Question

The spring block system as shown in the figure is in equilibrium. The string connecting blocks A and B is cut. The mass of all the three blocks is m and the spring constant of both the springs is k. The amplitude of resulting oscillation of block A is :

Answer 1

Hint: Here in this question we will use the Hooke’s law i.e. $mg + mg + mg = kx$ and we will draw a diagram by identifying the part of the spring block system .
Formula used: $mg + mg + mg = kx$

Complete step-by-step answer:

Here , let $x$ is the extension of spring just after cutting.

Applying Hooke’s law , $mg + mg + mg = kx,\;x = \dfrac{{3mg}}{k}$

The spring extension, when block A is in the mean position
 $ = \dfrac{{mg}}{k}$
Amplitude of oscillation of A
 $ = \dfrac{{3mg}}{k} - \dfrac{{mg}}{k} = \dfrac{{2mg}}{k}$

Note: Hooke’s law in physics states that the force required to extinguish or compress a spring linearly with respect to ${F_8} = kx$ distance, where k is characteristic of the spring constant factor. In many other situations where an elastic body is deformed. Hooke 's equation holds such as wind blowing on a tall building, and a musician plucking a guitar string. An elastic body or material; this equation is said to be linear elastic or hookean for which one would presume.