Question

A particle is performing linear S.H.M . What fraction of the total energy is potential, with the displacement is half the amplitude?

• A. $0.33$
• B. $0.75$
• C. $0.5$
• D. $0.25$

Answer 1

The correct option is D $0.25$

Total energy of particle executing simple harmonic motion is,

$E=\frac{1}{2}m{\omega }^2{A}^2$

Therefore, kinetic energy at position is,

$K=\frac{1}{2}m{\omega }^2(A^2-V^2)$

Kinetic energy at position half the amplitude of oscillation is,

$K=\frac{1}{2}m{\omega }^2[A^2-(\frac{A}{2}{)}^2]$

$=\frac{1}{2}m{\omega }^2\left(\frac{3}{4}{A}^2\right)$

$K=\frac{3}{4}E$

$P.E=E-K=E-\frac{3}{4}E=\frac{E}{4}$

Ratio$=\frac{P.E}{E}=\frac{E/4}{4}=0.25$

Now, Potential energy will be $0.25$ of the total energy.

Option D