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A block of mass m attached with a spring...

A block of mass m attached with a spring and held by a person such that the spring is in natural length `l_(0)`. Now the man releases the block, the ratio of maximum compression in the spring in the given situation , to that of the compression at equilibrium position.

A

`2 : 1`

B

`1 : 1`

C

`1 : 2`

D

`sqrt2 : 1`

Text Solution

Verified by Experts

The correct Answer is:
A
The loss of potential energy by the gravity = gain in potential energy in the spring for maximum compression

`mg x = (1)/(2) kx^(2) rArr x = (2mg)/(k)`
At equilibrium `Sigma f_(x) = 0`
`mg- kx_(0) =0`
`rArr x_(0) = (mg)/(k)`
So, required ratio `= 2 : 1`
Hence (A) is correct
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Knowledge Check

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