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A particle performing SHM has time period $\frac{2 \pi}{\sqrt{3}}$ and path length $4 \mathrm{~cm}$. The displacement from mean position at which acceleration is equal to velocity is
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Verified Answer
The correct answer is:
$1 \mathrm{am}$
Velocity $v=\omega \sqrt{A^{2}-x^{2}}$
and acceleration $=\omega^{2} x$ Given, $\omega \sqrt{A^{2}-x^{2}}=\omega^{2} x$
or $\sqrt{A^{2}-x^{2}}=\omega x$
Given,
$$
T=\frac{2 \pi}{\sqrt{3}}
$$
and
$$
\omega=\frac{2 \pi}{T}=\sqrt{3}
$$
Substituting the value of $\omega$ in Eq (i), we get $\sqrt{A^{2}-x^{2}}=\sqrt{3} x$
$$
\Rightarrow \quad A=2 x
$$
$$
\begin{aligned}
\text { As amplitude } &=\frac{\text { path length }}{2}=2 \mathrm{~cm} \\
\Rightarrow \quad x=1 \mathrm{~cm}
\end{aligned}
$$
and acceleration $=\omega^{2} x$ Given, $\omega \sqrt{A^{2}-x^{2}}=\omega^{2} x$
or $\sqrt{A^{2}-x^{2}}=\omega x$
Given,
$$
T=\frac{2 \pi}{\sqrt{3}}
$$
and
$$
\omega=\frac{2 \pi}{T}=\sqrt{3}
$$
Substituting the value of $\omega$ in Eq (i), we get $\sqrt{A^{2}-x^{2}}=\sqrt{3} x$
$$
\Rightarrow \quad A=2 x
$$
$$
\begin{aligned}
\text { As amplitude } &=\frac{\text { path length }}{2}=2 \mathrm{~cm} \\
\Rightarrow \quad x=1 \mathrm{~cm}
\end{aligned}
$$
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