Question

A horizontal spring is connected to a mass M. It executes simple harmonic motion. When the mass M passes through its mean position, an object of mass m is put on it and the two move together. The ratio of frequencies before and after will be-

• A. ${\left(1+\frac{m}{M}\right)}^{1/2}$
• B. $\left(1+\frac{m}{M}\right)$
• C. ${\left(\frac{M}{M+m}\right)}^{1/2}$
• D. $\left(\frac{M}{M+m}\right)$

Answer 1

The correct option is A ${\left(1+\frac{m}{M}\right)}^{1/2}$

At mean position $F_{net}=0$

By conservation of linear momentum,

$m{V}_1=(M+m)V_{2}m{\omega }_1{A}_1=(M+m){\omega }_2{A}_2$

But ${\omega }_1=\sqrt{\frac{h}{m}}$ and ${\omega }_2=\sqrt{\frac{h}{M+m}}$

On solving $\frac{A_1}{A_2}=\sqrt{M+mM}$

At mean position $F_{net}=0$

By conservation of linear momentum

$M{V}_1=(M+m)V_{2}M{\omega }_1{A}_1=(M+m){\omega }_2{A}_2\Rightarrow \left(\frac{M}{M+m}\right)=\frac{{\omega }_2{A}_2}{{\omega }_1{A}_1}$ But ${\omega }_1=\sqrt{\frac{h}{m}}{\omega }_2\sqrt{\frac{h}{M+m}}$

$\frac{A_1}{A_2}=\sqrt{\frac{M+m}{M}}=(1+\frac{m}{M}{)}^{1/2}$