MCQ
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
    $A,B$
  • B
    $B,D$
  • $A,B,D$
  • D
    $A,B,C$

Answer

Correct option: C.
$A,B,D$
c
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 $D$ .

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