Question

A body executes S.H.M with an amplitude A. At what displacement from the mean position, is the potential energy of the body one-fourth of its total energy?
$A.\dfrac{A}{4}$
$B.\dfrac{A}{2}$
$C.\dfrac{3A}{4}$
D. Some other fraction of A.

Answer 1

We have provided a body which is performing simple harmonic motion with the same amplitude A. the question has started a condition that potential energy of the body is one-fourth of its total energy. Therefore use expression of the potential energy which will give relation between frequency, mass and displacement as well as use expression for total energy of the particle performing simple harmonic motion.
Formula used:
Potential energy is given as,
$DE=\dfrac{1}{2}m{{\omega }^{2}}{{y}^{2}}$
Whereas, m=mass, $\omega$= frequency, y=displacement.

Complete answer:
A body is performing simple harmonic motion with amplitude A. Now we need to find the displacement from the mean position. Let ‘m’ is the mass of the body and y is the displacement with an angular frequency then potential energy of the body executing S.H.M is given by,
$P.E=\dfrac{1}{2}m{{\omega }^{2}}{{y}^{2}}..........(1)$
We know that the expression for total energy of the particle performing S.H.M is given as.
$P.E=\dfrac{1}{2}m{{\omega }^{2}}{{A}^{2}}........(2)$
Where A= Amplitude of the body
Now according to the condition stated in question,
$P.E=\dfrac{1}{4}(T.E)$
Then from equation (1) and (2), we get,
$\dfrac{1}{2}m{{\omega }^{2}}{{y}^{2}}=\dfrac{1}{4}\left( \dfrac{1}{2}m{{\omega }^{2}}{{A}^{2}} \right)$
$\begin{aligned} & {{y}^{2}}=\dfrac{{{A}^{2}}}{4} \\\ & \therefore y=\dfrac{A}{4} \\\ \end{aligned}$
Hence, $\dfrac{A}{4}$ is the displacement from the mean position.

Therefore the correct option is (A).

Additional information:
When a particle of the mass M performs simple harmonic motion at distance x from its mean position, then restoring force acting on the particle is given by $f=-kx$ where k is the force constant. Total energy of the particle performing simple harmonic motion is the sum of its potential energy and kinetic energy. Therefore, $T.E=K.E+P.E.$
Total energy of particles performing S.H.M is always proportional to its mass and to square of the frequency of oscillation and square of amplitude of the oscillation.

Note:
If the particle is having the displacement then work done to displace the body is stored in the form of potential. If a particle is performing S.H.M then at mean position, potential energy will be zero and hence total energy is possessed by only kinetic energy while at extreme position potential energy will be maximum and kinetic energy will be zero. Hence total energy will be possessed by only potential energy. Other than these two positions, kinetic and potential energies are enter convertible such that their sum i.e., total energy is constant. Total energy of particles performing S.H.M remains constant if it is independent of displacement y.