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A body of mass m is released from a heig...

A body of mass `m` is released from a height h to a scale pan hung from a spring. The spring constant of the spring is `k`, the mass of the scale pan is negligible and the body does not bounce relative to the pan, then the amplitude of vibration is

A

`mg/k`

B

`mg/ksqrt((1+(2hk)/mg))`

C

`mg/k+mg/ksqrt((1+(2hk)/mg))`

D

`mg/ksqrt((1+2hk)/(mg) (mg)/k))`

Text Solution

Verified by Experts

The correct Answer is:
B
Let x be the maximum elongation of spring. From conservation of energy,
`(1)/(2)kx^(2)=mgx+(1)/(2)mV^(2)` And `V=sqrt(2gh)`
Solving `x=(mg)/(k)+(mg)/(k)sqrt(1+(2hk)/(mg))`
Let `x_(0)` be the elongation in equilibrium position, then `x_(0)=(mg)/(k)`
So `A=x-x_(0)=(mg)/(k)+(mg)/(k)sqrt(1+(2hk)/(mg))`
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