10.
Picture the Problem: A mass oscillates on a spring.
Strategy: Use the formula for the period of a mass on a spring together with the given mass and frequency to find the spring
constant.
Solution: 1. Rearrange the formula:
2. Solve for k:
3. Substitute the numerical values:
f 
1
1
1


T 2 m k 2
k
m
1 k
4 2 m
2
2
4 m f  k
f2 
k  4 2 0.32 kg 1.6 Hz   32 N/m
2
Insight: A spring constant that is 4 times larger (128 N/m) would be required to double the frequency to 3.2 Hz.
11.
Picture the Problem: A mass oscillates on a spring.
Strategy: Use the formula for the period of a mass on a spring together with the given spring constant and period to find the
mass.
m
k
Solution: 1. Rearrange the formula:
T  2
2. Solve for m:
4 2 m  k T 2
m
 T 2  4 2
m
k
kT2
4 2
k T 2  22 N/m0.95 s 

 0.50 kg
4 2
4 2
2
3. Substitute the numerical values:
m
Insight: A mass that is four times larger (2.0 kg) would be required to double the period to 1.9 s.
16.
Picture the Problem: A mass oscillates on a spring, and then the spring constant is increased by a factor of 4.
Strategy: Use the formula for the period of a mass on a spring to set up a ratio that can be used to predict the new period.
Solution: Use a ratio to show the period is cut in half:
Tnew

Told
2
2
m
knew

m
kold
kold

knew
kold
4 kold

1
 Tnew  12 Told
2
Insight: The stiffer spring produces a larger force and a larger acceleration, resulting in more rapid change in motion and a
shorter period.
17.
Picture the Problem: A mass moves back and forth in simple harmonic motion.
Strategy: The frequency and period are inversely proportional. Use the formula for the period of a mass on a spring to set up a
ratio between the frequency of oscillation and the mass of the object.
Solution: Suppose the mass is increased by
a factor of 4, and use a ratio to show the
frequency will decrease:
f new 1 Tnew


f old
1 Told
1
2
k
mnew
1
2
k
mold

mold

mnew
mold
4 mold

1
 f new 
2
1
2
f old
Insight: The heavier mass will accelerate at a lower rate given the spring force remains the same. The slower acceleration results
in a longer period and a lower frequency of oscillation.
18. The frequency of oscillation for a mass-spring system is inversely proportional to the square root of the mass. Therefore, in order
to increase the frequency of oscillation you should decrease the mass. A smaller mass will experience a greater acceleration for
the same amount of spring force, resulting in faster oscillation.
19.
Picture the Problem: Four masses move back and forth in simple harmonic motion.
Strategy: The formula for the period of a mass on a spring can be used to predict the periods of the four masses using the given
spring constant and mass values. Then the periods can be ranked.
Solution: 1. Use the formula to
find the four periods:
TA  2
mA
0.1 kg
 2
 1.3 s
kA
10 N/m
TB  2
mB
0.4 kg
 2
 1.3 s
kB
40 N/m
TC  2
mC
0.4 kg
 2
 2.5 s
kC
10 N/m
TD  2
mD
0.1 kg
 2
 0.63 s
kD
40 N/m
2. Rank the periods: TD  TA  TB  TC
Insight: Increasing both the mass and the spring constant by the same factor produces no change in the period.