Two blocks of masses \(m\) and \(M,(M \gt m)\), 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 \(\mathrm{T}=2 \pi \sqrt{\frac{(\mathrm{~m}+\mathrm{M})}{\mathrm{k}}}\)
(B) The acceleration of the blocks is \(\mathrm{a}=\frac{\mathrm{kx}}{\mathrm{M}+\mathrm{m}}\) (\(\mathrm{x}=\) 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:

(A) As both blocks moving together so Time period \(=2 \pi \sqrt{\frac{\mathrm{~m}}{\mathrm{~K}}} ;\) where \(\mathrm{m}=\mathrm{M}+\mathrm{m}\)
\(\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{M}+\mathrm{m}}{\mathrm{~K}}}\)
(B) Let block is displaced by x in \((+\mathrm{ve})\) direction so force on block will be in(-ve) direction
\(\begin{aligned}
& \mathrm{F}=-\mathrm{Kx} \\
& (\mathrm{M}+\mathrm{m}) \mathrm{a}=-\mathrm{Kx} \\
& \mathrm{a}=-\frac{\mathrm{Kx}}{(\mathrm{M}+\mathrm{m})}
\end{aligned}\)
(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.

\(\begin{aligned}
& K A=(M+m) a \\
& a=\frac{K A}{(M+m)} \\
& f=m a=\frac{m K A}{(M+m)} \\
& f_{\max }=f_L=\mu \mathrm{mg} \\
& f=\mu m g \\
& \frac{m K A}{(M+m)}=\mu m g \\
& A=\frac{\mu(M+m) g}{K}
\end{aligned}\)
(E) Maximum friction can be \(\mu \mathrm{mg}\) as force is acting between blocks & normal force here is mg.