Question

In a grandfather clock, the second hand moves forward by one second for each half period of the clock’s pendulum.

Answer 1

To solve the problem it is necessary to take into account the concepts related to simple pendulum, i.e., a point mass that is suspended from a weightless string. Such a pendulum moves in a harmonic motion -the oscillations repeat regularly, and kineticenergy is transformed into potntial energy and vice versa.

In the given problem half of the period is equivalent to 1 second so the pendulum period is,

$T= 2s$

From the equations describing the period of a simple pendulum you have to

$T = 2\pi \sqrt{\frac{L}{g}}$

Where

g= gravity

L = Length

T = Period

Re-arrange to find L we have

$L = \frac{gT^2}{4\pi^2}$

Replacing the values,

$L = \frac{(9.8)(2)^2}{4\pi^2}$

$L = 0.99m$

In the case of the reduction of gravity because the pendulum is in another celestial body, as the moon for example would happen that,

$T = 2\pi \sqrt{\frac{L}{g/6}}$

$T = 2\pi\sqrt{\frac{6L}{g}}$

$T = 2\pi \sqrt{\frac{6*0.99}{9.8}}$

$T = 4.89s$

In this way preserving the same length of the rope but decreasing the gravity the Period would increase considerably.