Physics II
Homework III
CJ
Chapter 14: 7, 14, 22, 26, 33, 38, 54, 63
14.7. Visualize: The phase constant 23  has a plus sign, which implies that the object undergoing simple harmonic
motion is in the second quadrant of the circular motion diagram. That is, the object is moving to the left.
Solve: The position of the object is
x  t   A cos  t  0   A cos  2 ft  0    4.0 cm  cos  4 rad s  t  23  rad 
The amplitude is A  4 cm and the period is T  1 f  0.50 s. A phase constant 0  2 3 rad 120 (second quadrant)
means that x starts at  21 A and is moving to the left (getting more negative).
Assess: We can see from the graph that the object starts out moving to the left.
14.14. Model: The oscillating mass is in simple harmonic motion.
Solve: (a) The amplitude A  2.0 cm.
(b) The period is calculated as follows:

2
2
 10 rad s  T 
 0.628 s
T
10 rad s
(c) The spring constant is calculated as follows:

2
k
 k  m 2   0.050 kg 10 rad s   5.0 N m
m
(d) The phase constant 0   14  rad.
(e) The initial conditions are obtained from the equations
x  t    2.0 cm cos10t  14   and vx  t     20.0 cm s sin 10t  14  
At t  0 s, these equations become
x0   2.0 cm cos  14    1.414 cm and v0 x    20.0 cm ssin   14    14.14 cm s
In other words, the mass is at 1.414 cm and moving to the right with a velocity of 14.14 cm/s.
(f) The maximum speed is vmax  A   2.0 cm10 rad s  20.0 cm s .
(g) The total energy E  21 kA2 
1
2
 5.0 N m  0.02 m 
2
 1.0  10 3 J .
(h) At t  0.40 s, the velocity is
v0 x    20.0 cm s  sin 10 rad s  0.40 s   14    1.46 cm s
14-22. Model: Assume a small angle of oscillation so there is simple harmonic motion.
Solve:
The period of the pendulum is
T0  2
L0
 4.0 s
g
(a) The period is independent of the mass and depends only on the length. Thus T  T0  4.0 s.
(b) For a new length L  2L0,
2 L0
 2T0  5.66 s
g
T  2
(c) For a new length L  L0/2,
T  2
L0 2
1

T0  2.83 s
g
2
(d) The period is independent of the amplitude as long as there is simple harmonic motion. Thus T  4.0 s.
14.26. Model: The pendulum undergoes simple harmonic motion.
Solve: (a) The amplitude is 0.10 rad.
(b) The frequency of oscillations is
f 

5

Hz  0.796 Hz
2 2
(c) The phase constant    rad.
(d) The length can be obtained from the period:
2
2


 1 
g
1
L
 9.8 m s2  0.392 m
 g  

L
2

f
2

0.796
Hz






  2 f 


(e) At t  0 s,  0   0.10 rad  cos   0.10 rad . To find the initial condition for the angular velocity we take the
derivative of the angular position:
  t    0.10 rad  cos  5t    
d  t 
dt
   0.10 rad  5 sin  5t   
At t 0 s,  d dt 0   0.50 rad  sin    0 rad s .


(f) At t  2.0 s,  2.0   0.10 rad  cos 5  2.0 s     0.084 rad.
14.33. Visualize: Please refer to Figure P14.33.
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 
(a) At t  0 s, the equation for x yields
 5.0 cm  10.0 cm cos0   0  cos1  0.5   13  rad
Because the particle is moving to the right at t  0 s, it is in the lower half of the circular motion diagram, and the phase
constant is between  and 2 radians. Thus, 0   13  rad.
(b) At t  0 s,
 2    
v0 x   A sin0   10.0 cm  
 sin     6.80 cm
 T   3
(c) The maximum speed is
 2 
vmax   A  
 10.0 cm   7.85 cm s
 8.0 s 
Assess: The positive velocity at t  0 s is consistent with the position-versus-time graph and the negative sign of the
phase constant.
14-38. Solve: The object’s position as a function of time is x  t   A cost  0  . Letting x  0 m at t  0 s, gives
0  A cos0  0   21 
Since the object is traveling to the right, it is in the lower half of the circular motion diagram, giving a phase constant
between  and 0 radians. Thus, 0   21  and
x  t   A cost  21    x  t   Asint   0.10 m sin  21  t 
where we have used A  0.10 m and

2 2 rad 


rad s
T
4.0 s
2
Let us now find t where x  0.060 m:
2
 
 0.060 m 
0.060 m   0.10 m  sin  t   t  sin 1 
  0.410 s

2 
 0.10 m 
Assess: The answer is reasonable because it is approximately
1
8
of the period.
14.54. Model: The two blocks are in simple harmonic motion, without the upper block slipping. We will also apply
the model of static friction between the two blocks.
Visualize:
Solve: The net force acting on the upper block m1 is the force of friction due to the lower block m2. The model of static
friction gives the maximum force of static friction as
fs max  s n  s  m1g  m1amax  amax  sg


Using s  0.5 , amax  S g   0.5 9.8 m s2  4.9 m s2 . That is, the two blocks will ride together if the maximum
acceleration of the system is equal to or less than amax. We can calculate the maximum value of A as follows:
amax   Amax
2


4.9 m s2 1.0 kg  5.0 kg 
amax  m1  m2 
k

Amax  Amax 

 0.588 m
m1  m2
k
50 N m
14.63. Model: A completely inelastic collision between the bullet and the block resulting in simple harmonic motion.
Visualize:
Solve: (a) The equation for conservation of energy after the collision is
1 2 1
kA   mb  mB  vf2  vf 
2
2
k
A
mb  mB
2500 N m
 0.10 m   4.975 m s
1.010 kg
The momentum conservation equation for the perfectly inelastic collision pafter  pbefore is
 mb  mB  vf  mb vb  mBvB
1.010 kg  4.975 m s   0.010 kg vb  1.00 kg 0 m s  vb  502
ms
(b) No. The oscillation frequency k  mb  mB  depends on the masses but not on the speeds.