Discussion Forum



1)

A mass m1 connected to a horizontal  spring performs SHM with amplitude A. While mass m1 is passing through mean position, another mass m2 is placed on it so that both the masses move together  with amplitude A1.The ratio 

$\frac{A_{1}}{A}is (m_{2}<m_{1})$


A) $\left[\frac{m_{1}}{m_{1}+m_{2}}\right]^{1/2}$

B) $\left[\frac{m_{1}+m_{2}}{m_{1}}\right]^{1/2}$

C) $\left[\frac{m_{2}}{m_{1}+m_{2}}\right]^{1/2}$

D) $\left[\frac{m_{1}+m_{2}}{m_{2}}\right]^{1/2}$

Answer:

Option A

Explanation:

 The potential energy of oscillating mass is given by

 PE= $\frac{1}{2}kx^{2}$

When  only one block is oscillating at x= A

Emax= $\frac{1}{2} kA^{2}$ and at  mean position x=0  , So E=0

  $\therefore$   $A\propto\frac{1}{\sqrt{m}}$ ..........(i)

 when mass m2 is placed on top of the mass m. Then , total mass is (m1+m2) and E =0 at this point , as x=0

When they reaches at x= A,

then  $A_{1}\propto\frac{1}{\sqrt{m_{1}+m_{2}}}$  ......(ii)

Dividing Eq.(ii)  by Eq.(i),

 $\frac{A_{1}}{A}=\left(\frac{m_{1}}{m_{1}+m_{2}}\right)^{1/2}$