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. (a) Find the amplitudes of the simple harmonic motion of the block and of the car as seen from the road. (b) Find the time period (s) of the two simple harmonic motions.

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Solution

Let the amplitude of oscillation of 'm' and 'M' be $x_{1}$ and $x_{2}$ respectively.

(a) From law of conservation of momentum, $m x_{1} = M x_{2} . . . (1)$

[because only internal forces are present]

Again $\frac{1}{2} k x_{0}^{2} = \frac{1}{2} k (x_{1} + x_{2})^{2} . . . (2)$

[Block and mass oscillates in opposite directions. But $x p \to$ stretched part]. From equation (1) and equation (2),

$∴ x_{0} = x_{1} + \frac{m}{M} x_{1}$

$= (\frac{M + m}{M}) x$

So, $x_{2} = x_{0} - x_{1}$

$= x_{0} [1 - \frac{M}{M + m}]$

$= \frac{m x_{0}}{M + m}$ respectively

(b) 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

$\frac{1}{2} m v_{2}^{2} + \frac{1}{2} m (v_{1} - v_{2})^{2} + \frac{1}{2} k (x_{1} + x_{2})^{2} = c o n s t n a t . . . (3)$

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

Again from law of conservation of momentum,

$M x_{2} = m x_{1}$

$\Rightarrow x_{1} = (\frac{M}{m}) x_{2} . . . (4)$

$M v_{2} = m (v_{1} - v_{2})$

$\Rightarrow (v_{1} - v_{2}) = (\frac{M}{m}) v_{2} . . . (5)$

Putting the above values in equation (3), we get,

$\frac{1}{2} M v_{2}^{2} + \frac{1}{2} m \frac{M}{m^{2}} v_{2}^{2} + \frac{1}{2} k x_{2}^{2} (1 + \frac{M}{m}) =$ constant

$∴ M (1 + \frac{M}{m}) v_{2}^{2} + k (\frac{M}{m}) x_{2}^{2} =$ constant

$\Rightarrow M v_{2}^{2} + k (1 + \frac{M}{m}) x_{2}^{2} =$ constant

Taking derivative of both sides,

$M_{2} v_{2} \frac{d v_{2}}{d t} = k (\frac{M + m}{m}) 2 x_{2} \frac{d x_{2}}{d t} = 0$

$\Rightarrow m a_{2} + k (\frac{M + m}{m}) x_{2} = 0$

$[b e c a u s e, v_{2} = \frac{d x_{2}}{d t}]$

$\frac{a_{2}}{x_{2}} = \frac{- k M + m}{M m} ω^{2}$

$∴ ω = \frac{- k (M + m)}{M m}$

So, Time period,

$T = 2 π \sqrt{\frac{M m}{(M + m)}}$

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