Answer:
c)by a factor of four
Explanation:
The total energy of a simple harmonic oscillator is given by
[tex]E=\frac{1}{2}kA^2[/tex]
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
[tex]E'=\frac{1}{2}k(2A)^2=4(\frac{1}{2}kA^2)=4E[/tex]
So, the total energy increases by a factor 4.
