Question

Two identical blocks each of mass M are connected with works of spring constant K and placed on smooth surface as shown in figure. When the blocks are in contact the springs are in natural length. The collision between masses is elastic. The frequency of vibration on disturbing the masses symmetrically in the directions of arrows and releasing them is?

Answer 1

x = v √(m/2k)

Explanation:

The collision between C and A causes their velocities to be exchanged.

At the maximum compression, both the blocks move with the same velocity.

Momentum conservation between X and Y is

mv + m * 0 = (m+m)v'

So v' = v2

Energy conservation between X and Y is

1/2mv² =1/2(m+m)v'² + 12kx²

mv² = 2m(v/2)² + kx²

x = √(mv²/ 2k) = v √(m/2k) where x is the maximum compression.

Answer 2

The frequency of vibration is

$\frac{1}{\pi}\sqrt{\frac{K}{M}}$

Explanation:

• Since the collision between the masses is elastic

• Therefore, after collision the masses will exchange their velocities

• When the springs are compressed in the respective direction by x, either of the masses will come back to their original position, exchange the velocities which will be the same for both the blocks and go back to compress the spring again by a distance x

• Thus, we can see that the time period of the blocks are half of what would have been their natural time periods

• We know that the time period of an oscillating block of mass m attached with a spring constant k is given by $T=2\pi\sqrt{m/k}$

• Therefore, here the time period will be

$T=\frac{1}{2}\times 2\pi\sqrt{M/K}$

$\implies T=\pi\sqrt{M/K}$

Thus, the frequency of oscillation will be

$f=\frac{1}{T}$

$\implies f=\frac{1}{\pi}\sqrt{\frac{K}{M}}$

Hope this answer is helpful.

Know More:

Q: Two blocks each of mass m are connected to a spring of spring constant k. if both are given velocity v in opposite directions then the maximum elongation of spring is :

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