A particle executing simple harmonic motion h | vedclass

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Question

(A) Given,maximum speed of $SHM$ is $v_{\max} = A\omega = 40 \,ms^{-1}$ ... $(i)$Maximum acceleration of $SHM$ is $a_{\max} = A\omega^2 = 60 \,ms^{-2}

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$\frac{4 \pi}{3} \,s$

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(A) Given,maximum speed of $SHM$ is $v_{\max} = A\omega = 40 \,ms^{-1}$ ... $(i)$Maximum acceleration of $SHM$ is $a_{\max} = A\omega^2 = 60 \,ms^{-2}$ ... (ii)Dividing Eq. (ii) by Eq. $(i)$,we get:$\frac{A\omega^2}{A\omega} = \frac{60}{40}$$\omega = 1.5 \,rad/s = \frac{3}{2} \,rad/s$We know that the time period $T$ is related to angular frequency $\omega$ by the formula:$T = \frac{2\pi}{\omega}$Substituting the value of $\omega$:$T = \frac{2\pi}{3/2} = \frac{4\pi}{3} \,s$Hence,the time period of the oscillation is $\frac{4\pi}{3} \,s$.

$A$ particle executing simple harmonic motion has a maximum speed of $40 \,ms^{-1}$ and maximum acceleration of $60 \,ms^{-2}$. The period of oscillation is

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