# 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. - Physics

Sum

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?

#### Solution

(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 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 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

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

Is there an error in this question or solution?
Chapter 12: Simple Harmonics Motion - Exercise [Page 253]

#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 1
Chapter 12 Simple Harmonics Motion
Exercise | Q 14 | Page 253

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