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A particle of mass $5 \mathrm{~g}$ is executing S.H.M. with an amplitude $0.3 \mathrm{~m}$ and period $\frac{\pi}{5} \mathrm{~s}$. The maximum value of the force acting on the particle is
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Verified Answer
The correct answer is:
$0.15 \mathrm{~N}$
$$
\begin{aligned}
& \mathrm{m}=5 \mathrm{~g}=5 \times 10^{-3} \mathrm{~kg}, \mathrm{~A}=0.3 \mathrm{~m}, \mathrm{~T}=\frac{\pi}{5} \mathrm{~s} \\
& \omega=\frac{2 \pi}{\mathrm{T}}=10 \mathrm{rad} / \mathrm{s}
\end{aligned}
$$
Maximum force, $\mathrm{F}=\mathrm{m} \omega^2 \mathrm{~A}=5 \times 10^{-3} \times(10)^2 \times 0.3=0.15 \mathrm{~N}$
\begin{aligned}
& \mathrm{m}=5 \mathrm{~g}=5 \times 10^{-3} \mathrm{~kg}, \mathrm{~A}=0.3 \mathrm{~m}, \mathrm{~T}=\frac{\pi}{5} \mathrm{~s} \\
& \omega=\frac{2 \pi}{\mathrm{T}}=10 \mathrm{rad} / \mathrm{s}
\end{aligned}
$$
Maximum force, $\mathrm{F}=\mathrm{m} \omega^2 \mathrm{~A}=5 \times 10^{-3} \times(10)^2 \times 0.3=0.15 \mathrm{~N}$
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