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) From the free body diagram

$\therefore R+m{\omega }^{2}x-mg=0...\left(1\right)$

Resultant force = $m{\omega }^{2}x=mg-R$

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

$=\frac{mkx}{M+m}$

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

[for spring mass system ]

$\left(b\right)R=mg-m{\omega }^{2}x$

$=mg-m\frac{mk}{M+n}x$

$=mg-\frac{mkx}{M+N}$

For R to be smallest, $m{\omega }^{2}x$ should be max, i.e. x is maximum.

The particle should be at the highest point.

(c) We have, $R=mg-m{\omega }^{2}x$

The two blocks may oscillate together in such a way that R is greater than 0.

At limiting condition, R = 0

$mg=m{\omega }^{2}x$

$x=\frac{mg}{m{\omega }^2}=\frac{mg.(M+m)}{mk}$

So, the maximum amplitude

$=\frac{g(M+m)}{k}$