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A body is executing S.H.M. under the action of force having maximum magnitude $50 \mathrm{~N}$. When its energy is half kinetic and half potential, the magnitude of the force acting on the particle is
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Verified Answer
The correct answer is:
$25 \sqrt{2} \mathrm{~N}$
Potential energy is a half of the total energy
$$
\begin{aligned}
& \therefore \frac{1}{2} \mathrm{~m} \omega^2 \mathrm{x}^2=\frac{1}{2}\left[\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{~A}^2\right] \\
& \therefore \mathrm{x}^2=\frac{\mathrm{A}^2}{2} \text { or } \mathrm{x}=\frac{\mathrm{A}}{\sqrt{2}}
\end{aligned}
$$
Maximum force $F_m=m \omega^2 A$
Force at a distance $\mathrm{x}, \mathrm{F}^{\prime}=\mathrm{m} \omega^2 \mathrm{x}$
$$
\begin{aligned}
& \therefore \frac{F^{\prime}}{F_m}=\frac{x}{A}=\frac{1}{\sqrt{2}} \\
& \therefore F^{\prime}=\frac{F_m}{\sqrt{2}}=\frac{50}{\sqrt{2}}=25 \sqrt{2} N
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \frac{1}{2} \mathrm{~m} \omega^2 \mathrm{x}^2=\frac{1}{2}\left[\frac{1}{2} \mathrm{~m} \omega^2 \mathrm{~A}^2\right] \\
& \therefore \mathrm{x}^2=\frac{\mathrm{A}^2}{2} \text { or } \mathrm{x}=\frac{\mathrm{A}}{\sqrt{2}}
\end{aligned}
$$
Maximum force $F_m=m \omega^2 A$
Force at a distance $\mathrm{x}, \mathrm{F}^{\prime}=\mathrm{m} \omega^2 \mathrm{x}$
$$
\begin{aligned}
& \therefore \frac{F^{\prime}}{F_m}=\frac{x}{A}=\frac{1}{\sqrt{2}} \\
& \therefore F^{\prime}=\frac{F_m}{\sqrt{2}}=\frac{50}{\sqrt{2}}=25 \sqrt{2} N
\end{aligned}
$$
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