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(A) The equation of motion is $x(t) = A \sin(\omega t)$,where $\omega = \sqrt{k/m}$.The velocity is $v(t) = A\omega \cos(\omega t)$. At $t=0$,$v(0) =

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$(A,D)$

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(A) The equation of motion is $x(t) = A \sin(\omega t)$,where $\omega = \sqrt{k/m}$.The velocity is $v(t) = A\omega \cos(\omega t)$. At $t=0$,$v(0) = u_0 = A\omega$,so $A = u_0/\omega$.When speed is $0.5 u_0$,$A\omega \cos(\omega t) = 0.5 u_0 \implies u_0 \cos(\omega t) = 0.5 u_0 \implies \cos(\omega t) = 1/2$.Thus,$\omega t_1 = \pi/3$,so $t_1 = \frac{\pi}{3} \sqrt{\frac{m}{k}}$.At this time,the particle hits the wall elastically. The speed remains $0.5 u_0$ but the direction reverses.$(A)$ Since the collision is elastic and the spring potential energy is conserved,the total energy remains constant. When it returns to equilibrium $(x=0)$,the potential energy is zero,so kinetic energy is $1/2 m u_0^2$. Thus,speed is $u_0$. (Correct)$(B)$ After collision at $t_1$,the particle moves back. It reaches equilibrium when the phase $\omega t$ reaches $\pi$. The time taken from $t_1$ to reach equilibrium is $\Delta t = (\pi - \pi/3)/\omega = (2\pi/3)/\omega$. Total time $t = \pi/3\omega + 2\pi/3\omega = \pi\omega = \pi \sqrt{m/k}$. (Correct)$(C)$ Maximum compression occurs at the extreme position. After passing equilibrium at $t = \pi\sqrt{m/k}$,it reaches the other extreme at $t = \pi\sqrt{m/k} + \pi/2\sqrt{m/k} = 1.5\pi\sqrt{m/k}$. (Incorrect)$(D)$ The particle passes equilibrium for the second time after the collision when the phase reaches $2\pi$. Time $t = 2\pi/\omega = 2\pi\sqrt{m/k}$. (Incorrect)Wait,re-evaluating $(D)$: The particle hits the wall at $\omega t = \pi/3$. It reflects and moves towards equilibrium. It reaches equilibrium at $\omega t = \pi$. Then it moves to the other side and returns to equilibrium at $\omega t = 2\pi$. Total time $t = 2\pi\sqrt{m/k}$.Given the options,$(A)$ and $(B)$ are correct.

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