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A point mass of $400 \mathrm{~g}$ executes S.H.M. under a force $\mathrm{F}=-\left(10 \mathrm{Nm}^{-1}\right) \mathrm{x}$. If it crosses the centre of oscillation with a speed of $10 \mathrm{~ms}^{-1}$, the amplitude of motion is
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The correct answer is:
$4 \mathrm{~m}$
We have, $\mathrm{F}=-10 \mathrm{x}$ Comparing it with $\mathrm{F}=-\mathrm{m} \omega^2 \mathrm{x}$, we get $\mathrm{m} \omega^2=10$
$$
\Rightarrow \omega=\sqrt{\frac{10}{\mathrm{~m}}}=\sqrt{\frac{10}{0.4}}=5 \mathrm{rad} / \mathrm{sec}
$$
Now, $\mathrm{V}_0=\omega \mathrm{A}$
$$
\begin{aligned}
& \Rightarrow 10=5 \mathrm{~A} \\
& \Rightarrow \mathrm{A}=2 \mathrm{~m}
\end{aligned}
$$
$$
\Rightarrow \omega=\sqrt{\frac{10}{\mathrm{~m}}}=\sqrt{\frac{10}{0.4}}=5 \mathrm{rad} / \mathrm{sec}
$$
Now, $\mathrm{V}_0=\omega \mathrm{A}$
$$
\begin{aligned}
& \Rightarrow 10=5 \mathrm{~A} \\
& \Rightarrow \mathrm{A}=2 \mathrm{~m}
\end{aligned}
$$
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