Question

All the surfaces shown in figure are frictionless. The mass of the car is M, that of the block is m and the spring has spring constant k. initially, the car and the block are at rest and the spring is stretched through a length $x_0$ when the system is released. Find the time period(s) of the two simple harmonic motions.

• A. $2\pi \sqrt{\frac{Mm}{k(M+m)}}$
• B. $2\pi \sqrt{\frac{k(M+m)}{Mm}}$
• C. $2\pi \sqrt{\frac{mM}{k(M+m)}}$
• D. None of these

Answer 1

The correct option is A

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

At any position, Let the velocities be $v_1$ and $v_2$ respectively.

Here, $v_1$= velocity of 'm' with respect to M.

By energy method

Total energy = Constant

$\left(\frac{1}{2}\right)M{v}^2+\left(\frac{1}{2}\right)m(v_1-v_2{)}^2+(\frac{1}{2}k(x_1+x_2{)}^2)=Constant...........\left(i\right)$

[$v_1-v_2=$ Absolute velocity of mass 'm' as seen from the road.]

Again, from law of conservation of momentum,

$m{x}_2=m{x}_1\Rightarrow x_1=\frac{M}{m}{x}_2$ .....(1)

$m{v}_2=m(v_1-v_2)\Rightarrow (v_1-v_2)=\frac{M}{m}{v}_2$ .....(2)

Putting the above values i equation (1), we get

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

$\therefore M(1+\frac{M}{m})v_2+k{(1+\frac{M}{m})}^2{x}_2^2$

$\Rightarrow m{v}_2^2+k(1+\frac{M}{m})x_2^2$=constant

Taking derivative of both sides,

$M\times 2v_2\frac{d{v}_2}{dt}+k\frac{(M+m)}{m}-e{x}_2^2\frac{d{x}_2}{dt}=0$

$\Rightarrow m{a}_2+k\left(\frac{M+m}{m}\right)x_2=0$ [because, $v_2=\frac{d{x}_2}{dt}$]

$\Rightarrow \frac{a_2}{x_2}=-\frac{k(M+m)}{Mm}={\omega }^2$

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

So, Time period, $T=2\pi \sqrt{\frac{Mm}{k(M+m)}}$