A particle is performing $S.H.M.$ with energy of vibration $90 \,J$ and amplitude $6 \,cm$. When the particle reaches at distance $4 \,cm$ from mean position, it is stopped for a moment and then released. The new energy of vibration will be ........... $J$
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(a)
Energy $=90 \,J \quad$ Amplitude $=6 \,cm$
Maximum energy $=\frac{1}{2} m A^2 \omega^2=90$
$\text { or } m \omega^2=\frac{180}{36 \times 10^{-4}}$
$\therefore m \omega^2=30 \times 10^2$
When particle is stopped the point where it is stopped is the new amplitude but angular velocity will remain same.
$E=\frac{1}{2} m A_2^2 \omega^2$
$\text { or } E=3000 A_2^2$
$A_2=4 \times 10^{-2}$
$A_2=3000$
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