Question

A spring - block pendulum is shown in the figure. The system is hanging in equilibrium. A bullet of mass m/2 moving with a speed u hits the block from down and gets embedded as shown in the figure. Find the amplitude of oscillation now.

• A. $\sqrt{\frac{m{u}^2}{8k}+{\left(\frac{mg}{2k}\right)}^2}$
• B. $\sqrt{\frac{m{u}^2}{k}+{\left(\frac{mg}{k}\right)}^2}$
• C. $\sqrt{\frac{m{u}^2}{6k}+{\left(\frac{mg}{2k}\right)}^2}$
• D. $\sqrt{\frac{m{u}^2}{8k}+{\left(\frac{mg}{2k}\right)}^2}$

Answer 1

The correct option is C $\sqrt{\frac{m{u}^2}{6k}+{\left(\frac{mg}{2k}\right)}^2}$


In equilibrium position, $mg=k{x}_0$
$\Rightarrow x_0=\frac{mg}{k}$
When the bullet gets embedded, the new equilibrium position changes.
Let it is a distance x from initial equilibrium position.
$\therefore (m+\frac{m}{2})g=k(x_0+x)$
$\Rightarrow \frac{3}{2}mg=k(\frac{mg}{k}+x)$
$x=\frac{3}{2}\frac{mg}{k}-\frac{mg}{k}$
$\Rightarrow x=\frac{mg}{k}$
Applying conservation of linear momentum,
$\frac{m}{2}u+0=(m+\frac{m}{2})v$
$v=\frac{m}{2}\times \frac{2}{3m}u=\frac{u}{3}$
So, at t = 0, mass $\frac{3m}{2}$ is at a distance $\frac{mg}{2k}$ from its mean position and moving up with velocity $\frac{u}{3}$
$\omega =\sqrt{\frac{k}{\left(3m/2\right)}}=\sqrt{\frac{2k}{3m}}$
$v=\omega \sqrt{A^2-x^2}$
$\Rightarrow \frac{u}{3}=\sqrt{\frac{2k}{3m}[A^2-{\left(\frac{mg}{2k}\right)}^2]}$
$\frac{u^2}{9}=\frac{2k}{3m}[A^2-{\left(\frac{mg}{2k}\right)}^2]$
$A^2-{\left(\frac{mg}{2k}\right)}^2=\frac{m{u}^2}{6k}$
$\Rightarrow A=\sqrt{\frac{m{u}^2}{6k}+{\left(\frac{mg}{2k}\right)}^2}$