A particle executing SHM has a maximum speed | vedclass

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Question

(A) Given: Maximum speed,$v_{\max} = 0.5 \ m s^{-1}$; Maximum acceleration,$a_{\max} = 1.0 \ m s^{-2}$.We know that for $SHM$,the maximum speed is giv

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$2 \ rad \ s^{-1}$

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(A) Given: Maximum speed,$v_{\max} = 0.5 \ m s^{-1}$; Maximum acceleration,$a_{\max} = 1.0 \ m s^{-2}$.We know that for $SHM$,the maximum speed is given by $v_{\max} = \omega A$,where $\omega$ is the angular frequency and $A$ is the amplitude.Also,the maximum acceleration is given by $a_{\max} = \omega^2 A$.Dividing the expression for maximum acceleration by the expression for maximum speed:$\frac{a_{\max}}{v_{\max}} = \frac{\omega^2 A}{\omega A} = \omega$.Substituting the given values:$\omega = \frac{1.0 \ m s^{-2}}{0.5 \ m s^{-1}} = 2 \ rad \ s^{-1}$.Thus,the angular frequency of oscillation is $2 \ rad \ s^{-1}$.

$A$ particle executing $SHM$ has a maximum speed of $0.5 \ m s^{-1}$ and a maximum acceleration of $1.0 \ m s^{-2}$. The angular frequency of oscillation is

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