Question

The amplitude of a system moving in simple harmonic motion is doubled. Determine by what factor the following change.

Answer 1

(a) The total energy increases by a factor 4

The total energy of a simple harmonic system is given by:

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

where

k is the spring constant

A is the amplitude of the motion

In this part of the problem, the amplitude is doubled:

A' = 2A

So the new total energy is

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

So, the energy quadruples.

(b) The maximum speed increases by a factor 2

The maximum speed in a simple harmonic motion is given by

$v=\omega A$

where

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

A is the amplitude

In this part of the problem, k and m do not change, so the angular frequency does not change. Instead, the amplitude is doubled:

A' = 2A

So the new maximum speed is

$v'=\omega (A')=\omega (2A)=2 (\omega A)=2 v$

so, the maximum speed doubles.

(c) The maximum acceleration increases by a factor 2

The maximum acceleration in a simple harmonic motion is given by

$a=\omega^2 A$

where

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

A is the amplitude

In this part of the problem, k and m do not change, so the angular frequency does not change. Instead, the amplitude is doubled:

A' = 2A

So the new maximum acceleration is

$a'=\omega^2 (A')=\omega^2 (2A)=2 (\omega^2 A)=2 a$

so, the maximum acceleration doubles.

(d) The period does not change

The period in a simple harmonic motion is given by

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

where m is the mass and k is the spring constant.

In this problem, the amplitude is doubled:

A' = 2A

However, we notice that the period does not depend on the amplitude, and since both m and k do not change, then the period will remain constant.