A mass ‘ \mathrm{m}_{1} ‘ connected to a horizontal spring performs S.H.M. with amplitude ‘A‘. While mass ‘ \mathrm{m}_{1} ‘ is passing through mean position another mass ‘ \mathrm{m}_{2} ‘ is placed on it so that both the masses move together with amplitude \mathrm{A}_{1}, The ratio of \frac{\mathrm{A}_{1}}{\mathrm{~A}} is \left(\mathrm{m}_{2}<\mathrm{m}_{1}\right)
A) \left[\frac{m_{1}}{m_{1}+m_{2}}\right]^{\frac{1}{2}}
B) \left[\frac{\mathrm{m}_{1}+\mathrm{m}_{2}}{\mathrm{~m}_{1}}\right]^{\frac{1}{2}}
C) \left[\frac{m_{2}}{m_{1}+m_{2}}\right]^{\frac{1}{2}}
D) \left[\frac{\mathrm{m}_{1}+\mathrm{m}_{2}}{\mathrm{~m}_{2}}\right]^{\frac{1}{2}}
A mass ‘ \mathrm{m}_{1} ‘ connected to a horizontal spring performs S.H.M. with amplitude ‘A‘. While mass ‘ \mathrm{m}_{1} ‘ is passing through mean position another mass ‘ \mathrm{m}_{2} ‘ is placed on it so that both the masses move together with amplitude \mathrm{A}_{1}, The ratio of \frac{\mathrm{A}_{1}}{\mathrm{~A}} is \left(\mathrm{m}_{2}<\mathrm{m}_{1}\right)
A) \left[\frac{m_{1}}{m_{1}+m_{2}}\right]^{\frac{1}{2}}
B) \left[\frac{\mathrm{m}_{1}+\mathrm{m}_{2}}{\mathrm{~m}_{1}}\right]^{\frac{1}{2}}
C) \left[\frac{m_{2}}{m_{1}+m_{2}}\right]^{\frac{1}{2}}
D) \left[\frac{\mathrm{m}_{1}+\mathrm{m}_{2}}{\mathrm{~m}_{2}}\right]^{\frac{1}{2}}

Answer is (A)
P.E of the oscillating mass is given by
E=\frac{1}{2} m \omega x^{2}=\frac{1}{2} k x^{2} \quad \text { Where } k=m \omega^{2}
When only black is oscillating, at x=A
E=E_{\max }=\frac{1}{2} m \omega A^{2} and at mean position i.e. x=0
\therefore \quad A \propto \frac{1}{\sqrt{\mathrm{m}}}
E=0
When mass \mathrm{m}_{2} is placed on top of the mass \mathrm{m}_{1}
Then, total mass is \left(m_{1}+m_{2}\right) and E=0 at this point, as x=0.
When the combination reaches x=A,
Then A_{1} \propto \frac{1}{\sqrt{m_{1}+m_{2}}}
\therefore \frac{\mathrm{A}_{1}}{\mathrm{~A}}=\sqrt{\frac{\mathrm{m}_{1}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}}