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?
Figure

Answer 1

Explanation:

(a) As the block moves vertically in upward direction, the resultant force is $F = mkx/(M+m)$ as $R+m \omega^{2} x-m g=0$  and  $\omega=\sqrt{\frac{k}{M+m}}$ where m is mass of small block and M is mass of bigger block and x is the distance of the block from its equilibrium position.  

(b) The normal force on the smaller block at the same position $R=m g-m \omega^{2} x=m g-\frac{m k x}{M+N}$.  

This shows that for R to be small, the value of $m \omega^{2} x$ should be bigger which can be attained only when the particle is at highest point.

(c) When the blocks oscillate together, R = 0, i.e. $m g=m \omega^{2} x$.  Thus, on solving we get the maximum amplitude $x=\frac{\mathrm{s}(\mathrm{M}+\mathrm{m})}{\mathrm{k}}$.