Question 54: Oscillations

Question

Two blocks of masses $m$ and $M,(M>m)$, are placed on a frictionless table as shown in figure. A massless spring with spring constant k is attached with the lower block. If the system is slightly displaced and released, then ( $\mu=$ coefficient of friction between the two blocks)
A. The time period of small oscillation of the two blocks is $T=2 \pi \sqrt{\frac{(m+M)}{k}}$
B. The acceleration of the blocks is $a=-\frac{k x}{M+m}$ ( $x=$ displacement of the blocks from the mean position)
C. The magnitude of the frictional force on the upper block is $\frac{m \mu|x|}{M+m}$
D. The maximum amplitude of the upper block, if it does not slip, is $\frac{\mu(M+m) g}{k}$
E. Maximum frictional force can be $\mu(\mathrm{M}+\mathrm{m}) \mathrm{g}$.
Choose the correct answer from the options given below :

Accepted Answer

$$\begin{aligned}
&	ext { (A) As both blocks moving together so }\
&\begin{aligned}
& 	ext { Time period }=2 \pi \sqrt{\frac{\mathrm{~m}}{\mathrm{~K}}} ; 	ext { where } \mathrm{m}=\mathrm{M}+\mathrm{m} \
& \qquad \mathrm{~T}=2 \pi \sqrt{\frac{\mathrm{M}+\mathrm{m}}{\mathrm{~K}}}
\end{aligned}
\end{aligned}$$
(B) Let block is displaced by x in $(+\mathrm{ve})$ direction so force on block will be in(-ve) direction
$$\begin{aligned}
& F=-K x \
& (M+m) a=-K x
\end{aligned}$$
$$a=-\frac{K x}{(M+m)}$$
(C) As upper block is moving due to friction thus
$$\mathrm{f}=\mathrm{ma}=\frac{\mathrm{mKx}}{(\mathrm{M}+\mathrm{m})}$$
(D) This option is like two block problem in friction for maximum amplitude, force on block is also maximum, for which both blocks are moving together.
$$\begin{aligned}
& K A=(M+m) a \
& a=\frac{K A}{(M+m)} \
& f=m a=\frac{m K A}{(M+m)} \
& f_{\max }=f_L=\mu m g \
& f=\mu m g \
& \frac{m K A}{(M+m)}=\mu m g \
& A=\frac{\mu(M+m) g}{K}
\end{aligned}$$
(E) Maximum friction can be $\mu \mathrm{mg}$ as force is acting between blocks & normal force here is mg .

Answer

B, C, D Only

Answer

C, D, E Only

Answer

A, B, D Only

Answer

A, B, C Only

Oscillations

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