Q.Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of 2° to the right with the vertical, the other pendulum makes an angle of 1° to the left of the vertical. What is the phase difference between the pendulums?

The motion of a simple pendulum for small angles is an example of Simple Harmonic Motion (SHM).

Let the angular displacement of a pendulum be given by $\theta(t) = A \cos(\omega t + \phi)$, where $A$ is the amplitude and $\phi$ is the initial phase.

1. For the first pendulum, at a certain instant (let $t=0$), it is at its extreme position of $2^\circ$ to the right. So, $\theta_1(0) = +2^\circ$. Since the amplitude $A = 2^\circ$, we have $2^\circ = 2^\circ \cos(\phi_1)$, which implies $\cos(\phi_1) = 1$. Thus, we can take $\phi_1 = 0$.
2. At the same instant ($t=0$), the second pendulum is at $1^\circ$ to the left of the vertical. So, $\theta_2(0) = -1^\circ$. …

The motion of a simple pendulum, when its angular displacement from the vertical is small (typically less than about 10-15 degrees), can be accurately approximated as Simple Harmonic Motion (SHM). In SHM, the position of an oscillating object can be described by a sinusoidal function of time.

For two identical pendulums oscillating independently, their angular frequency ($\omega$) will be the same. If they also have equal amplitudes, the only difference in their motion will be their initial phase, which determines their relative positions and velocities at any given time. The problem asks for this phase difference.

We can represent the angular displacement $\theta(t)$ of a pendulum undergoing SHM as:

where $A$ is the amplitude, $\omega$ is the angular frequency, $t$ is time, and $\phi$ is the initial phase constant.

Let's break down the problem:

1. Define the SHM equations for each pendulum: Let $\theta_1(t)$ and $\theta_2(t)$ be the angular displacements of the two pendulums. Since they are identical and oscillate with equal amplitude, their amplitude $A$ and angular frequency $\omega$ are the same. We can write their equations as:

The phase difference between the pendulums is $|\phi_1 - \phi_2|$.

2. Apply the given conditions at a specific instant:

The problem states: "when one pendulum is at its extreme position making an angle of 2° to the right with the vertical, the other pendulum makes an angle of 1° to the left of the vertical."

Let's define "right" as positive displacement and "left" as negative displacement.

The amplitude $A$ is the maximum displacement, which is $2^\circ$. So, $A = 2^\circ$.

Let's consider the instant described as $t=0$ for convenience.
*   **For the first pendulum (P1):** It is at its extreme position, $2^\circ$ to the right.
    So, $\theta_1(0) = +2^\circ$.
    Substituting into its SHM equation:

    A simple choice for $\phi_1$ is $0$ radians. So, we can write $\theta_1(t) = A \cos(\omega t)$.

*   **For the second pendulum (P2):** At the same instant ($t=0$), it is at $1^\circ$ to the left.
    So, $\theta_2(0) = -1^\circ$.
    Substituting into its SHM equation:

3. Determine the possible phase angles for P2: We need to find $\phi_2$ such that $\cos(\phi_2) = -1/2$. …

Step 1: Model each pendulum's angular displacement as $\theta(t) = A\cos(\omega t + \phi)$, both with the same amplitude $A$ and frequency $\omega$ (identical pendulums).

Step 2: The first pendulum is at its extreme position ($2°$), so this value equals the amplitude: $A = 2°$. Taking this instant as the reference, $\cos\phi_1 = 1 \Rightarrow \phi_1 = 0$.

Step 3: At the same instant the second pendulum reads $\theta_2 = -1°$, so $2\cos\phi_2 = -1 \Rightarrow \cos\phi_2 = -\tfrac12 \Rightarrow \phi_2 = \dfrac{2\pi}{3}$ (taking the principal/smaller solution). …