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Simple Harmonic Motion

Section: Physics

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Question : 9 of 34

Marks: +1, -0

If the maximum velocity and maximum acceleration of a particle executing simple harmonic motion are respectively

5\,\mathrm{m}\,\mathrm{s}^{-1}

10\,\mathrm{m}\,\mathrm{s}^{-2}

, then the time period of oscillation of the particle is

[AP EAPCET 22 May 2025 S2]

$\pi\,\mathrm{s}$
$2\pi\,\mathrm{s}$
2s
$1\,\mathrm{s}$

Solution:

Maximum velocity

v_{\max}=5\,\mathrm{m/s}

Maximum acceleration

a_{\max}=10\,\mathrm{m/s}^2

For SHM relationships:

\: v_{\max}= \omega A

\: a_{\max}= \omega^2 A

where

\omega

is the angular frequency, and

A

is the amplitude.

Step 1: Divide the two equations

\: \: \frac{a_{\max}}{v_{\max}} = \: \frac{\omega^2 A}{\omega A} = \omega

\: \Rightarrow \omega = \: \frac{a_{\max}}{v_{\max}} = \: \frac{10}{5} = 2\,\mathrm{rad/s}

Step 2: Find the time period

\: T = \: \frac{2\pi}{\omega} = \: \frac{2\pi}{2} = \pi\,\mathrm{s}

Final Answer:

T = \pi\,\mathrm{s}

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