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

Question

The spring block system as shown in figure is in equilibrium. The string connecting blocks A and B is cut. The mass of all the three blocks is m and spring constant of both the spring is k. The amplitude of resulting oscillation of block A is: <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/71f6ebafb165c7026dc2d1b9676143e5-.png" /> (A) (m g/k) (B) (2 m g/k) (C) (3 m g/k) (D) (4 m g/k)

Tardigrade Admin · Accepted Answer

Just after cutting the string extension in spring

= \frac{3 m g}{k}

and extension in the spring when block is in mean position

= \frac{m g}{k}

\Rightarrow

Amplitude of oscillation is given as

A = \frac{3 m g}{k} - \frac{m g}{k} = \frac{2 m g}{k}

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

$\frac{m g}{k}$

$\frac{2 m g}{k}$

$\frac{3 m g}{k}$

$\frac{4 m g}{k}$

Just after cutting the string extension in spring $= \frac{3 m g}{k}$ and extension in the spring when block is in mean position $= \frac{m g}{k}$

$\Rightarrow$ Amplitude of oscillation is given as
$A = \frac{3 m g}{k} - \frac{m g}{k} = \frac{2 m g}{k}$

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

kmg​

k2mg​

k3mg​

k4mg​