Question

A simple pendulum is made of a mass mounted onto a rod. It oscillates at a frequency f. By what factor would the frequency of the pendulum change if the distance between the mass and the pendulum's pivot doubled, assuming the pendulum rod has negligible mass?

a) 1/2 and 2/1
b)1/2
c) 2
d) √2

Answer 1

The frequency of a pendulum would change by a factor of √2 if the distance between the mass and the pivot point is doubled.

Step-by-step explanation:

In a simple pendulum, the frequency of oscillation is determined by the length of the pendulum and the acceleration due to gravity. According to the formula for the frequency of a simple pendulum, f = (1/2π) √(g/L), where g is the acceleration due to gravity and L is the length of the pendulum.

If the distance between the mass and the pendulum's pivot doubles, the length of the pendulum L will also double. Plugging this new value of L into the formula, we can find the new frequency f'. By using the formula for the ratio of two frequencies, f'/f = √(L'/L), we can find that f'/f = √(2), which is approximately 1.414. Therefore, the frequency of the pendulum would change by a factor of √2.