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(C) Case $(i)$: The block is at $x_0$, where its velocity is maximum $(v_{max} = \omega A = \sqrt{k/M} A)$. When mass $m$ is placed, momentum is conse

Vedclass · Accepted Answer

$A, B, D$

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(C) Case $(i)$: The block is at $x_0$, where its velocity is maximum $(v_{max} = \omega A = \sqrt{k/M} A)$. When mass $m$ is placed, momentum is conserved: $M v_{max} = (M + m) v'_{max}$. Thus, $v'_{max} = \frac{M}{M+m} \sqrt{\frac{k}{M}} A$. The new angular frequency is $\omega' = \sqrt{k/(M+m)}$. Since $v'_{max} = \omega' A'$, we have $A' = \frac{v'_{max}}{\omega'} = \sqrt{\frac{M}{M+m}} A$. The amplitude changes by a factor of $\sqrt{\frac{M}{M+m}}$.Case $(ii)$: The block is at $x = x_0 + A$, where its velocity is $0$. Placing mass $m$ does not change the velocity $(0)$. The new equilibrium position remains $x_0$, and the block starts oscillating from the same amplitude $A$ with a new frequency $\omega'$. Thus, the amplitude remains unchanged.Time Period: In both cases, the new time period is $T' = 2\pi \sqrt{\frac{M+m}{k}}$, which is the same.Energy: In case $(i)$, kinetic energy decreases due to the inelastic collision. In case $(ii)$, potential energy remains the same, but total energy is conserved as no work is done by external forces during the placement.Speed: In case $(i)$, the speed at $x_0$ decreases. In case $(ii)$, the speed at $x_0$ (which is $v'_{max}$) is less than the original $v_{max}$ because the new amplitude is the same but the frequency is lower. Thus, $A, B, D$ are correct.

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