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 $fA$. The value of $f$ is:

• A. $\frac{1}{\sqrt{2}}$
• B. 1
• C. $\frac{1}{2}$
• D. $\sqrt{2}$

Answer 1

The correct option is A $\frac{1}{\sqrt{2}}$

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

Here, $x$ = distance of block from mean position,
$k=\text{spring constant}$

At extreme position,

$P.E=\frac{1}{2}k{A}^2\text{and}K.E=0$

At equilibrium position,

$K.E=\frac{1}{2}k{A}^2\text{and}P.E=0$

When half of the mass of the block breaks off at equilibrium position, its kinetic energy becomes half.

$K.E^{\prime }=\frac{1}{2}\left(\frac{1}{2}k{A}^2\right)$

Now, at extreme position,

$P.E^{\prime }=\frac{1}{2}k{A}^{\prime 2}=\frac{1}{2}\left(\frac{1}{2}k{A}^2\right)$

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

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

$\Rightarrow f=\frac{1}{\sqrt{2}}$

Hence, option $\left(A\right)$ is correct.