First slide

Simple harmonic motion

Question

For any S.H.M., amplitude is 6 cm. If instantaneous potential energy is half the total energy then distance of particle from its mean position is

Easy

A

3 cm
B

4.2 cm
C

5.8 cm
D

6 cm

Solution

If at any instant displacement is y then it is given that
$U = \frac{1}{2} \times E$ $\Rightarrow \frac{1}{2} {mω}^{2} y^{2} = \frac{1}{2} \times (\frac{1}{2} {mω}^{2} a^{2})$
$\Rightarrow y = \frac{a}{\sqrt{2}} = \frac{6}{\sqrt{2}} = 4 .2 cm$

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