Question

Two clocks are taken from the Earth to the Moon. The first clock uses a pendulum mechanism, while the second uses a spring-loaded mechanism. What happens with the time displayed by the pendulum clock? chiu (jc87778) – Springs and Pendulums – schmidt – (SCI402 AC) 2 1. faster 2. slower 3. no change 010 (part 2 of 2) 10.0 points What happens with the time displayed by the spring-loaded clock?
1. slower

2. faster

3. no change

Answer 1

Answer:

Part 1. Pendulum clock is slower.

Part 2. Spring-loaded clock remains the same.

Explanation:

The period of a simple pendulum is given by

$T=2\pi\sqrt{\dfrac{l}{g}}$

where $l$ = length of pendulum and $g$ = acceleration due to gravity.

It is seen that the period is inversely proportional to the square root of the gravitational acceleration. So if gravity increases, period decreases and vice versa.

$g$ on the moon is about one-fifth that of the Earth. Hence, the pendulum will have a larger period, about twice ($\sqrt{5} = 2.24$). A larger period means it takes longer to finish an oscillation, so the pendulum clock is slower.

The period of a loaded spring is given by

$T=2\pi\sqrt{\dfrac{k}{m}}$

where $k$ = the spring constant and $m$ = mass of load on the spring.

It is seen that this relation does not depend on gravity nor does it have any parameter that depends on gravity: k is a constant of the spring that does not change while mass is independent of location.

Hence, the spring-loaded clock will remain the same.

As a note, one might assume that gravity affects the loaded spring because the load is 'pushed' down by gravity. In fact, only the equilibrium position is affected by gravity; it only determines where the oscillation starts from, not how long it takes.