2.
Picture the Problem: A bird flaps its wings with a constant frequency.
Strategy: The period is the time for one cycle. The frequency is the inverse of the period, or the number of cycles per second. In
this problem the frequency is doubled.
Solution: 1. Use a ratio to show the period is halved:
f old
Tnew 1 f new
f
1

 old 

 Tnew  12 Told
Told 1 fold
f new 2 f old
2
1
1

 0.1 s
f 10 Hz
which is half of answer to problem 1.
Insight: Likewise, when the frequency is cut in half, the period is doubled.
2. Invert the frequency to determine the period:
3.
T
Picture the Problem: A violin string oscillates 440 times per second.
Strategy: The period is the time for one cycle. The frequency is the inverse of the period, or the number of cycles per second. In
this problem the frequency is increased.
Solution: Use a ratio to show the period is decreased by a
factor of n if the frequency is increased by a factor of n:
f old
Tnew 1 f new
f
1

 old 

 Tnew  Told n
Told 1 f old
f new n f old
n
Insight: Likewise, when the frequency is decreased, the period is increased.
4.
Picture the Problem: A tennis ball is hit back and forth between two players with a constant period.
Strategy: The period is the time for one cycle. In this case, a cycle is complete when the ball goes to the second player and then
returns to the first player. The frequency is the inverse of the period.
Solution: 1. The period is twice the time required to go
to the other player:
2. Invert the period to determine the frequency:
7.
T  2t  2 2.3 s   4.6 s
f 
1
1

 0.22 Hz
T 4.6 s
Picture the Problem: A mass moves back and forth in simple harmonic motion with an amplitude of 0.25 m and a period of 1.2
s
Strategy: The amplitude of the motion is the maximum distance the mass travels away from the equilibrium point.
In a single cycle the object travels a distance of four amplitudes: two amplitudes from maximum to minimum, and
two amplitudes during its return.
Solution: Because the period is 1.2 s, after 2.4 s two periods have elapsed. The mass will therefore have traveled
eight amplitudes during that time. The total distance is thus 8×0.25 m = 2.0 m.
Insight: It will take another three periods or 3.6 s for the cart to travel 3.0 meters more.
14.
Picture the Problem: A mass moves back and forth in simple harmonic motion.
Strategy: The frequency and period are inversely proportional.
Solution: The frequency and period of an oscillator are related by f  1 T . Therefore, if the period T is increased,
the frequency f will decrease.
Insight: Likewise, an increase in frequency (faster oscillation) results in a decrease in the period of oscillation.
15.
Picture the Problem: A mass moves back and forth in simple harmonic motion.
Strategy: There must always be a restoring force in order for oscillation to occur.
Solution: The restoring force always pushes the mass back toward the equilibrium point. The farther the mass is
located away from the equilibrium point, the harder the force pushes. These are the characteristics of a restoring
force that lead to simple harmonic motion.
Insight: Suppose the force pushed back toward the equilibrium point, but was constant in magnitude on either side
of the equilibrium point. In this case the object would still oscillate back and forth, but its motion would be not be described by a
simple sine or cosine function.
21.
Picture the Problem: The processing clock speed of a computer is described in terms of a frequency.
Strategy: The period is the time required to complete one cycle, and in this case it is the time elapsed as a computer completes
one binary operation. The period is also the inverse of the frequency.
1
1
1


 5.56  1010 s  0.556 ns
f 1.80 GHz 1.80  109 Hz
Insight: A higher clock speed implies the computer can perform operations in a shorter period of time.
Solution: Invert f to find T:
T