One end of a spring of force constant k is fi | vedclass

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(C) Given:Maximum compression (Amplitude $A$) $= 2 x_0$Distance of the block from the other wall $= x_0$Mass of the block $= m$Since the block is rele

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$\frac{2 \pi}{3} \sqrt{\frac{m}{k}}$

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(C) Given:Maximum compression (Amplitude $A$) $= 2 x_0$Distance of the block from the other wall $= x_0$Mass of the block $= m$Since the block is released from the extreme position,its displacement $x(t)$ as a function of time can be represented as:$x(t) = A \cos(\omega t)$where $\omega = \sqrt{\frac{k}{m}}$ is the angular frequency.Taking the equilibrium position as $x = 0$,the block is initially at $x = -2 x_0$ (compressed position). When it moves towards the other wall,it passes through the equilibrium position and hits the wall at $x = +x_0$.Substituting $x = x_0$ and $A = 2 x_0$ into the equation:$x_0 = 2 x_0 \cos(\omega t)$$\cos(\omega t) = \frac{1}{2}$Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$,we have:$\omega t = \frac{\pi}{3}$$t = \frac{\pi}{3 \omega} = \frac{\pi}{3} \sqrt{\frac{m}{k}}$Wait,let's re-evaluate the motion. The block starts at $x = -2 x_0$ and moves to $x = +x_0$. The time taken to go from $x = -2 x_0$ to $x = 0$ is $T/4$. The time taken to go from $x = 0$ to $x = x_0$ is found by $\sin(\omega t) = \frac{x_0}{2 x_0} = \frac{1}{2}$,so $\omega t = \frac{\pi}{6}$,which means $t = \frac{\pi}{6 \omega}$.Total time $t = \frac{T}{4} + \frac{\pi}{6 \omega} = \frac{\pi}{2 \omega} + \frac{\pi}{6 \omega} = \frac{4 \pi}{6 \omega} = \frac{2 \pi}{3 \omega} = \frac{2 \pi}{3} \sqrt{\frac{m}{k}}$.

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