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(A) For a particle in simple harmonic motion,the maximum acceleration is given by $A_{max} = \omega^2 A$,where $\omega$ is the angular frequency and $

Vedclass · Accepted Answer

$32/3 \ m$

Vedclass · Answer

(A) For a particle in simple harmonic motion,the maximum acceleration is given by $A_{max} = \omega^2 A$,where $\omega$ is the angular frequency and $A$ is the amplitude.Given $A_{max} = 24 \ m/s^2$,so $\omega^2 A = 24$ --- $(i)$The maximum velocity is given by $V_{max} = \omega A$.Given $V_{max} = 16 \ m/s$,so $\omega A = 16$ --- $(ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{\omega^2 A}{\omega A} = \frac{24}{16}$$\omega = 1.5 \ rad/s = 3/2 \ rad/s$Substitute $\omega = 3/2$ into equation $(ii)$:$(3/2) A = 16$$A = 16 	imes (2/3) = 32/3 \ m$Therefore,the amplitude is $32/3 \ m$.

$A$ particle executing simple harmonic motion has a maximum acceleration of $24 \ m/s^2$ and a maximum velocity of $16 \ m/s$. What is its amplitude?

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