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 fig. 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

$\therefore$ $R+mw{y}^{2}x-mg=0$

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

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

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

$w=\sqrt{\frac{k}{M+m}}$

[for spring mas system]

b. $R=mg-m{w}^{2}x$

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

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

For $LR$ to be smallest $m{w}^{2}x$ should be max, i.e. $x$ is maximum.

The particle should be at the highest point.

c. We have $R=mg-m{w}^{2}x$

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

At limiting condition $R=0$

$mg=m{w}^{2}x$

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

So the maximum amplitude

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