Solution to Question 15
1. System Analysis & Angular Frequencies
We are given a system with two blocks and two springs fixed between rigid supports. Let the origin $(x=0)$ be the midpoint between the supports. The supports are at a distance $2l$, so the walls are at $x = -l$ and $x = +l$. The relaxed length of each spring is $l$.
Initially, the blocks are held at a separation of $l$, centered at the origin.
- Left Block: Mass $2m$, attached to spring of constant $k$. Initial position $x_L(0) = -l/2$.
- Right Block: Mass $m$, attached to spring of constant $2k$. Initial position $x_R(0) = +l/2$.
We calculate the angular frequency ($\omega$) for each block independently: $$ \omega_L = \sqrt{\frac{k}{2m}} $$ $$ \omega_R = \sqrt{\frac{2k}{m}} = \sqrt{4 \cdot \frac{k}{2m}} = 2\sqrt{\frac{k}{2m}} $$
Let $\omega_L = \omega$. Then $\omega_R = 2\omega$.
2. Collision Analysis
Both blocks are released from rest. They perform simple harmonic motion (SHM) towards the equilibrium position ($x=0$) until they collide. The equations of motion are: $$ x_L(t) = -\frac{l}{2} \cos(\omega t) $$ $$ x_R(t) = \frac{l}{2} \cos(2\omega t) $$
They collide when their positions are equal ($x_L = x_R$): $$ -\frac{l}{2} \cos(\omega t) = \frac{l}{2} \cos(2\omega t) $$ $$ \cos(2\omega t) + \cos(\omega t) = 0 $$
Using the identity $\cos(2\theta) = 2\cos^2\theta – 1$: $$ 2\cos^2(\omega t) – 1 + \cos(\omega t) = 0 $$ $$ (2\cos(\omega t) – 1)(\cos(\omega t) + 1) = 0 $$
The feasible solution for the first collision ($t > 0$) corresponds to: $$ \cos(\omega t) = \frac{1}{2} \implies \omega t = \frac{\pi}{3} = 60^\circ $$
Position of collision ($x_c$): $$ x_c = -\frac{l}{2} \cos(60^\circ) = -\frac{l}{2} \cdot \frac{1}{2} = -\frac{l}{4} $$ The blocks collide at $x = -l/4$ (in the left half).
3. Velocities & Momentum Before Collision
We calculate the velocity of each block just before impact ($t = \frac{\pi}{3\omega}$).
Left Block ($2m$): $$ v_L = \frac{dx_L}{dt} = \frac{l}{2}\omega \sin(\omega t) $$ $$ v_L = \frac{l\omega}{2} \sin(60^\circ) = \frac{l\omega}{2} \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}l\omega \quad (\text{Rightwards}) $$
Right Block ($m$): $$ v_R = \frac{dx_R}{dt} = -\frac{l}{2}(2\omega) \sin(2\omega t) = -l\omega \sin(120^\circ) $$ $$ v_R = -l\omega \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2}l\omega \quad (\text{Leftwards}) $$
Total Momentum ($P_{sys}$): $$ P_{sys} = (m_L \cdot v_L) + (m_R \cdot v_R) $$ $$ P_{sys} = \left(2m \cdot \frac{\sqrt{3}}{4}l\omega\right) + \left(m \cdot \left(-\frac{\sqrt{3}}{2}l\omega\right)\right) $$ $$ P_{sys} = \frac{\sqrt{3}}{2}ml\omega – \frac{\sqrt{3}}{2}ml\omega = 0 $$
Since the collision is perfectly inelastic (they stick together) and the initial total momentum is zero, the combined mass will stop instantaneously upon collision. $$ V_{final} = 0 $$
4. Oscillation After Collision
After the collision, we have a single system oscillating:
- Total Mass: $M_{total} = 2m + m = 3m$
- Effective Spring Constant: Since the springs act in parallel on the combined mass (both resist displacement from equilibrium), $K_{eff} = k + 2k = 3k$.
- New Angular Frequency: $\Omega = \sqrt{\frac{K_{eff}}{M_{total}}} = \sqrt{\frac{3k}{3m}} = \sqrt{\frac{k}{m}}$
Initial Conditions for New Motion: At $t=0$ (post-collision), the system is at position $x = -l/4$ with velocity $V = 0$. Since the velocity is zero, this position represents the extreme point (amplitude) of the new oscillation. $$ \text{New Amplitude } A’ = |x_c| = \frac{l}{4} $$
The maximum speed in SHM is given by $V_{max} = A’ \Omega$. Substituting the values: $$ V_{max} = \left( \frac{l}{4} \right) \sqrt{\frac{k}{m}} $$