Question

In a simple harmonic oscillation, what fraction of total mechanical energy is in the form of kinetic energy, when the particle is midway between mean and extreme position.

• A. $\frac{1}{4}$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{3}{4}$

Answer 1

The correct option is D $\frac{3}{4}$

The total mechanical energy of simple harmonic oscillation is given by,

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

The kinetic energy of simple harmonic oscillation at agiven position is given by,

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

Here, $x=\frac{A}{2}$

$\Rightarrow K.E=\frac{1}{2}m{\omega }^2(A^2-\frac{A^2}{4})$

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

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

$\Rightarrow \frac{K.E}{T.E}=\frac{3}{4}$

Hence, option $\left(B\right)$ is correct.