
(c) To calculate the phase constant  0 ,
x0  A cos0  0.05 m   0.10 m cos0  0  cos1  21    13  rad
Because the mass is moving to the right at t  0 s, it is in the lower half of the circular motion diagram. Hence, 0   13  rad.
(d) At t  0 s, vx  t     0.10 m 2 rad s sin  2 t  0  becomes
  0.10 m 2 rad s sin   13  rad   0.544 m s
(e) The maximum speed vmax   A   2 rad s 0.10 m  0.628 m s .
(g) At t  1.3 s, x1.3 s   0.10 m  cos 2 1.3 s   13  rad   0.0669 m .
(h) At t  1.3 s, v1.3 sx    0.10 m 2 rad s sin  2 rad s1.3 s  13  rad  0.467 m s .
14.20. Model: The vertical oscillations constitute simple harmonic motion.
Solve:
To find the oscillation frequency using   2 f  k m , we first need to find the spring constant k. In
equilibrium, the weight mg of the block and the spring force kL are equal and opposite. That is, mg  kL  k  mg L .
The frequency of oscillation f is thus given as
f 
1
2
k
1

m 2
mg L
1

m
2
g
1

L 2
9.8 m s2
 3.52 Hz
0.020 m
14.23. Model: Assume the small-angle approximation so there is simple harmonic motion.
Solve:
The period is T  12.0 s 10 oscillations  1.20 s and is given by the formula
2
T  2
2
L
 T 
 1.20 s 
2
L
 g   2  9.8 m s  35.7 cm
g
2







14.34. Visualize: Please refer to Figure P14.34.
Solve:
The position and the velocity of a particle in simple harmonic motion are
x  t   A cost  0  and vx  t    A sin t  0   vmax sin t  0 
From the graph, T  12 s and the angular frequency is

2
2



rad s
T
12 s 6
(a) Because vmax  A  60 cm s , we have
A
60 cm s


60 cm s
 6 rad s
 114.6 cm
(b) At t  0 s,
v0x   A sin0  30 cm    60 cm s  sin 0  30 cm
 0  sin 1  0.5 rad   61  rad  30  or 65  rad 150 
Because the velocity at t  0 s is negative and the particle is slowing down, the particle is in the second quadrant of the circular
motion diagram. Thus 0  65  rad.
(c) At t  0 s, x0  114.6 cm cos 65  rad   99.2 cm .
14.44. Solve: (a) The velocity is –0.25 m/s at times that satisfy the equation
0.25 m/s  (0.35 m/s) sin(20t   )  sin(20t   ) 
0.25 m/s
 0.7143
0.35 m/s
The time is
20t   rad  sin1 (0.7143)  0.7956 rad  t 
0.7956 rad   rad
 0.117 s
20 rad/s
This is a time at which v  –0.25 m/s, but it’s not a time after t  0 s. But the sine function is periodic with period 2 rad, so
another angle whose sine is 0.7143 is
20t   rad  sin1 (0.7143)  0.7956 rad  2 rad  t 
0.7956 rad   rad
 0.197 s
20 rad/s
This is a time t > 0 s at which v  –0.25 m/s.
(b) The amplitude is A  vmax/  (0.35 m/s)(20 rad/s)  0.0175 m  1.75 cm. Thus the position at t  0.197
s is
x  A cos( t  0 )  (1.75 cm)cos((20 rad/s)(0/197 s)   rad)  1.22 cm
14.51. Model: The compact car is in simple harmonic motion.
Solve: (a) The mass on each spring is 1200 kg 4  300 kg . The spring constant can be calculated as follows:
2
2
k
 k  m 2  m  2 f    300 kg 2  2.0 Hz   4.74  104 N m
m
(b) The car carrying four persons means that each spring has, on the average, an additional mass of 70 kg.
That is, m  300 kg  70 kg  370 kg . Thus,
2 
f 

1

2 2
k
1

m 2
4.74  104 N m
 1.80 Hz
370 kg
Assess: A small frequency change from the additional mass is reasonable because frequency is inversely proportional to the
square root of the mass.
14.76. Model: The two springs obey Hooke’s law. Assume massless springs.
Visualize: Each spring is shown separately. Note that x  x1  x2.
Solve: Only spring 2 touches the mass, so the net force on the mass is Fm  F2 on m. Newton’s third law tells us that
F2 on m  Fm on 2 and that F2 on 1  F1 on 2. From Fnet  ma, the net force on a massless spring is zero. Thus Fw on 1  F2 on 1  k1x1 and
Fm on 2  F1 on 2  k2x2. Combining these pieces of information,
Fm  k1x1  k2x2
The net displacement of the mass is x  x1  x2, so
x  x1  x2 
Fm Fm  1
k  k2
1

    Fm  1
Fm
k1
k2  k1 k2 
k1k2
Turning this around, the net force on the mass is
kk
kk
Fm  1 2 x  keff x where keff  1 2
k1  k2
k1  k2
keff, the proportionality constant between the force on the mass and the mass’s displacement, is the effective spring constant.
Thus the mass’s angular frequency of oscillation is

keff

m
1 k1k2
m k1  k2
Using 12  k1 / m and  22  k2 / m for the angular frequencies of either spring acting alone on m, we have

(k1 / m) (k2 /m)
1222

(k1 /m)  (k2 / m)
12  22
Since the actual frequency f is simply a multiple of , this same relationship holds for f:
f 
f 12 f 22
f  f 22
2
1