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(D) The maximum acceleration $(a_{\max})$ of a particle executing simple harmonic motion is given by the formula: $a_{\max} = A\omega^2$,where $A$ is

Vedclass · Accepted Answer

$0.61$

Vedclass · Answer

(D) The maximum acceleration $(a_{\max})$ of a particle executing simple harmonic motion is given by the formula: $a_{\max} = A\omega^2$,where $A$ is the amplitude and $\omega$ is the angular velocity.Given:Angular velocity $\omega = 3.5\, rad/s$Maximum acceleration $a_{\max} = 7.5\, m/s^2$Rearranging the formula to solve for amplitude $A$:$A = \frac{a_{\max}}{\omega^2}$Substituting the given values:$A = \frac{7.5}{(3.5)^2} = \frac{7.5}{12.25} \approx 0.61\, m$Therefore,the amplitude of oscillation is $0.61\, m$.

$A$ particle executes harmonic motion with an angular velocity and maximum acceleration of $3.5\, rad/s$ and $7.5\, m/s^2$ respectively. The amplitude of oscillation is .... $m$

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