Form the centre of a large concave mirror of radius of curvature 3 sm , a small sperical steel ball is placed at a little distance on the mirror itself . The ball is then released to execute oscillatory motion on the mirror . The time period of motion of the ball is (Neglect friction and take g=10m/s^(2)

Form the centre of a large concave mirror of radius of curvature 3 sm

1.	Form the centre of a large concave mirror of radius of curvature 3 sm , a small sperical steel ball is placed at a little distance on the mirror itself . The ball is then released to execute oscillatory motion on the mirror . The time period of motion of the ball is (Neglect friction and take g=10m/s^(2)
Answer» 2.38s 3.06s 3.44s 4.22s Solution :The given situation is The ball is PLACED at A, a small distance O from the centre of the mirror of radius of curvature R ( = OC) `angleACO = theta`, MASS of the ball = m Resolving weight mg into TWO components, we get `mgsin theta` providing the restoring force for S.H.M. `therefore F = -mg sin theta ` ` = -mg theta` (`theta` is small so x/R is very small) where `x = OA = Rtheta` `rArr F = -mg(x)/(R)` Also, `F = -kx`, on COMPARING `therefore k = (mg)/(R )` SO, the time period is `T = 2pi sqrt((m)/(k)) = 2pisqrt((R)/(g))` `=2pi XX sqrt((3)/(10)) = 2pisqrt(0.3)` = 3.44 s