Amplitude Ratio after Elastic Collision in Spring Mass System

To find the value of $\frac{4b^2}{a^2}$, we analyze the motion of the particles before, during, and after the collision.

1. State of the System just before Collision ($t = t_0^-$)

The positions of particles 1 and 2 at any time $t$ are given by: $x_1(t) = (x_0 + d) + a \sin(\omega t)$ $x_2(t) = (x_0 - d) - a \sin(\omega t)$

Differentiating with respect to $t$, the velocities of the particles are: $v_1(t) = \dot{x}_1(t) = a\omega \cos(\omega t)$ $v_2(t) = \dot{x}_2(t) = -a\omega \cos(\omega t)$

At time $t_0 = \frac{\pi}{2\omega}$, we have $\omega t_0 = \frac{\pi}{2}$. Therefore: $\sin(\omega t_0) = 1 \quad \text{and} \quad \cos(\omega t_0) = 0$

Thus, just before the collision ($t = t_0^-$):

• Positions: $x_1(t_0^-) = x_0 + d + a$ $x_2(t_0^-) = x_0 - d - a$
• Velocities: $v_1(t_0^-) = 0$ $v_2(t_0^-) = 0$

The elongation of the spring from its natural/equilibrium length $2d$ is: $\Delta x = x_1(t_0^-) - x_2(t_0^-) - 2d = 2a$

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2. Velocities immediately after Collision ($t = t_0^+$)

Particle 3 (mass $m$) approaches particle 2 with speed $u_0 = \frac{a\omega}{2}$ and undergoes a $1\text{D}$ instantaneous elastic collision with particle 2 (mass $m$).

Since both particles have equal mass $m$, they exchange their velocities during the elastic collision: $v_2(t_0^+) = u_0 = \frac{a\omega}{2}$ $v_1(t_0^+) = v_1(t_0^-) = 0$

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3. Center of Mass Frame and Internal Energy

The spring constant $k$ is related to the angular frequency $\omega$ for the relative oscillation of two masses connected by a spring: $\mu = \frac{m \cdot m}{m + m} = \frac{m}{2}$ $\omega = \sqrt{\frac{k}{\mu}} = \sqrt{\frac{2k}{m}} \implies m\omega^2 = 2k$

The total internal energy $E_{\text{int}}$ of oscillation of the two-particle system after the collision is the sum of the potential energy stored in the spring and the kinetic energy in the center of mass frame.

1. Potential Energy ($U$): $U = \frac{1}{2} k (\Delta x)^2 = \frac{1}{2} k (2a)^2 = 2 k a^2$

2. Kinetic Energy in CM Frame ($K_{\text{CM}}$): The velocity of the center of mass of the system of particles 1 and 2 is: $v_{\text{cm}} = \frac{v_1(t_0^+) + v_2(t_0^+)}{2} = \frac{0 + \frac{a\omega}{2}}{2} = \frac{a\omega}{4}$

   The relative velocities with respect to the center of mass are: $v_{1, \text{cm}} = 0 - \frac{a\omega}{4} = -\frac{a\omega}{4}$ $v_{2, \text{cm}} = \frac{a\omega}{2} - \frac{a\omega}{4} = \frac{a\omega}{4}$

   Thus, the kinetic energy in the CM frame is: $K_{\text{CM}} = \frac{1}{2} m \left(-\frac{a\omega}{4}\right)^2 + \frac{1}{2} m \left(\frac{a\omega}{4}\right)^2 = \frac{1}{16} m a^2 \omega^2$

Substituting $m\omega^2 = 2k$: $K_{\text{CM}} = \frac{1}{16} (2k) a^2 = \frac{1}{8} k a^2$

The total internal energy after the collision is: $E_{\text{int}} = U + K_{\text{CM}} = 2 k a^2 + \frac{1}{8} k a^2 = \frac{17}{8} k a^2$

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4. Determination of New Amplitude $b$

When particles 1 and 2 oscillate with a new amplitude $b$, the maximum relative displacement of each particle from the center of mass is $b$, making the amplitude of total relative stretch/compression equal to $2b$.

The total internal energy of the oscillating system in terms of amplitude $b$ is: $E_{\text{int}} = \frac{1}{2} k (2b)^2 = 2 k b^2$

Equating the internal energies: $2 k b^2 = \frac{17}{8} k a^2$

$b^2 = \frac{17}{16} a^2$

Multiplying both sides by 4: $\frac{4b^2}{a^2} = 4 \times \frac{17}{16} = \frac{17}{4} = 4.25$