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One end of an ideal spring is fixed to a...

One end of an ideal spring is fixed to a wall at origin `O` and axis of spring is parallel to x-axis. A block of mass `m = 1kg` is attached to free end of the spring and it is performing SHM. Equation of position of the block in co-ordinate system shown in figure is `x = 10 + 3 sin (10t)`. Here, t is in second and `x` in `cm`. Another block of mass `M = 3kg`, moving towards the origin with velocity `30 cm//s` collides with the block performing SHM at `t = 0` and gets stuck to it. Calculate

(a) new amplitude of oscillations,
(b) neweqution for position of the combined body,
( c) loss of energy during collision. Neglect friction.

A

`20 (rad)/(s)`

B

`5(rad)/(s)`

C

`100(rad)/(s)`

D

`50(rad)/(s)`

Text Solution

Verified by Experts

The correct Answer is:
B
`omega=sqrt((K)/(m))`, `K=omega^2m=(10)^2=100(N)/(m)`
Angular frequency of oscillatino of combined body is
`omega'=sqrt((K)/(m+M))=sqrt((100)/(4))=5(rad)/(s)`.
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