When the displacement of an object in simple harmonic motion is one-quarter of the amplitude A, the potential energy is what fraction of the total energy?

Question

Quizplus · Accepted Answer

The total mechanical energy of an object in simple harmonic motion is given by 
$$E= \frac{1}{2}kA^2$$
where $k$ is the spring constant and $A$ is the amplitude.

When the displacement of the object is one-quarter of the amplitude, the displacement is $d=\frac{1}{4}A$.
The potential energy of the object is given by 
$$U=\frac{1}{2}k(d-A)^2 = \frac{1}{2}k\left(\frac{3}{4}Aight)^2 = \frac{9}{32}kA^2$$
Therefore, the fraction of the total energy that is potential energy is 
$$\frac{U}{E}=\frac{\frac{9}{32}kA^2}{\frac{1}{2}kA^2}=\frac{9}{16}=0.5625$$

The closest fraction to this value is $\frac{11}{20}$, which is equivalent to $0.55$. Among the answer choices, the only fraction that is close to $0.55$ is $\boxed{	extbf{(C) } \frac{11}{20}}$.

When the Displacement of an Object in Simple Harmonic Motion