Question

If the amplitude of a simple harmonic oscillator is doubled, by what factor does the total energy increase?

Answer 1

Answer:

c)by a factor of four

Explanation:

The total energy of a simple harmonic oscillator is given by

$E=\frac{1}{2}kA^2$

where

k is the spring constant of the oscillator

A is the amplitude of the motion

In this problem, the amplitude of the oscillator is doubled, so

A' = 2A

Therefore, the new total energy is

$E'=\frac{1}{2}k(2A)^2=4(\frac{1}{2}kA^2)=4E$

So, the total energy increases by a factor 4.