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A block with mass $M$ is connected by a massless spring with stiffiess constant $k$ to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude $A$ about an equilibrium position $x_0$. Consider two cases: ($i$) when the block is at $x_0$; and ($ii$) when the block is at $x=x_0+A$. In both the cases, a perticle with mass $m$ is placed on the mass $M$ ?
($A$) The amplitude of oscillation in the first case changes by a factor of $\sqrt{\frac{M}{m+M}}$, whereas in the second case it remains unchanged
($B$) The final time period of oscillation in both the cases is same
($C$) The total energy decreases in both the cases
($D$) The instantaneous speed at $x_0$ of the combined masses decreases in both the cases
$A,B$
$B,D$
$A,B,D$
$A,B,C$
Solution
In case $I$,
From Conservation of momentum,
$MV _1=( M + m ) V _2$
$\frac{ MV _1}{ M + m }= V _2$
$\sqrt{\frac{k}{M+m}} A_2=\frac{M}{M+m} \sqrt{\frac{k}{M}} A_1$
$A_2=\sqrt{\frac{M}{M+m}} A_1$
In case $II$,
$A_2=A_1$
$T =2 \pi \sqrt{\frac{\overline{M+m}}{k}}$ in both the cases.
Total energy decreases in first case where as remain same in $2$ nd case. Instantaneous speed at $x_0$ decreases in both case.
Answer is $A , B$ and $D$ .
