A small marble of mass m oscillates simple harmonically inside a watch glass whose radius of curvature is 2.5m. What is its period of motion. (Take `g= 10m//s^(2))`

Options

A. 3 s
B. 3.142 s
C. 3.55 s
D. 3.75 s

Answer

B

Explanation

The component `mg "cos" theta` is balance by the normal reaction `vec(N)`. The component `mg "sin" theta` gives the restoring force. It brings back the marble to the position O. For small `theta`, expressed in radian,
R.F. `=ma = - mg "sin" theta = - mg theta`
`ma = - mg(x)/(R ) therefore a = - (gx)/( R)`
`therefore` acceleration per unit displacemtn `=|(a)/(x)|=(g)/(R )`
`therefore T=(2pi)/(sqrt((a)/(x)))=2pi sqrt((R )/(g))`
`=2pixxsqrt((2.5)/(10)=2pixx(1)/(2)=pi=3.142s`