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

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(C) Let the equilibrium position be $B$. The block is released from $A$ where the displacement is $2x_0$. The wall is at $C$ such that $BC = x_0$.The

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

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(C) Let the equilibrium position be $B$. The block is released from $A$ where the displacement is $2x_0$. The wall is at $C$ such that $BC = x_0$.The motion is simple harmonic with amplitude $A = 2x_0$. The position of the block at any time $t$ is given by $x(t) = A \sin(\omega t + \phi)$.Taking the equilibrium position $B$ as $x=0$,the block starts from $x = -2x_0$ at $t=0$. Thus,$x(t) = -2x_0 \cos(\omega t)$,where $\omega = \sqrt{k/m}$.The block strikes the wall at $C$ where $x = x_0$. So,$x_0 = -2x_0 \cos(\omega t) \implies \cos(\omega t) = -1/2$.The time taken to reach the equilibrium position $B$ from $A$ is $T/4$. The time taken to go from $B$ to $C$ is $t_{BC}$.Using $x = A \sin(\omega t)$,at $x = x_0$ and $A = 2x_0$,we have $x_0 = 2x_0 \sin(\omega t_{BC}) \implies \sin(\omega t_{BC}) = 1/2$.Thus,$\omega t_{BC} = \pi/6 \implies t_{BC} = \frac{\pi}{6\omega} = \frac{\pi}{6} \sqrt{\frac{m}{k}} = \frac{T}{12}$.The total time $t = T/4 + T/12 = 4T/12 = T/3$.Since $T = 2\pi \sqrt{m/k}$,the total time is $t = \frac{1}{3} (2\pi \sqrt{m/k}) = \frac{2\pi}{3} \sqrt{\frac{m}{k}}$.

$A$. $Pa \cdot s$	$(i)$ $[L^2 \ T^{-2} \ K^{-1}]$
$B$. $N \cdot m \cdot K^{-1}$	$(ii)$ $[M \ L \ T^{-3} \ K^{-1}]$
$C$. $J \cdot kg^{-1} \cdot K^{-1}$	$(iii)$ $[M \ L^{-1} \ T^{-1}]$
$D$. $W \cdot m^{-1} \cdot K^{-1}$	$(iv)$ $[M \ L^2 \ T^{-2} \ K^{-1}]$

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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