Question

Identical balls oscillate with the same period T on Earth. Ball A is attached to an ideal spring and ball B swings back and forth to form a simple pendulum. These systems are now taken to the Moon, where g = 1.6 m/s^2 and set into oscillation. Which of the following statements about these systems are true of the oscillations on the Moon?
a. Ball A will take longer to complete one cycle.
b. Ball B will take longer to complete one cycle.
c. Ball A and B will still take the same amount of time to complete one cycle as they did on Earth.
d. Ball A and B will still have equal periods, but they less than they were on Earth.

Answer 1

Answer:

B. Ball B will take longer to complete one cycle

Explanation:

This is simply because the period of a simple pendulum is affected by acceleration due to gravity, while the period of an ideal spring is not.

This can be clearly deduced by observing the formulas for the various systems.

Formula for period of simple pendulum:

T = 2π × $\sqrt{\frac{L}{g} }$

Formula for period of an oscillating spring:

T= 2π ×$\sqrt{\frac{m}{K} }$

The period of the simple pendulum is affected by the length of the string and the acceleration due to gravity as shown above. Thus, it will have a different period as the gravitational acceleration changes on the moon. Thus will be a larger period (slower oscillation) as the gravitational acceleration is smaller in this case

The period of the oscillating spring is only affected by the mass of the load an the spring constant as shown above. Thus, it will have a period similar to the one it had on the earth because the mass of the ball did not change as the setup was taken to the moon.

All these will make the ball on the spring (Ball B) oscillate faster than the ball swinging on the string (Ball A)