CSULA- PHYSICS 212
Fall 2012
Solution: Material to Study
Chapter 15
OBJECTIVE QUESTIONS
OQ-12 (7e- Q5)
(a) Yes. In simple harmonic motion, one-half of the time, the velocity is in the same direction as the displacement away
from equilibrium.
(b) Yes. Velocity and acceleration are in the same direction half the time.
(c) No. Acceleration is always opposite to the position vector, and never in the same direction.
OQ-14 (7e-Q7)
Answer (c). The equilibrium position is 15 cm below the starting point. The motion is symmetric about the equilibrium position, so
the two turning points are 30 cm apart.
CONCEPTUAL QUESTIONS
CQ-3 (7e-Q3)
(a) You can take    , or equally well,    .
(b) At t  0 , the particle is at its turning point on the negative side of equilibrium, at x   A .
CQ-5 (7e-Q11)
(a) No force is exerted on the particle. The particle moves with constant velocity.
(b) The particle feels a constant force toward the left. It moves with constant acceleration toward the left. If its initial push is toward
the right, it will slow down, turn around, and speed up in motion toward the left. If its initial push is toward the left, it will just speed
up.
(c) A constant force towards the right acts on the particle to produce constant acceleration toward the right.
(d) The particle moves in simple harmonic motion about the lowest point of the potential energy curve.
PROBLEMS
P5 (7e-P3)
x  (4.00 m) cos(3.00 t   )
& x  A cos(t   )
3.00
1
(a).   2f  3.00  f 
 1.50 Hz  T   0.667 s
2
f
(b). A  4.00 m
(c).   
(d ). x(t  0.250 s)  (4.00 m) cos(3.00 (0.250 )   )  (4.00 m) cos(0.75   ) 
x(t  0.250 )  (4.00 m) cos(1.75 )  2.83 m
P8 (7e-P6)
(a)
T
12.0 s
 2.40 s
5
(b)
f 
1
1

 0.417 Hz
T 2.40
(c)
  2 f  2  0.417  2.62 rad s
P14 (7e-P10)
m  1.00 kg , k  25.0 N m , and A  3.00 cm . At t  0 , x  3.00 cm
k
25.0

 5.00 rad s
m
1.00
2
2
so that, T 

 1.26 s
 5.00
(a)

(b)
vmax  A  3.00 102 m  5.00 rad s  0.150 m s
amax  A 2  3.00  102 m  5.00 rad s  0.750 m s2
2
(c)
Because x  3.00 cm and v  0 at t  0 , the required solution is x   A cos t
dx
dv
v
 15.0sin  5.00t  cm s
a
 75.0cos 5.00t  cm s2
Or x  3.00cos 5.00t  cm
dt
dt
P18 (7e-P14)
2
2

 25.1 rad s
T 0.250
m  200 g , T  0.250 s , E  2.00 J;  
(a)
k  m 2  0.200 kg  25.1 rad s  126 N m
(b)
E
2
kA 2
A
2
2E

k
2  2.00
 0.178 m
126
P21 (7e-None)
Recall that: Emech  U s  K  12 kA2  12 kx2  12 mv 2
If : x = (⅓)A, then:
(b). U s  12 k 13 A2 

1 1
9 2

kA2 
1
9
Emech
(a). Emech  U s  K  Emech  Emech  K  K  89 Emech
1
9
(c). Emech  U s  K  12 kA2  12 kx2  12

1
2

kx2  A2  x2  12 x2  x   2 / 3 A
(d). NO. The maximum value for the potential energy is: U s , max  12 kA2 . Since the total energy must remain constant at
Emech 
1
2
kA2 , therefore: The kinetic energy can never be greater than Umax.
P25 (7e-P23)
(a) The motion is simple harmonic because the tire is rotating with constant velocity and you are looking at the motion of the bump
projected in a plane perpendicular to the tire.
3.00 m s
 10.0 rad s.
(b) Since the car is moving with speed v  3.00 m s, and its radius is 0.300 m, we have:  
0.300 m
2
2
Therefore, the period of the motion is: T 

 0.628 s .
 10.0 rad s