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

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(C) The block is released from a compression of $A = 2x_0$. The motion is simple harmonic with time period $T = 2\pi \sqrt{\frac{m}{k}}$.The block sta

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

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(C) The block is released from a compression of $A = 2x_0$. The motion is simple harmonic with time period $T = 2\pi \sqrt{\frac{m}{k}}$.The block starts from the extreme position $(x = -A = -2x_0)$ and moves towards the equilibrium position $(x = 0)$.The time taken to reach the equilibrium position from the extreme position is $t_1 = \frac{T}{4}$.After reaching the equilibrium position,the block continues to move towards the other wall,which is at a distance $x_0$ from the equilibrium position.The displacement $x$ as a function of time is given by $x(t) = A \sin(\omega t)$,where $\omega = \sqrt{\frac{k}{m}}$.To reach $x = x_0$ from $x = 0$,we have $x_0 = (2x_0) \sin(\omega t_2)$,which gives $\sin(\omega t_2) = \frac{1}{2}$.Thus,$\omega t_2 = \frac{\pi}{6}$,so $t_2 = \frac{\pi}{6\omega} = \frac{\pi}{6} \sqrt{\frac{m}{k}} = \frac{T}{12}$.The total time taken to strike the wall is $t = t_1 + t_2 = \frac{T}{4} + \frac{T}{12} = \frac{3T + T}{12} = \frac{4T}{12} = \frac{T}{3}$.Substituting $T = 2\pi \sqrt{\frac{m}{k}}$,we get $t = \frac{1}{3} (2\pi \sqrt{\frac{m}{k}}) = \frac{2\pi}{3} \sqrt{\frac{m}{k}}$.

One end of a spring of force constant $k$ is fixed to a vertical wall and the other to a block of mass $m$ resting on a smooth horizontal surface. There is another wall at a distance $x_0$ from the block. The spring is then compressed by $2x_0$ and released. The time taken to strike the wall is

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