Question

A block of mass $\mathrm{m}$ attached to a massless spring is performing oscillatory motion of amplitude$‘\mathrm{A}’$ 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$\mathrm{fA}$. The value of $\mathrm{f}$ is:

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

Answer 1

The correct option is D

$\sqrt{2}$

Step 1: Given data and diagram

Mass of a block$=\mathrm{m}$

Amplitude of an oscillatory motion$=\mathrm{A}$

When mass $\mathrm{m}$ breaks to $\frac{\mathrm{m}}{2}$ its amplitude becomes $\mathrm{fA}$

The diagram for the given situation is shown below:

Step 2: Finding the value of $\mathrm{f}$

In spring block system the mechanical energy remains conserved, hence the kinetic energy at mean position is equal to potential energy at extreme position.

The distance between the mean position and extreme position is called amplitude.

Let velocity at mean position$={\mathrm{v}}_{\mathrm{mean}}$

Therefore, $\frac{1}{2}\mathrm{m}{\left({\mathrm{v}}_{\mathrm{mean}}\right)}^2=\frac{1}{2}\mathrm{k}{\left(\mathrm{A}\right)}^2\left(1\right)$

Here, $\mathrm{k}$ is the spring constant

Now let velocity of mass $\frac{\mathrm{m}}{2}$ at mean position$=\mathrm{v}{\text{'}}_{\mathrm{mean}}$

Apply conservation of momentum we get,

$$\begin{gathered} {\mathrm{mv}}_{\mathrm{mean}}=\frac{\mathrm{m}}{2}{{\mathrm{v}}^{\text{'}}}_{\mathrm{mean}} \\ \Rightarrow {{\mathrm{v}}^{\text{'}}}_{\mathrm{mean}}=2{\mathrm{v}}_{\mathrm{mean}}\left(\mathrm{P}\right) \end{gathered}$$

Now,

$$\begin{gathered} \frac{1}{2}\times \frac{\mathrm{m}}{2}{\left({{\mathrm{v}}^{\text{'}}}_{\mathrm{mean}}\right)}^2=\frac{1}{2}\mathrm{k}{\left({\mathrm{A}}^{\text{'}}\right)}^2 \\ \Rightarrow \frac{\mathrm{m}}{2}{\left({{\mathrm{v}}^{\text{'}}}_{\mathrm{mean}}\right)}^2=\mathrm{k}{\left({\mathrm{A}}^{\text{'}}\right)}^2\left(2\right) \end{gathered}$$

Divide equation $\left(2\right)$ by $\left(1\right)$ we get,

$$\begin{gathered} \frac{1}{2}{\left(\frac{{{\mathrm{v}}^{\text{'}}}_{\mathrm{mean}}}{{\mathrm{v}}_{\mathrm{mean}}}\right)}^2={\left(\frac{{\mathrm{A}}^{\text{'}}}{\mathrm{A}}\right)}^2 \\ \Rightarrow \frac{1}{2}{\left(2\right)}^2={\left(\frac{{\mathrm{A}}^{\text{'}}}{\mathrm{A}}\right)}^2\because \mathrm{v}{\text{'}}_{\mathrm{mean}}=2{\mathrm{v}}_{\mathrm{mean}} \\ \Rightarrow \left(\frac{{\mathrm{A}}^{\text{'}}}{\mathrm{A}}\right)=\sqrt{2} \\ \Rightarrow {\mathrm{A}}^{\text{'}}=\sqrt{2}\mathrm{A} \\ \Rightarrow {\mathrm{A}}^{\text{'}}=\mathrm{fA} \end{gathered}$$

Comparing the equation, we get $\mathrm{f}=\sqrt{2}$

Hence, option(D) is the correct answer.