Question

All the surfaces shown in figure (12-E15) 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.

Answer 1

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

$$(1/2)Mv^{2}+(1/2)m(v_{1}-v_{2})^{2}+(1/2)k(x_{1}+x_{2})^{2}=$$ Constant ..(i)

[$$v_{1}-v_{2}=$$ Absolute velocity of mass $$m$$ as seen from the road.]

Again, from law of conservation of momentum

$$mx_{2}=mx_{1}\Rightarrow x_{1}=\dfrac{M}{m}x_{2}$$ $$....(1)$$

$$mv_{2}=m(v_{1}-v_{2})\Rightarrow (v_{1}-v_{2})=\dfrac{M}{m}v_{2}$$ $$.....(2)$$

Putting the above values in equation $$(1)$$, we get

$$\dfrac{1}{2}Mv_{2}^{2}+\dfrac{1}{2}m\dfrac{M^{2}}{m^{2}}v_{2}^{2}+\dfrac{1}{2}kx_{2}^{2}\left(1+\dfrac{M}{m}\right)^{2}=$$ constant

$$\therefore M\left(1+\dfrac{M}{m}\right)v_{2}+k\left(1+\dfrac{M}{m}\right)^{2}x_{2}^{2}=$$ Constant.

$$mv_{2}^{2}+l\left(1+\dfrac{M}{m}\right)x_{2}^{2}=$$ constant

Taking derivative of both sides,

$$M\times 2v_{2}\dfrac{dv_{2}}{dt}+k\dfrac{(M+m)}{m}--ex_{2}^{2}\dfrac{dx_{2}}{dt}=0$$

$$\Rightarrow ma_{2}+k\left(\dfrac{M+m}{m}\right)x_{2}=0$$ [because, $$c_{2}=\dfrac{dx_{2}}{dt}$$]

Taking derivative of both sides,

$$M\times 2v_{2}\dfrac{dv_{2}}{}+k\dfrac{(M+m)}{m}-ex_{2}^{2}\dfrac{dx_{2}}{dt}=0$$

$$\Rightarrow ma_{2}+k\left(\dfrac{M+m}{m}\right)x_{2}=0$$ [because, $$v_{2}=\dfrac{dx_{2}}{dt}$$]

$$\Rightarrow \dfrac{a_{2}}{x_{2}}=-\dfrac{k(M+m)}{Mm}=\omega^{2}$$

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

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