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Simple Harmonic Motion

Section: Physics

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Question : 1 of 34

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A particle is executing simple harmonic motion starting from its mean position. If the time period of the particle is 1.5 s , then the minimum time at which the ratio of the kinetic and total energies of the particle becomes

3 : 4

[AP EAPCET 24 May 2025 S1]

$\frac{1}{4}\ \mathrm{s}$
$\frac{1}{12}\ \mathrm{s}$
$\frac{1}{8}\ \mathrm{s}$
$\frac{1}{6}\ \mathrm{s}$

Solution:

For particle in SHM,

T = 1.5\ \mathrm{sso}

\omega = \frac{2\pi}{T} = \frac{4\pi}{3}\ \mathrm{rad/s}

Let oscillations starts from

t = 0

and

x = A \sin \omega t

is displacement equation Then, KE at time

t

is,

K = \frac{1}{2} m v^{2} = \frac{1}{2} m \left(\frac{dx}{dt}\right)^{2}

= \frac{1}{2} m \cdot A^{2} \omega^{2} \cos^{2} \omega t \cdots (i)

And total energy of particle is,

E = \frac{1}{2} k A^{2}

Here,

k = m \omega^{2}

= \frac{1}{2} m A^{2} \omega^{2} \cdots (ii)

Given, ratio,

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

\Rightarrow \frac{ \frac{1}{2} m A^{2} \omega^{2} \cos^{2} \omega t }{ \frac{1}{2} m A^{2} \omega^{2} } = \frac{3}{4}

\Rightarrow \cos^{2} \omega t = \frac{3}{4}

\Rightarrow \cos \omega t = \sqrt{ \frac{3}{2} }

\Rightarrow \omega t = \frac{\pi}{6}

\Rightarrow t = \frac{\pi}{6\omega} = \frac{\pi \times 3}{6 \times 4\pi} = \frac{1}{8}\ \mathrm{s}

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