A particle performing linear S.H.M. has maximum velocity of 25 cm/s and maximum acceleration of 100 cm/s$^2$. Find the amplitude and period of oscillation. ($\pi = 3.142$)

A particle performing linear S.H.M. has maximum velocity of 25 cm/s and maximum acceleration of 100 cm/s². Fin

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Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2018Subjective· 2mImportance★★★★★

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For SHM, $v_{\max}=A\omega$ and $a_{\max}=A\omega^2$ ; dividing the two directly gives $\omega$ , from which $A$ and $T$ follow.

For a particle performing linear SHM with amplitude $A$ and angular frequency $\omega$ :

$v_{\max} = A\omega, \qquad a_{\max} = A\omega^2.$

Dividing the second by the first eliminates $A$ :

$\frac{a_{\max}}{v_{\max}} = \omega \;\Rightarrow\; \omega = \frac{100}{25} = 4\ \text{rad/s}.$

Then the amplitude is …