A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant 100 N / m A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, the amplitude of the motion is

Question

A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant 100 N / m A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together, the amplitude of the motion is (A) 1 cm (B) 2 cm (C) 3 cm (D) 5 cm

Tardigrade Admin · Accepted Answer

At mean position, the speed gets altered by conservation of momentum,

(1 k g) (v_{1}) = (1 + 3) k g (v_{2})

v_{2} = \frac{v _{1}}{4}

Now,

\frac{1}{2} (1 k g) v_{1}^{2} = \frac{1}{2} k A^{2} \dots (i)

\frac{1}{2} (4 k g) v_{2}^{2} = \frac{1}{2} K A^{2} \dots (ii)

Dividing (ii) by (i), we get

\frac{4 v _{2}^{2}}{v _{1}^{2}} = (\frac{A ^{'}}{A})^{2} \dots (ii)

\frac{2 v _{2}}{v _{1}} = \frac{A ^{'}}{A} \Rightarrow \frac{A ^{'}}{A} = 2 (\frac{1}{4}) = \frac{1}{2}

\Rightarrow A^{'} = \frac{A}{2} = \frac{10 c m}{2} = 5 c m