A bead of mass m is sliding down the fixed inclined rod without friction. It is connected to a point P on the horizontal surface with a light spring of spring constant k. The bead is initially released from rest and the spring is initially unstressed and vertical. The bead just stops at the bottom of the inclined rod. Find the angle which the inclined rod makes with horizontal.

\[{{\cot }^{-1}}\left( 1+\sqrt{\frac{2mg}{kh}} \right)\]

[tan^-1left( 1+sqrtfrac2mgkh right)]

[cot ^-1left( 1+sqrtfracmgkh right)]

[tan^-1left( 1+sqrtfracmgkh right)]

A mass M is suspended from a light spring. An additional mass m added displaces the spring further by a distance x. Now the combined mass will oscillate on the spring with period [CPMT 1989 1998 ; UPSEAT 2000]

\[T=2\pi \sqrt{\left( (M+m)x/mg \right)}\]

[T=2pi sqrtleft( mgx(M+m) right)]

[T=(pi 2)sqrtleft( mgx(M+m) right)]

[T=2pi sqrtleft( (M+m)mgx right)]

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