Download Simple Harmonic Motion

Homework #1:
1. A 0.40-kg object is attached to a spring with force constant 160 N/m so that the object is allowed to
move on a horizontal frictionless surface. The object is released from rest when the spring is compressed
0.15 m. Find (a) the force on the object and (b) its acceleration at that instant.
2. A spring is hung from a ceiling, and an object attached to its lower end stretches the spring by a distance
of 5.00 cm from its unstretched position when the system is in equilibrium. If the spring constant is 47.5
N/m, determine the mass of the object.
3. A slingshot consists of a light leather cup containing a stone. The cup is pulled back against two parallel
rubber bands. It takes a force of 15 N to stretch either one of these bands 1.0 cm. (a) What is the potential
energy stored in the two bands together when a 50-g stone is placed in the cup and pulled back 0.20 m from
the equilibrium position? (b) With what speed does the stone leave the slingshot?
4. A child’s toy consists of a piece of plastic attached to a spring (see figure below). The spring is
compressed against the floor a distance of 2.00 cm, and the toy is released. If the toy has a mass of 100 g
and rises to a maximum height of 60.0 cm, estimate the force constant of the spring.
5. An automobile having a mass of 1,000 kg is driven into a brick wall in a safety test. The bumper
behaves like a spring with constant 5.00 × 106 N/m and is compressed 3.16 cm as the car is brought to rest.
What was the speed of the car before impact, assuming that no energy is lost in the collision with the wall?
6. The frequency of vibration of an object–spring system is 5.00 Hz when a 4.00-g mass is attached to the
spring. What is the force constant of the spring?
7. A spring stretches 3.9 cm when a 10-g object is hung from it. The object is replaced with a block of
mass 25 g that oscillates in simple harmonic motion. Calculate the period of motion.
8. A man enters a tall tower, needing to know its height. He notes that a long pendulum extends from the
ceiling almost to the floor and that its period is 15.5 s. (a) How tall is the tower? (b) If this pendulum is
taken to the Moon, where the free-fall acceleration is 1.67 m/s2, what is the period there?
9. A simple 2.00-m-long pendulum oscillates at a location where g = 9.80 m/s2. How many complete
oscillations does it make in 5.00 min?
10. The free-fall acceleration on Mars is 3.7 m/s2. (a) What length of pendulum has a period of 1 s on
Earth? What length of pendulum would have a 1-s period on Mars? (b) An object is suspended from a
spring with force constant 10 N/m. Find the mass suspended from this spring that would result in a period
of 1 s on Earth and on Mars.
Solutions:
1.
(a) The force exerted on the block by the spring is
Fs   kx   160 N m
  0.15 m    24 N
or Fs  24 N directed tow ard equilibrium position
(b) From Newton's second law, the acceleration is
a
2.
Fs  24 N
m
m

  60 2  60 2 tow ard equilibrium position
m 0.40 kg
s
s
When the system is in equilibrium, the tension in the spring F  k x must equal
the weight of the object. Thus,


2
k x  47.5 N  5.00  10 m

 0.242 kg
k x  m g giving m 
g
9.80 m s2
3.
(a) Assume the rubber bands obey Hooke’s law. Then, the force constant of
each band is
F
15 N
k s 
 1.5  103 N m
-2
x 1.0  10 m
Thus, when both bands are stretched 0.20 m, the total elastic potential
energy is
1 
PEs  2 kx2   1.5  103 N m
2 

 0.20 m 
2
 60 J
(b) Conservation of mechanical energy gives  KE  PEs f   KE  PEsi , or
2 60 J
 49 m s
50  10-3 kg
1 2
m v  0  0  60 J, so v 
2
4.
From conservation of mechanical energy,
 KE  PE
g
 PEs
giving
k
2m ghf
2
i
x

   KE  PE
g
f

1
 PEs or 0  m ghf  0  0  0  kxi2
i
2


m
2 0.100 kg 9.80 m s2  0.600 m

2
2.00  10
2

 2.94  103 N m
5.

Conservation of mechanical energy, KE  PEg  PEs
gives
or
6.
   KE  PE
f
g

 PEs ,
i
1 2
1
m vi  0  0  0  0  kx2f ,
2
2
vi 
k
xi 
m
5.00  106 N m
3.16  102 m  2.23 m s
1000 kg


The period of oscillations of a mass-spring system is given by T  2 m k and
the frequency is
f
1
1

T 2
k
m

Thus, k  4 2 f2m  4 2 5.00 s-1
  4.00  10
2
3

kg  3.95 N m
7. The spring constant is found from


2
Fs m g  0.010 kg 9.80 m s
k 

 2.5 N m
x
x
3.9  10-2 m
When the object attached to the spring has mass m  25 g , the period of
oscillation is
T  2
8. (a)
0.025 kg
m
 2
 0.63 s
k
2.5 N m
The height of the tower is almost the same as the length of the pendulum. From
T  2 L g , we obtain


9.80 m s2 15.5 s
gT 2
L

 59.6 m
4 2
4 2
2
(b) On the Moon, where g  1.67 m s2 , the period will be
T  2
L
59.6 m
 2
 37.5 s
g
1.67 m s2
9.
The period of a simple pendulum is T  2 L g where L is its length. The
number of complete oscillations per second (that is, the frequency) for this
pendulum is then given by
f
9.80 m s2
 0.352 s-1
2.00 m
g
1

L 2
1 1

T 2
Hence, the number of oscillations in a time t 5.00 m in  300 s is


N  f t  0.352 s-1  300 s  105.7 or 105 com plete oscillations
10. (a) From T  2 L g , the length of a pendulum with period T is L 
 9.80 m s  1.0 s
On Earth, with T  1.0 s, L 
2
4 2
 3.7 m s  1.0 s
If T  1.0 s on Mars, L 
2
4 2
gT 2
.
4 2
2
 0.25 m  25 cm
2
 0.094 m  9.4 cm
(b) The period of an object on a spring is T  2 m k , which is independent of
the local free-fall acceleration. Thus, the same mass will work on Earth and
on Mars. This mass is
kT 2 10 N m  1.0 s

 0.25 kg
4 2
4 2
2
m