A 75-kg diver is standing at the end of a diving board while it is vibrating up and down in simple harmonic motion, as indicated in Figure 10.35. The diving board has an effective spring constant of k=4100 N/m, and the vertical distance between the highest and lowest points in the motion is 0.30 m. (a) What is the amplitude of the motion? (b) Starting when the diver is at the highest point, what is his speed one-quarter of a period later? (c) If the vertical distance between his highest and lowest points were doubled to 0.60 m, what would be the time required for the diver to make one complete motional cycle?

Figure 10.35  A diver at the end of a diving board is bouncing up and down in simple harmonic motion.

Concept Questions and Answers How is the amplitude A related to the vertical distance between the highest and lowest points of the diver’s motion?

Answer   The amplitude is the distance from the midpoint of the motion to either the highest or the lowest point. Thus, the amplitude is one-half the vertical distance between the highest and lowest points in the motion.

Starting from the top, where is the diver located one-quarter of a period later, and what can be said about his speed at this point?

Answer   The time for the diver to complete one motional cycle is defined as the period. In one cycle, the diver moves downward from the highest point to the lowest point and then moves upward and returns to the highest point. In a time equal to one-quarter of a period, the diver completes one-quarter of this cycle and, therefore, is halfway between the highest and lowest points. His speed is momentarily zero at the highest and lowest points and is a maximum at the halfway point.

If the amplitude of the motion were to double, would the period also double?

Answer   No. The period is the time to complete one cycle, and it is equal to the distance traveled during one cycle divided by the average speed. If the amplitude doubles, the distance also doubles. However, the average speed also doubles. We can verify this by examining Equation 10.7, which gives the diver’s velocity as . The speed is the magnitude of this value, or . Since the speed is proportional to the amplitude A, the speed at every point in the cycle also doubles when the amplitude doubles. Thus, the average speed doubles. However, the period, being the distance divided by the average speed, does not change.

Solution

(a)	Since the amplitude A is one-half the vertical distance between the highest and lowest points in the motion, .
(b)	When the diver is halfway between the highest and lowest points, his speed is a maximum. The maximum speed of an object vibrating in simple harmonic motion is given by Equation 10.8 as vmax=Aw, where A is the amplitude of the motion and w is the angular frequency. The angular frequency can be determined from Equation 10.11 as , where k is the effective spring constant of the diving board and m is the mass of the diver. The maximum speed is
(c)	The period is the same, regardless of the amplitude of the motion. From Equation 10.4 we know that the period T and the angular speed w are related by T=2p/w, where . Thus, the period can be written as As expected, the period does not depend on the amplitude of the motion.

A 68.0-kg bungee jumper is standing on a tall platform (h₀=46.0 m), as indicated in Figure 10.36. The bungee cord has an unstrained length of L₀=9.00 m, and when stretched, behaves like an ideal spring with a spring constant of k=66.0 N/m. The jumper falls from rest, and the only forces acting on him are his weight and, for the latter part of the descent, the elastic force of the bungee cord. What is his speed (it is not zero) when he is at the following heights above the water (see the drawing): (a) h_A=37.0 m and (b) h_B=15.0 m?

Figure 10.36  A bungee jumper jumps from a height of h 0 = 46.0 m. The length of the unstrained bungee cord is L 0 = 9.00 m.

Concept Questions and Answers Can we use the conservation of mechanical energy to find his speed at any point during the descent?

Answer   Yes. His weight and the elastic force of the bungee cord are the only forces acting on him and are conservative forces. Therefore, the total mechanical energy remains constant (is conserved) during his descent.

What types of energy does he have when he is standing on the platform?

Answer   Since he’s at rest, he has neither translational nor rotational kinetic energy. The bungee cord is not stretched, so there is no elastic potential energy. Relative to the water, however, he does have gravitational potential energy, since he is 46.0 m above it.

What types of energy does he have at point A?

Answer   Since he’s moving downward, he possesses translational kinetic energy. He is not rotating though, so his rotational kinetic energy is zero. Because the bungee cord is still not stretched at this point, there is no elastic potential energy. But he still has gravitational potential energy relative to the water, because he is 37.0 m above it.

What types of energy does he have at point B?

Answer   He has translational kinetic energy, because he’s still moving downward. He’s not rotating, so his rotational kinetic energy is zero. The bungee cord is stretched at this point, so there is elastic potential energy. He has gravitational potential energy, because he is still 15.0 m above the water.

Solution

(a)

The total mechanical energy is the sum of the kinetic and potential energies, as expressed in Equation 10.14: The conservation of mechanical energy states that the total mechanical energy at point A is equal to that at the platform: While standing on the platform, the jumper is at rest, so v0=0 m/s and w0=0 rad/s. The bungee cord is not stretched, so x0=0 m. At point A, the jumper is not rotating, wA= 0 rad/s, and the bungee cord is still not stretched, xA=0 m. With these substitutions, the conservation of mechanical energy becomes Solving for the speed at point A yields

(b)

At point B the total mechanical energy is the same as it was on the platform, so We set wB=0 rad/s, since there is no rotational motion. Furthermore, the bungee cord stretches by an amount xB=h0 –L0 –hB (see the drawing). Therefore, we have