JEE MAIN - Physics (2025 - 3rd April Morning Shift - No. 15)
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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 :
Explanation
$$\begin{aligned} &\text { (A) As both blocks moving together so }\\ &\begin{aligned} & \text { Time period }=2 \pi \sqrt{\frac{\mathrm{~m}}{\mathrm{~K}}} ; \text { where } \mathrm{m}=\mathrm{M}+\mathrm{m} \\ & \qquad \mathrm{~T}=2 \pi \sqrt{\frac{\mathrm{M}+\mathrm{m}}{\mathrm{~K}}} \end{aligned} \end{aligned}$$
(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.
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$$\begin{aligned} & K A=(M+m) a \\ & a=\frac{K A}{(M+m)} \\ & f=m a=\frac{m K A}{(M+m)} \\ & f_{\max }=f_L=\mu m g \\ & f=\mu m g \\ & \frac{m K A}{(M+m)}=\mu m g \\ & A=\frac{\mu(M+m) g}{K} \end{aligned}$$
(E) Maximum friction can be $\mu \mathrm{mg}$ as force is acting between blocks & normal force here is mg .
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