A block of mass m attached to massless spring | vedclass

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Question

At equilibrium position

$V _{0}=\omega_{0} A =\sqrt{\frac{ K }{ m }} A....(i)$

$V =\omega A '=\sqrt{\frac{ K }{\frac{ m }{2}}} A '...(ii

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

At equilibrium position

$V _{0}=\omega_{0} A =\sqrt{\frac{ K }{ m }} A....(i)$

$V =\omega A '=\sqrt{\frac{ K }{\frac{ m }{2}}} A '...(ii)$

$	herefore \quad A'= \sqrt{2} A$

$P_{i}=P_{f}$

$mV_0=\frac m{2} V$

$m(A \omega)=\frac{m}{2}\left(A^{\prime} \omega^{\prime}ight)$

$m A \sqrt{\frac{k}{m}}=\frac{m}{2} A^{\prime} \sqrt{\frac{k}{m / 2}}$

$m^{2} A^{2}\left[\frac{K}{m}ight]=\frac{m^{2}}{42}\left(A^{\prime}ight)^{2} \frac{k}{m} 	imes v$

$2 A^{2}=\left(A^{\prime}ight)^{2} \Rightarrow A^{\prime}=\sqrt{2}$

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