Question

If the collision occurs at time \( t_0 = \frac{\pi}{2\omega} \), then the value of \( \frac{4b^2}{a^2} \) will be:

Hint:

For oscillations with energy conservation, consider potential energy in the spring and kinetic energy of the system.

Answer 1

At \( t = \frac{\pi}{2\omega} \):
\( v_1 = 0 \), \( v_2 = 0 \),
the extension of the spring is: \[ x_1 - x_2 = 2d + 2a. \] The velocity of the center of mass is: \[ (V_{\text{cm}})(2m) = m \left( \frac{a\omega}{2} \right) \] \[ \Rightarrow V_{\text{cm}} = \frac{a\omega}{4}. \] Spring constant: \[ k = m_{\text{r}}\omega^2 = \frac{m}{2}\omega^2. \] Using energy conservation: \[ \frac{1}{2} m \left(\frac{a\omega}{2}\right)^2 + \frac{1}{2} k (2a)^2 = \frac{1}{2} (2m) \left(\frac{a\omega}{4}\right)^2 + \frac{1}{2} k (2b)^2. \] Simplify: \[ \frac{4b^2}{a^2} = 4.25. \]