Q: A horizontal spring-block system of mass 2 kg executes SHM. When the block is passing through its equilibrium position, an object of mass 1kg is put on it and the two move together. The new amplitude of vibration is (A being its initial amplitude)

(a) $\displaystyle \sqrt{\frac{2}{3}} A$

(b) $ \displaystyle \sqrt{\frac{3}{2}} A $

(c) $ \displaystyle \sqrt{2} A$

(d) $ \displaystyle \frac{A}{\sqrt2} $

Ans: (a)

Sol: Applying momentum conservation

2v = (2+1)v’

=> v’ = 2v/3

$\large E_i = \frac{1}{2}\times 2 v^2 = v^2$

$\large E_f = \frac{1}{2}\times 3 v’^2 $

$\large E_f = \frac{1}{2}\times 3 (\frac{2v}{3})^2 = \frac{2}{3}v^2$

$\large E_f = \frac{2}{3} E_i $

$\large \frac{1}{2}k A’^2 = \frac{2}{3} \times \frac{1}{2}k A^2 $

$\large A’ = \sqrt{\frac{2}{3}} A$