First slide

Simple hormonic motion

Question

The amplitude of oscillation of particle in SHM is √3cm. The displacement from mean position at which its potential and kinetic energies are in the ratio 1 : 2 is

Easy

A

1cm
B

0.866 cm
C

$\frac{1}{\sqrt{3}} cm$
D

√2 cm

Solution

$P.E. = \frac{1}{2} {mω}^{2} x^{2}; K.E. = \frac{1}{2} {mω}^{2} (A^{2} - x^{2})$

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