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An object of mass $3 \mathrm{~kg}$ is executing simple harmonic motion with an amplitude $\frac{2}{\pi} \mathrm{m}$. If the kinetic energy of the object when it crosses the mean position is $6 \mathrm{~J}$, the time period of oscillation of the object is
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The correct answer is:
$2 \mathrm{~s}$
Mass of object, $\mathrm{m}=3 \mathrm{~kg}$
Amplitude, $A=\frac{2}{\pi} \mathrm{m}$
Kinetic energy $=6 \mathrm{~J}$
Kinetic energy $=\frac{1}{2} m \omega^2 A^2=6$
$\begin{aligned}
& \frac{1}{2} \times 3 \times \frac{4 \pi^2}{\mathrm{~T}^2} \times \frac{4}{\pi^2}=6 \\
& \Rightarrow \mathrm{T}^2=4 ; \mathrm{T}=2 \mathrm{~s}
\end{aligned}$
Amplitude, $A=\frac{2}{\pi} \mathrm{m}$
Kinetic energy $=6 \mathrm{~J}$
Kinetic energy $=\frac{1}{2} m \omega^2 A^2=6$
$\begin{aligned}
& \frac{1}{2} \times 3 \times \frac{4 \pi^2}{\mathrm{~T}^2} \times \frac{4}{\pi^2}=6 \\
& \Rightarrow \mathrm{T}^2=4 ; \mathrm{T}=2 \mathrm{~s}
\end{aligned}$
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