Two blocks with masses $m_{1} = 1kg\ and\ m_{2} = 2$ kg are connected by a spring constant $k = 24\ Nm^{- 1}$ and placed on a frictionless horizontal surface. The block $m_{1}$ is imparted an initial velocity $v_{0} = 12\ cms^{\ - 1}$ to the right, the amplitude of oscillation is

Question

Dr. Abubacker · Accepted Answer

(b) The amplitude of oscillations will be the maximum compression is the spring. At the time of maximum compression velocities of both the blocks are equal say $v$, then using law of conservation of momentum,
$m_{1}v_{0} = \left( m_{1} + m_{2} ight)v$
or $1 	imes 12 = (1 + 2)v\ \ or\ \ v = 4\ cms^{- 1}$
Using law of conservation of energy, we have
$\frac{1}{2}m_{1}v_{0}^{2} = \frac{1}{2}kx^{2} + \frac{1}{2}\left( m_{1} + m_{2} ight)v^{2}$
Putting the value and solving we get $x = 2$ cm

Q. Two blocks with masses $m_{1} = 1kg\ and\ m_{2} = 2$ kg are connected by a spring constant $k = 24\ Nm^{- 1}$ and placed on a frictionless horizontal surface. The block $m_{1}$ is imparted an initial velocity $v_{0} = 12\ cms^{\ - 1}$ to the right, the amplitude of oscillation is

Explanation

Reviewed by Dr. Abubacker

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