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Check Your Understanding

15.1

The ruler is a stiffer system, which carries greater force for the same amount of displacement. The ruler snaps your hand with greater force, which hurts more.

15.2

You could increase the mass of the object that is oscillating. Other options would be to reduce the amplitude, or use a less stiff spring.

15.3

A ketchup bottle sits on a lazy Susan in the center of the dinner table. You set it rotating in uniform circular motion. A set of lights shine on the bottle, producing a shadow on the wall.

15.4

The movement of the pendulums will not differ at all because the mass of the bob has no effect on the motion of a simple pendulum. The pendulums are only affected by the period (which is related to the pendulum’s length) and by the acceleration due to gravity.

15.5

Friction often comes into play whenever an object is moving. Friction causes damping in a harmonic oscillator.

15.6

The performer must be singing a note that corresponds to the natural frequency of the glass. As the sound wave is directed at the glass, the glass responds by resonating at the same frequency as the sound wave. With enough energy introduced into the system, the glass begins to vibrate and eventually shatters.

Conceptual Questions

1.

The restoring force must be proportional to the displacement and act opposite to the direction of motion with no drag forces or friction. The frequency of oscillation does not depend on the amplitude.

3.

Examples: Mass attached to a spring on a frictionless table, a mass hanging from a string, a simple pendulum with a small amplitude of motion. All of these examples have frequencies of oscillation that are independent of amplitude.

5.

Since the frequency is proportional to the square root of the force constant and inversely proportional to the square root of the mass, it is likely that the truck is heavily loaded, since the force constant would be the same whether the truck is empty or heavily loaded.

7.

In a car, elastic potential energy is stored when the shock is extended or compressed. In some running shoes elastic potential energy is stored in the compression of the material of the soles of the running shoes. In pole vaulting, elastic potential energy is stored in the bending of the pole.

9.

The overall system is stable. There may be times when the stability is interrupted by a storm, but the driving force provided by the sun bring the atmosphere back into a stable pattern.

11.

The maximum speed is equal to vmax=Aωvmax=Aω and the angular frequency is independent of the amplitude, so the amplitude would be affected. The radius of the circle represents the amplitude of the circle, so make the amplitude larger.

13.

The period of the pendulum is T=2πL/g.T=2πL/g. In summer, the length increases, and the period increases. If the period should be one second, but period is longer than one second in the summer, it will oscillate fewer than 60 times a minute and clock will run slow. In the winter it will run fast.

15.

A car shock absorber.

17.

The second law of thermodynamics states that perpetual motion machines are impossible. Eventually the ordered motion of the system decreases and returns to equilibrium.

19.

All harmonic motion is damped harmonic motion, but the damping may be negligible. This is due to friction and drag forces. It is easy to come up with five examples of damped motion: (1) A mass oscillating on a hanging on a spring (it eventually comes to rest). (2) Shock absorbers in a car (thankfully they also come to rest). (3) A pendulum is a grandfather clock (weights are added to add energy to the oscillations). (4) A child on a swing (eventually comes to rest unless energy is added by pushing the child). (5) A marble rolling in a bowl (eventually comes to rest). As for the undamped motion, even a mass on a spring in a vacuum will eventually come to rest due to internal forces in the spring. Damping may be negligible, but cannot be eliminated.

Problems

21.

Proof

23.

0.400 s/beat

25.

12,500 Hz

27.

a. 340 km/hr; b. 11.3×10311.3×103 rev/min

29.

f = 1 3 f 0 f = 1 3 f 0

31.

0.009 kg; 2%

33.

a. 1.57×105N/m1.57×105N/m; b. 77 kg, yes, he is eligible to play

35.

a. 6.53×103N/m6.53×103N/m; b. yes, when the man is at his lowest point in his hopping the spring will be compressed the most

37.

a. 1.99 Hz; b. 56.9 cm; c. 77.6 cm

39.

a. 0.335 m/s; b. 5.61×10−4J5.61×10−4J

41.

a. x(t)=2mcos(0.52s−1t)x(t)=2mcos(0.52s−1t); b. v(t)=(−1.05m/s)sin(0.52s−1t)v(t)=(−1.05m/s)sin(0.52s−1t)

43.

24.8 cm

45.

4.01 s

47.

1.58 s

49.

9.82002 m/s 2 9.82002 m/s 2

51.

6%

53.

141 J

55.

a. 4.90×10−3m4.90×10−3m; b. 1.15×10−2m1.15×10−2m

Additional Problems

57.

94.7 kg

59.

a. 314 N/m; b. 1.00 s; c. 1.25 m/s

61.

ratio of 2.45

63.

The length must increase by 0.0116%.

65.

θ = ( 0.31 rad ) sin ( 3.13 s −1 t ) θ = ( 0.31 rad ) sin ( 3.13 s −1 t )

67.

a. 0.99 s; b. 0.11 m

Challenge Problems

69.

a. 3.95×106N/m3.95×106N/m; b. 7.90×106J7.90×106J

71.

F constant r F constant r

73.

a. 7.54 cm; b. 3.25×104N/m3.25×104N/m

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