JEE Main · 2011 · 10 Apr 2011 · Offline · Elasticity & SHM

A mass M , attached to a horizontal spring, executes S.H.M. with amplitude

A_{1}

. When the mass M passes through its mean position then a smaller mass

m

is placed over it and both of them move together with amplitude

A_{2}

. The ratio of

(\frac{A _{1}}{A _{2}})

is :

Answer: (C)

Solution

At the mean position the whole energy is kinetic and the speed is

v = ω A

. When the extra mass is dropped on there, no impulsive spring force acts, so linear momentum is conserved,

M v_{1} = (M + m) v_{2}

, i.e.

M ω_{1} A_{1} = (M + m) ω_{2} A_{2}

. The spring constant is unchanged, so

ω_{1} = k / M

and

ω_{2} = k / (M + m)

. Substituting,

M k / M A_{1} = (M + m) k / (M + m) A_{2}

, which reduces to

M A_{1} = M + m A_{2}

. Hence

\frac{A _{1}}{A _{2}} = (\frac{M + m}{M})^{1/2}

JEE Main · 2011 · 10 Apr 2011 · Offline · Single correct option