Question

Time periods of pendulums \( A \) and \( B \) are \( T \) and \( \frac{5T}{2} \). If they start executing S.H.M. at the same time from the mean position, the phase difference between them after the bigger pendulum has completed one oscillation is

• A. \( \frac{\pi}{4} \)
• B. \( \frac{\pi}{2} \)
• C. \( \frac{\pi}{8} \)
• D. \( \frac{\pi}{16} \)
• E. \( \pi \)

Hint:

Phase difference = difference in angular frequencies × time.

Answer 1

Answer: (E)

Note: The OCR'd question seems to have a typo. A standard version of this problem would use T and 5T/4. Let's assume the time periods are T and 5T/4. The image actually shows 5T/2. Let's solve with that. Step 1: Understanding the Concept:
The phase of an oscillator describes its position in the cycle. The phase is given by \(\phi(t) = \omega t + \phi_0\). If they start from the mean position at \(t=0\), the initial phase \(\phi_0\) is zero for both. The phase difference at a later time `t` is the difference between their individual phases.
Step 2: Key Formula or Approach:
1. The phase of an oscillator is \(\phi(t) = \omega t = \frac{2\pi}{T_{period}} t\). 2. Identify the time periods: \(T_A = T\) and \(T_B = 5T/2\). The "bigger pendulum" is B, as it has a longer period. 3. The time `t` at which we need to find the phase difference is the time it takes for pendulum B to complete one oscillation, which is \(t = T_B = 5T/2\). 4. Calculate the phase of each pendulum at this time: \(\phi_A(t)\) and \(\phi_B(t)\). 5. The phase difference is \(\Delta\phi = |\phi_A(t) - \phi_B(t)|\).
Step 3: Detailed Explanation:
Given time periods: \(T_A = T\) and \(T_B = 5T/2\). The corresponding angular frequencies are: \[ \omega_A = \frac{2\pi}{T_A} = \frac{2\pi}{T} \] \[ \omega_B = \frac{2\pi}{T_B} = \frac{2\pi}{5T/2} = \frac{4\pi}{5T} \] We are interested in the time when the bigger pendulum (B) completes one oscillation. This time is \(t = T_B = 5T/2\). Now, calculate the phase of each pendulum at this time `t`: Phase of A: \[ \phi_A(t) = \omega_A t = \left(\frac{2\pi}{T}\right) \left(\frac{5T}{2}\right) = 5\pi \] Phase of B: \[ \phi_B(t) = \omega_B t = \left(\frac{4\pi}{5T}\right) \left(\frac{5T}{2}\right) = \frac{4\pi}{2} = 2\pi \] The phase difference is the difference between their phases: \[ \Delta\phi = |\phi_A - \phi_B| = |5\pi - 2\pi| = 3\pi \] Since phase is typically represented within a \(2\pi\) interval, a phase difference of \(3\pi\) is equivalent to a phase difference of \(\pi\) (because \(3\pi = 2\pi + \pi\)). A phase difference of \(\pi\) means they are in opposite phase. The options do not include \(3\pi\), but \(\pi\) is an option. This is the correct interpretation.
Step 4: Final Answer:
The phase difference is \(\pi\).