Question

When the displacement of a particle executing SHM is one-fourth of its amplitude, what fraction of its total energy is the kinetic energy?

• A. $\frac{16}{15}$
• B. $\frac{15}{16}$
• C. $\frac{3}{4}$
• D. $\frac{4}{3}$

Answer 1

Answer: (D)

$\frac{4}{3}$

Answer 2

Let the displacement of the particle executing SHM at any instant of time t from its equilibrium position is given by

$x = A\cos(\omega t + \varphi)$

Velocity $v = \frac{dx}{dt} = - \omega A\sin(\omega t + \varphi)$

Kinetic energy of the particle is

$k = \frac{1}{2}mv^{2} = \frac{1}{2}m\omega^{2}A^{2}\sin^{2}(\omega t + \varphi)$

Potential energy of the particle is

$U = \frac{1}{2}m\omega^{2}x^{2} = \frac{1}{2}m\omega^{2}A^{2}\cos^{2}(\omega t + \varphi)$

Average value of kinetic energy over a period is

$< K > = \frac{1}{T}\int_{0}^{T}{Kdt} = \frac{1}{T}\int_{0}^{T}{\frac{1}{2}m\omega^{2}A^{2}\sin^{2}(\omega t + \varphi)dt}$

$= \frac{1}{2T}m\omega^{2}A^{2}\int_{0}^{T}\left\lbrack \frac{1 - \cos 2(\omega t + \varphi)}{2} \right\rbrack dt$

Since the average value of both a sine and a cosine function for a complete cycle or over a time period T is 0.

$\therefore < K > = \frac{1}{4T}m\omega^{2}A^{2}\lbrack t\rbrack_{0}^{T} = \frac{1}{4T}m\omega^{2}A^{2}T = \frac{1}{4}m\omega^{2}A^{2}$average value of potential energy over a period is

$< U > = \frac{1}{T}\int_{0}^{T}{\frac{1}{2}m\omega^{2}A^{2}\cos^{2}(\omega t + \varphi)dt}$

$= \frac{1}{2T}m\omega^{2}A^{2}\int_{0}^{T}\left\lbrack \frac{1 + \cos 2(\omega t + \varphi)}{2} \right\rbrack dt$

Since the average value of both a sine and a cosine function for a complete cycle or over a time period T is zero.

$\therefore < U > = \frac{1}{4T}m\omega^{2}A^{2}\lbrack t\rbrack_{0}^{T} = \frac{1}{4T}m\omega^{2}A^{2}T = \frac{1}{4}m\omega^{2}A^{2}$