Question

A small block of mass m is kept on a bigger block of mass M which is attached to a vertical spring of spring constant k as shown in the figure. The system oscillates vertically. (a) Find the resultant force on the smaller block when it is displaced through a distance x above its equilibrium position. (b) Find the normal force on the smaller block at this position. When is this force smallest in magnitude? (c) What can be the maximum amplitude with which the two blocks may oscillate together?

Answer 1

(a) Consider the free body diagram.
     Weight of the body, W = mg
     Force, F = ma = mω²x

x is the small displacement of mass m.
As normal reaction R is acting vertically in the upward direction, we can write:
R + mω²x − mg = 0                ....(1)

Resultant force = mω²x = mg − R

\[\Rightarrow m \omega^2 x = m\left( \frac{k}{M + m} \right)x\] 

\[                           = \frac{mkx}{M + m}\] 

\[\text { Here }, \] 

\[\omega = \sqrt{\left\{ \frac{k}{M + m} \right\}}\]

(b) R = mg − mω²x

\[= mg - m\frac{k}{M + N}x\] 

\[ = mg - \frac{mkx}{M + N}\]

It can be seen from the above equations that, for R to be smallest, the value of mω²xshould be maximum which is only possible when the particle is at the highest point.

(c) R = mg − mω²x
     As the two blocks oscillate together R becomes greater than zero.
    When limiting condition follows, 
   i.e. R = 0
      mg = mω²x

\[x = \frac{mg}{m \omega^2} = \frac{mg \cdot \left( M + m \right)}{mk}\]

Required maximum amplitude

\[= \frac{g\left( M + m \right)}{k}\]