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A mass of 1 kg is suspended from a spring of force constant 400 N , executing SHM total energy of the body is 2 J , then maximum acceleration of the spring will be
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$40 \mathrm{~m} / \mathrm{s}^2$
$T=2 \pi \sqrt{\frac{m}{R}}=2 \pi \sqrt{\frac{1}{400}}=\frac{2 \pi}{20}$
Frequency $n=\frac{20}{2 \pi}$
$\omega=2 \pi n=2 \pi \times \frac{20}{2 \pi}=20$
Total energy
$\begin{aligned}& E=\frac{1}{2} m \omega^2 r^2 \\& 2=\frac{1}{2} \times 1 \times 400 \times r^2\end{aligned}$
$\begin{gathered}r=\frac{1}{10} \mathrm{~m} \\ \text { Acceleration }=\omega^2 r=400 \times \frac{1}{10}=40 \mathrm{~m} / \mathrm{s}^2\end{gathered}$
Frequency $n=\frac{20}{2 \pi}$
$\omega=2 \pi n=2 \pi \times \frac{20}{2 \pi}=20$
Total energy
$\begin{aligned}& E=\frac{1}{2} m \omega^2 r^2 \\& 2=\frac{1}{2} \times 1 \times 400 \times r^2\end{aligned}$
$\begin{gathered}r=\frac{1}{10} \mathrm{~m} \\ \text { Acceleration }=\omega^2 r=400 \times \frac{1}{10}=40 \mathrm{~m} / \mathrm{s}^2\end{gathered}$
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