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A bead of mass m slides down a frictionl...

A bead of mass m slides down a frictionless thin fixed wire held on the vertical plane and then performs small oscillations at the lowest point O of the wire. The wire takes the shape of a parabola near O and potential energy is given by U (x)=`cx^(2)`, where c is a constant. The period of the small oscillations will be :

A

`T =2pisqrt((m)/(2c))`

B

`T =pisqrt((m)/(c))`

C

`T =2pisqrt((2m)/(c))`

D

`T =2pisqrt((m)/(c))`

Text Solution

Verified by Experts

The correct Answer is:
A
Potential energy =`U(x)=cx^(2)`
Hence, force`F=-(dU(x))/(dx)=-2cx`
The equation of motion is `m(d^(2)x)/(dt^(2))=-2cx therefore omega^(2)=(2c)/(m) Rightarrow T=2pisqrt((m)/(2c))`
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