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

Question

Quantrex Academy · Accepted Answer

Correct answer: A. A, B, D Only

(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}<br/>& \mathrm{F}=-\mathrm{Kx} \\ & (\mathrm{M}+\mathrm{m}) \mathrm{a}=-\mathrm{Kx} \\ & \mathrm{a}=-\frac{\mathrm{Kx}}{(\mathrm{M}+\mathrm{m})}<br/>\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.
Two blocks of masses m and M,(M m) , are placed on a frictionless table as shown in figure. A massless spring with

Two blocks of masses m and M,(M m) , are placed on a frictionless table as shown in figure. A massless spring with

\begin{aligned}<br/>& 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}<br/>\end{aligned}

(E) Maximum friction can be

\mu \mathrm{mg}

as force is acting between blocks \& normal force here is mg.

Two blocks of masses m and M,(M m) , are placed on a frictionless...

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