Two blocks with masses m1=1 kg and m2=2 kg are connected by a spring constant k=24 N m- 1 and placed on a frictionless horizontal surface. The block m1 is imparted an initial velocity v0=12 cm s - 1 to the right, the value of amplitude of oscillation (in cm ) of the block with respect to other is

Question

Two blocks with masses m1=1 kg and m2=2 kg are connected by a spring constant k=24 N m- 1 and placed on a frictionless horizontal surface. The block m1 is imparted an initial velocity v0=12 cm s - 1 to the right, the value of amplitude of oscillation (in cm ) of the block with respect to other is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

The amplitude of oscillations will be the maximum compression in the spring. At the time of maximum compression velocities of both the blocks are equal say

v

, then using law of conservation of momentum,

m_{1} v_{0} = (m_{1} + m_{2}) v

or

1 \times 12 = (1 + 2) v or v = 4 c m s^{- 1}

Using law of conservation of energy, we have

\frac{1}{2} m_{1} v_{0}^{2} = \frac{1}{2} k x^{2} + \frac{1}{2} (m_{1} + m_{2}) v^{2}

Putting the value and solving we get

x = 2

cm