Question 17

Step 1: Find state of system at $t_0 = \frac{\pi}{2\omega}$. Velocities: $v_1 = a\omega \cos(\pi/2) = 0$ and $v_2 = -a\omega \cos(\pi/2) = 0$. Positions: $x_1 - x_2 = (x_0 + d + a \sin(\pi/2)) - (x_0 - d - a \sin(\pi/2)) = 2d + 2a$. The equilibrium separation is $2d$, so the displacement from equilibrium is $\delta = 2a$.

Step 2: Collision effects. Particle 3 hits particle 2 elastically. Velocity of particle 2 after collision is $v'_2 = u_0 = \frac{a\omega}{2}$. Velocity of particle 1 remains $v_1 = 0$.

Step 3: Energy calculation in CM frame. Relative velocity $v_{rel} = v_1 - v'_2 = -\frac{a\omega}{2}$. Reduced mass $\mu = \frac{m}{2}$ and spring constant $k = \frac{m\omega^2}{2}$ (since $\omega = \sqrt{k/\mu}$). Energy of oscillation $E = \frac{1}{2} \mu v_{rel}^2 + \frac{1}{2} k \delta^2$ $E = \frac{1}{2} \left(\frac{m}{2}\right) \left(-\frac{a\omega}{2}\right)^2 + \frac{1}{2} \left(\frac{m\omega^2}{2}\right) (2a)^2 = \frac{ma^2\omega^2}{16} + ma^2\omega^2 = \frac{17}{16} ma^2\omega^2$

Step 4: Relate energy to new amplitude $b$. The energy for amplitude $b$ of each mass (relative amplitude $2b$) is: $E = \frac{1}{2} k (2b)^2 = \frac{1}{2} \left(\frac{m\omega^2}{2}\right) (4b^2) = mb^2\omega^2$ Equating the energies: $mb^2\omega^2 = \frac{17}{16} ma^2\omega^2 \implies \frac{b^2}{a^2} = \frac{17}{16}$ $4 \frac{b^2}{a^2} = 4 \times \frac{17}{16} = \frac{17}{4} = 4.25$