Animated Solution for Physics - Oscillations: Comprehension Passage

A Dance of Oscillators

Imagine two identical masses, connected by a massless spring, performing a perfectly synchronized dance on a frictionless floor. Their center of mass is stationary at $x_0$, and they oscillate with an amplitude $a$ and angular frequency $\omega$.

Before we dive into the collisions, we need to understand the spring itself. For a two-body system, the effective mass (or reduced mass) is $\mu =\frac{m\cdot m}{m+m}=\frac{m}{2}$. Using the standard relation $\omega =\sqrt{\frac{k}{\mu }}$, we can deduce that the spring constant is $k=\frac{m{\omega }^2}{2}$.

By differentiating their position equations, we find their velocities at any time $t$: $v_1(t)=a\omega \cos (\omega t)$ and $v_2(t)=-a\omega \cos (\omega t)$.

The High-Speed Collision

In the first scenario, a third mass (Particle 3) comes sliding in with speed $u_0=\frac{a\omega }{2}$ and strikes Particle 2 exactly at $t_0=0$.

Let's freeze time right before the impact. At $t=0$, $\cos (0)=1$. This means Particle 1 is moving right with $v_1=a\omega =2u_0$, and Particle 2 is moving left with $v_2=-a\omega =-2u_0$.

Now, the collision happens. Because Particle 3 and Particle 2 have the exact same mass and the collision is perfectly elastic, they simply exchange their velocities. It's like a perfect cue ball strike in billiards! Particle 3 bounces back with $-2u_0$, and Particle 2 is suddenly jolted to the right with $+u_0$. Particle 1, far away on the other end of the spring, is completely unaware and continues moving right at $2u_0$.

We need the new center of mass velocity of the 1-2 system. We just average their momentums:

Substituting $u_0=\frac{a\omega }{2}$, we get $v_{cm}=\frac{3}{4}a\omega =0.75a\omega$. Thus, the ratio is 0.75.

Striking at the Extremes

In the second scenario, the collision happens at $t_0=\frac{\pi }{2\omega }$.

At this exact moment, $\omega t_0=\frac{\pi }{2}$, making the cosine terms zero. Both oscillating masses are momentarily at rest at their extreme positions. The spring is stretched to its absolute maximum, with an extension of $2a$.

Particle 3 strikes Particle 2 with speed $u_0$. Again, they swap velocities. Particle 2 takes off with $u_0$, while Particle 1 remains at rest. The new center of mass velocity is $v_{cm}=\frac{u_0}{2}$.

Energy Conservation

To find the new amplitude $b$, we must use the conservation of mechanical energy. The total energy of the system right after the collision must equal the total energy when it reaches its new maximum amplitude.

Right after the collision, the energy consists of the spring's massive potential energy (stretched by $2a$) and Particle 2's kinetic energy:

When the system reaches its new amplitude $b$, the spring is stretched by $2b$. But wait! The entire system is now drifting with $v_{cm}$. We must include this translational kinetic energy:

Equating the two and substituting $v_{cm}=\frac{u_0}{2}$:

Rearranging to solve for $b^2$:

Now, we substitute our knowns: $k=\frac{m{\omega }^2}{2}$ and $u_0=\frac{a\omega }{2}$. The $m$ and ${\omega }^2$ terms beautifully cancel out, leaving us with:

Finally, we want the value of $\frac{4b^2}{a^2}$: