Physics 103
Spring 2008
Homework 1 Solutions
Giancoli Chapter 11
Problems:
7.
The relationship between frequency, mass, and spring constant is f 
(a)
(b)
1
2
1
f 
2
f
1
2
k
.
m
k
2
 k  4 2 f 2 m  4 2 4.0 Hz  2.5  104 kg  0.1579 N m  0.16 N m
m
k
1 0.1579 N m

 2.8 Hz
m 2 5.0  10 4 kg


10. The relationship between the velocity and the position of a SHO is given by Equation (11-5). Set
that expression equal to half the maximum speed, and solve for the displacement.
v  vmax 1  x 2 A 2  12 vmax   1  x 2 A 2 
1
2
 1  x 2 A2 
1
4
 x 2 A2 
3
4

x   3 A 2  0.866 A
13. (a)
The total energy of an object in SHM is constant. When the position is at the amplitude, the
speed is zero. Use that relationship to find the amplitude.
Etot  12 mv 2  12 kx 2  12 kA 2
A
(b)
m 2
v  x2 
k

3.0 kg
0.55 m s2  0.020 m 2  6.034  10 2 m  6.0  10 2 m
280 N m
Again use conservation of energy. The energy is all kinetic energy when the object has its
maximum velocity.
2
Etot  12 mv 2  12 kx 2  12 kA 2  12 mvmax
vmax  A
15. (a)
k
 6.034  10 2 m
m

280 N m
 0.5829 m s  0.58 m s
3.0 kg
The work done to compress a spring is stored as potential energy.
W  12 kx 2
b)



k
2 3.0 J 
2W

 416.7 N m  4.2  10 2 N m
x2
0.12 m 2
The distance that the spring was compressed becomes the amplitude of its motion. The
maximum acceleration is given by amax 
amax 
k
A
m
 m
k
amax
k
A. Solve this for the mass.
m
 4.167  10 2 N m 
A
 0.12 m   3.333 kg  3.3 kg
15 m s2

21. The equation of motion is x  0.38 sin 6.50 t  A sin t.
(a) The amplitude is A  xmax  0.38 m .
1
(b)
The frequency is found by   2 f  6.50 s1  f  6.50 s  1.03 Hz
(c)
The period is the reciprocal of the frequency. T  1 f  1 1.03 Hz  0.967 s .
The total energy is given by
(d)
2
2
Etotal  12 mvmax
 12 m  A  
2
(e)
1
2
0.300 kg 6.50 s1 0.38 m 
2
 0.9151 J  0.92 J .
The potential energy is given by
Epotential  12 kx 2  12 m 2 x 2 
1
2
0.300 kg 6.50 s1  0.090 m 2  0.0513 J 
2
5.1  10 2 J .
The kinetic energy is given by
Ekinetic  Etotal  Epotential  0.9151 J  0.0513 J  0.8638 J  0.86 J .
(f)
23. (a)
Find the period and frequency from the mass and the spring constant.
T  2 m k  2 0.755 kg 124 N m  0.490 s
f  1 T  1 0.490 s  2.04 Hz
(b)
The initial speed is the maximum speed, and that can be used to find the amplitude.
(c)
The maximum acceleration can be found from the mass, spring constant, and amplitude
Vmax  A k m  A  vmax m k  2.96 m s 0.755 kg 124 N m  0.231m
amax  A k m  0.231 m 124 N m 0.755 kg   37 .9 m s 2
(d)
Because the mass started at the equilibrium position of x  0, the position function will be
proportional to the sine function.
x  0.231 m  sin  2 2.04 Hz t 
(e)
(b)
x  0.231 m  sin 4.08 t 
The maximum energy is the kinetic energy that the object has when at the equilibrium position.
2
E  12 mvmax

28. (a)

1
2
0.755 kg2.96
m s  3.31 J
2
60 s
 1.7 s cycle .
36 cycles
The frequency is given by f  36 cycles  0.60 Hz .
60 s
The period is given by T 
29. The period of a pendulum is given by T  2 L g. Solve for the length using a period of 2.0
seconds.

L


2
T 2 g 2.0 s  9.8 m s

 0.99 m
4 2
4 2
2
T  2 L g
32. (a)
The frequency can be found from the length of the pendulum,
and the acceleration due to gravity.
f 
(b)
1
2
9.80 m s 2
 0.57151 Hz  0.572 Hz
0.760 m
g
1

L 2
To find the speed at the lowest point, use the conservation of
energy relating the lowest point to the release point of the
pendulum. Take the lowest point to be the zero level of
gravitational
potential energy.
Etop  Ebottom

KEtop  PEtop  KEbottom  PEbottom
2
0  mg L  L cos  o   12 mvbottom
0


vbottom  2gL 1  cos o  2 9.80 m s2 0.760 m 1  cos 12.0º   0.571 m s
(c)
The total energy can be found from the kinetic energy at the bottom of the motion.
2
Etotal  12 mvbottom

1
2
0.365 kg0.571 m s2 
5.95  102 J
37. The distance between wave crests is the wavelength of the wave.
  v f  343m s 262 Hz  1.31 m
41. To find the time for a pulse to travel from one end of the cord to the other, the velocity of the pulse
on the cord must be known. For a cord under tension, we have v 
v
44. (a)
x

t
FT
 t 
m L
x
FT
m L

28 m
FT
.
m L
 0.35 s
150 N
0.65 kg  28 m
Both waves travel the same distance, so x  v1t1  v2t 2 . We let the smaller speed be v1 , and the
larger speed be v2 . The slower wave will take longer to arrive, and so t1 is more than t 2 .
t1  t 2  2.0 min  t 2  120 s  v1 t 2  120 s   v2 t 2 
t2 
v1
5.5 km s
120 s  
120 s   220 s
v2  v1
8.5 km s  5.5 km s
x  v2 t 2  8.5 km s 220 s   1.9 10 3 km
(b)
This is not enough information to determine the epicenter. All that is known is the distance of
the epicenter from the seismic station. The direction is not known, so the epicenter lies on a
circle of radius 1.9 103 km from the seismic station. Readings from at least two other seismic
stations are needed to determine the epicenter’s position.