 # A block of mass m attached to a massless spring is performing oscillatory motion of amplitude ' A ' on a frictionless horizontal plane. Question:

A block of mass $m$ attached to a massless spring is performing oscillatory motion of amplitude ' $A$ ' on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become $f A$. The value of $f$ is :

1.  $\frac{1}{\sqrt{2}}$

2. $\frac{2}{3}$

3.  $\frac{3}{\sqrt{2}}$

4. $\frac{\sqrt{2}}{3}$

Correct Option: 1,

Solution:

(1) Potential energy of spring $=\frac{1}{2} k x^{2}$

Here, $x=$ distance of block from mean position, $k=$ spring constant

At mean position, potential energy $=\frac{1}{2} k A^{2}$

At equilibrium position, half of the mass of block breaks off, so its potential energy becomes half.

Remaining energy $=\frac{1}{2}\left(\frac{1}{2} k A^{2}\right)=\frac{1}{2} k A^{\prime 2}$

Here, $A^{\prime}=$ New distance of block from mean position

$\Rightarrow A^{\prime}=\frac{A}{\sqrt{2}}$  