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A small mass m attached to one end of a ...

A small mass m attached to one end of a spring with a negligible mass and an unstretched length L, executes vertical oscillations with angular frequency `omega_(0)`. When the mass is rotated with an angular speed `omega` by holding the other end of the spring at a fixed point, the mass moves uniformly in a circular path in a horizontal plane. Then the increase in length of the spring during this rotation is

A

`(omega^(2)L)/(omega_(0)^(2)-omega^(2))`

B

`(omega_(0)^(2)L)/(omega^(2)-omega_(0)^(2))`

C

`(omega^(2)L)/(omega_(0)^(2))`

D

`(omega_(0)^(2)L)/(omega^(2))`

Text Solution

Verified by Experts

The correct Answer is:
A

`kx sin theta=momega^(2)(L+x)sintheta`
`implies kx=momega^(2)(L+x)`
Since, `omega_(0)=sqrt(k/m)`
`k=momega_(0)^(2)`
`momega_(0)^(2)x=momega^(2)(L+x)`
`implies x=(omega^(2)L)/(omega_(0)^(2)-omega^(2))`.
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