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A particle executing SHM maximum speed of $30 \mathrm{~cm} / \mathrm{s}$ and a maximum acceleration of $60 \mathrm{~cm} / \mathrm{s}^2$. The period of oscillation is
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The correct answer is:
$\pi \mathrm{sec}$
$\pi \mathrm{sec}$
Let us consider a equation of an SHM is represented by $y=a \sin \omega t$
$$
v=\frac{d y}{d t}=a \omega \cos \omega t
$$
$(v)_{\max }=a \omega=30 \mathrm{~cm} / \mathrm{sec} \quad$ (given) ...(i)
Acceleration $(a)=\frac{d v}{d t}=-a \omega^2 \sin \omega t$
$$
a_{\max }=\omega^2 a=60
$$
Eqs. (i) and (ii), we get
$$
\begin{aligned}
&\omega(\omega a)=60 \Rightarrow \omega(30)=60 \\
&\omega=2 \mathrm{rad} / \mathrm{s} \\
&\frac{2 \pi}{T}=2 \mathrm{rad} / \mathrm{s} \\
&T=\pi \mathrm{sec}
\end{aligned}
$$
$$
v=\frac{d y}{d t}=a \omega \cos \omega t
$$
$(v)_{\max }=a \omega=30 \mathrm{~cm} / \mathrm{sec} \quad$ (given) ...(i)
Acceleration $(a)=\frac{d v}{d t}=-a \omega^2 \sin \omega t$
$$
a_{\max }=\omega^2 a=60
$$
Eqs. (i) and (ii), we get
$$
\begin{aligned}
&\omega(\omega a)=60 \Rightarrow \omega(30)=60 \\
&\omega=2 \mathrm{rad} / \mathrm{s} \\
&\frac{2 \pi}{T}=2 \mathrm{rad} / \mathrm{s} \\
&T=\pi \mathrm{sec}
\end{aligned}
$$
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