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A particle executing simple harmonic motion of amplitude $5 \mathrm{~cm}$ has maximum speed of $31.4 \mathrm{~cm} / \mathrm{s}$. The frequency of its oscillation is:
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The correct answer is:
$1 \mathrm{~Hz}$
Here $a=5 \mathrm{~cm}, V_{\max }=\frac{31.4 \mathrm{~cm}}{\mathrm{~s}}$
$\begin{aligned}
& V_{\max }=\omega a=31.4=2 \pi \mathrm{v} \times 5 \\
& \Rightarrow 31.4=10 \times 3.14 \times V \\
& \Rightarrow \quad V=1 Hz
\end{aligned}$
$\begin{aligned}
& V_{\max }=\omega a=31.4=2 \pi \mathrm{v} \times 5 \\
& \Rightarrow 31.4=10 \times 3.14 \times V \\
& \Rightarrow \quad V=1 Hz
\end{aligned}$
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