The displacement of a simple harmonic motion | vedclass

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(C) Given,amplitude of oscillation,$a = 6 	ext{ cm}$.Let the displacement be $x 	ext{ cm}$.When the kinetic energy $(K)$ is equal to the potential e

Vedclass · Accepted Answer

$3 \sqrt{2} 	ext{ cm}$

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(C) Given,amplitude of oscillation,$a = 6 	ext{ cm}$.Let the displacement be $x 	ext{ cm}$.When the kinetic energy $(K)$ is equal to the potential energy $(U)$,we have $K = U$.The kinetic energy of a simple harmonic oscillator is given by $K = \frac{1}{2} m \omega^2 (a^2 - x^2)$.The potential energy is given by $U = \frac{1}{2} m \omega^2 x^2$.Equating the two:$\frac{1}{2} m \omega^2 (a^2 - x^2) = \frac{1}{2} m \omega^2 x^2$$a^2 - x^2 = x^2$$2x^2 = a^2$$x^2 = \frac{a^2}{2}$$x = \frac{a}{\sqrt{2}}$Substituting $a = 6 	ext{ cm}$:$x = \frac{6}{\sqrt{2}} = 3 \sqrt{2} 	ext{ cm}$.

The displacement of a simple harmonic motion of amplitude $6 \text{ cm}$ when its kinetic energy is equal to its potential energy is:

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