# Find the range of velocities of the blocks shown in the figure below during the motion? How do we solve this problem without seeing from the center of mass frame?

Mar 24, 2018

Just take the reduced mass of the system,which will give you a single block with a spring attached to it.

Here the reduced mass is $\frac{2 \cdot 3}{2 + 3} = \frac{6}{5} K g$

So,angular frequency of the motion is, $\omega = \sqrt{\frac{K}{\mu}} = \sqrt{\frac{500}{6}} = 9.13 r a {\mathrm{ds}}^{-} 1$ (given, $K = 100 N {m}^{-} 1$)

Given,velocity in mean position is $3 m {s}^{-} 1$ and it is the maximum velocity of its motion.

So,range of velocity i.e amplitude of motion will be $A = \frac{v}{\omega}$

so,$A = \frac{3}{9.13} = 0.33 m$