A particle executing simple harmonic motion h | vedclass

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(B) The acceleration $a$ of a particle in simple harmonic motion is given by $a = \omega^2 y$,where $\omega$ is the angular frequency and $y$ is the d

Vedclass · Accepted Answer

$12$

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(B) The acceleration $a$ of a particle in simple harmonic motion is given by $a = \omega^2 y$,where $\omega$ is the angular frequency and $y$ is the displacement from the mean position.Given: Amplitude $A = 6\, cm$,displacement $y = 2\, cm$,and acceleration $a = 8\, cm/s^2$.Using the formula $a = \omega^2 y$:$8 = \omega^2 	imes 2$$\omega^2 = 4$$\omega = 2\, rad/s$.The maximum speed $v_{\max}$ of a particle in simple harmonic motion is given by $v_{\max} = A\omega$.Substituting the values: $v_{\max} = 6\, cm 	imes 2\, rad/s = 12\, cm/s$.

$A$ particle executing simple harmonic motion has an amplitude of $6\, cm$. Its acceleration at a distance of $2 \,cm$ from the mean position is $8\, cm/s^2$. The maximum speed of the particle is ... $cm/s$.

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$Reason :$ Displacement and velocity of $S.H.M.$ differ in phase by $\frac{\pi }{2}$.

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