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The potential energy of a simple harmonic oscillator of mass $2 \mathrm{~kg}$ at its mean position is $5 \mathrm{~J}$. If its total energy is $9 \mathrm{~J}$ and amplitude is $1 \mathrm{~cm}$, then its time period is
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Verified Answer
The correct answer is:
$\frac{\pi}{100} \mathrm{~s}$
Given, total energy $=9 \mathrm{~J}$
$\mathrm{PE}$ at mean position $=5 \mathrm{~J}$
So, maximum $\mathrm{KE}=9 \mathrm{~J}-5 \mathrm{~J}=4 \mathrm{~J}$
Now, in SHM
Maximum (at mean) $\mathrm{KE}=$ Maximum PE (at extremes)
$$
\begin{aligned}
& \therefore \quad \frac{1}{2} k a^2=4 \mathrm{~J} \\
& \Rightarrow \quad k=\frac{8}{a^2}=\frac{8}{10^{-4}}=8 \times 10^4 \mathrm{~J} / \mathrm{m}^2 \\
&
\end{aligned}
$$
Now, time period
$$
T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \times \sqrt{\frac{2}{8 \times 10^4}} \Rightarrow T=\frac{\pi}{100} \mathrm{~s}
$$
$\mathrm{PE}$ at mean position $=5 \mathrm{~J}$
So, maximum $\mathrm{KE}=9 \mathrm{~J}-5 \mathrm{~J}=4 \mathrm{~J}$
Now, in SHM
Maximum (at mean) $\mathrm{KE}=$ Maximum PE (at extremes)
$$
\begin{aligned}
& \therefore \quad \frac{1}{2} k a^2=4 \mathrm{~J} \\
& \Rightarrow \quad k=\frac{8}{a^2}=\frac{8}{10^{-4}}=8 \times 10^4 \mathrm{~J} / \mathrm{m}^2 \\
&
\end{aligned}
$$
Now, time period
$$
T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \times \sqrt{\frac{2}{8 \times 10^4}} \Rightarrow T=\frac{\pi}{100} \mathrm{~s}
$$
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