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"Two pendulums with the same mass and length $R$ are released from rest. The first pendulum is released from an angle of $\theta_1 = -2º$, and the second pendulum is released from an angle $\theta_2 = +5º$. Calculate the angle at which they collide with each other."

Now the solution states that the frequency of oscillation (for small $\theta$) for each pendulum is $w = \sqrt{\frac{g}{R}}$ hence $T_1 = T_2$ and therefore they arrive at the same time at $\theta = 0º$. Therefore the collision happens at $\theta = 0º$.

I know how to come up with the frequency of oscillation $w$ but I do not quite understand the implications for the collision angle. Why is that when $T_1 = T_2$ that implies they meet necessarily at $\theta = 0º$? Is that a general rule? If $\theta_1$ started at $-30º$ instead, would it still be true that they meet at $0º$?

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  • $\begingroup$ The result relies on the "small angle" approximation - since the pendulums have the same length and mass, they take the same amount of time to go from the extreme to the center of the swing. This approximation gets progressively worse as the initial angle increases, though, so it shouldn't be applied for initial angles of more than about 20 degrees. A pendulum released from 30 degrees is a bit too far to use the approximation well, so it wouldn't have the same period as a pendulum of identical length/mass that was released from 2 degrees, and the collision would not be at 0 degrees. $\endgroup$ Feb 5 at 17:38
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This is due to the symmetry of the situation. We can divide a pendulum swing into four parts:

  1. Pendulum swings from the highest point on the left down to the bottom.
  2. Pendulum swings from the bottom up to the highest point on the right.
  3. Pendulum swings from the highest point on the right down to the bottom.
  4. Pendulum swings from the bottom up to the highest point on the left.

If you think about it, each of these steps takes equivalent time: $T/4$, where $T$ is the period of pendulum. Since $T$ is the same for all pendulums with length $R$ (in the small angle limit), then starting steps 1 and 3 for two pendulums (that are point masses, so they don't collide too early) at the same time will always cause them to conclude steps 1 and 3 at the same time, precisely at the point where they collide.

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Jonathan's solution is the most elegant I think. Here's a more mathematical one, assuming you know what the equation of motion is for the pendulum.

The angle $\theta(t)$ evolves according to the equation: \begin{equation} \ddot\theta(t)=-\omega^2\theta(t) \end{equation}

The solutions are given by: \begin{equation} \theta(t)=\theta_0\cos(\omega t) \end{equation} where $\theta_0$ is the initial position. To solve the problem, the angles of the two pendulums have to be the same and that can only happen if the cos function is zero, which implies that the angle must be zero. The time at which they collide is then: \begin{equation} t=\frac{\pi}{2\omega} \end{equation} which, as Jonathan said, is $T/4$.

Regarding your last question, if a pendulum started at $30°$, the approximation of small angles wouldn't be valid anymore and you wouldn't have a simple harmonic oscillator.

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