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 :

Options

A. `T =2pisqrt((m)/(2c))`
B. `T =pisqrt((m)/(c))`
C. `T =2pisqrt((2m)/(c))`
D. `T =2pisqrt((m)/(c))`

Answer

A

Explanation

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))`