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…

(A) As both blocks moving together so Time period $=2 \pi \sqrt{\frac{\mathrm{~m}}{\mathrm{~K}}} ;$=2π K m​​; where $\mathrm{m}=\mathrm{M}+\mathrm{m}$m=M+m

$$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{M}+\mathrm{m}}{\mathrm{~K}}}$$T=2π KM+m​​

(B) Let block is displaced by x in $(+\mathrm{ve})$(+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

$$\mathrm{f}=\mathrm{ma}=\frac{\mathrm{mKx}}{(\mathrm{M}+\mathrm{m})}$$f=ma=(M+m)mKx​

(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 $\mu \mathrm{mg}$μmg as force is acting between blocks & normal force here is mg.