Question

The spring shown in figure is kept I a stretched position with extension $x_0$ when the system is released. Assuming the horizontal surface to be frictionless, find the frequency of oscillation.

• A. $\frac{1}{2\pi }\sqrt{\frac{kMm}{(M+m)}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{k(M+m)}{Mm}}$
• C. $2\pi \sqrt{\frac{kMm}{(M+m)}}$
• D. $2\pi \sqrt{\frac{k(M+m)}{Mm}}$

Answer 1

The correct option is B

$\frac{1}{2\pi }\sqrt{\frac{k(M+m)}{Mm}}$

No external force on the system. So center of mass at rest

If block m moves by x towards right and block M moves by X towards left then mx = MX

Also the spring is totally compressed by (x + X). Applying conservation of energy method.

$\frac{1}{2}k(x+X{)}^2+\frac{1}{2}m{v}_1^2+\frac{1}{2}M{v}_2^2=constant$

Taking derivative with respect to time on both sides

$-k(x+X)(v_1+v_2)+m{v}_1{a}_1+M{v}_2{a}_2=0$

$-Kx(1+\frac{m}{M})(v_1+v_2)=(m{v}_1{a}_1+M{a}_2{v}_2)$

$\therefore mx=Mx$

$\frac{dx}{dt}=M\frac{dx}{dt}$

$\Rightarrow m{v}_1=M{v}_2$

$\Rightarrow m{a}_1=M{a}_2$

$-kx(1+\frac{m}{M})=m{a}_1$

$\omega =m\sqrt{\frac{k(M+m)}{Mm}}$

frequency $f=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{k(M+m)}{Mm}}$