Question

Two particles 'A' and 'B' perform S.H.M. starting from mean position, with periodic time T and \(\frac{3T}{2}\) respectively. The phase difference between the particles when particle 'A' completes one oscillation will be

• A. \(\left(\frac{\pi}{4}\right)\)
• B. \(\left(\frac{\pi}{3}\right)\)
• C. \(\left(\frac{\pi}{2}\right)\)
• D. \(\left(\frac{2\pi}{3}\right)\)

Hint:

Find the time taken for A to complete one oscillation and use it to find the phase of each particle.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For a particle in S.H.M. starting from the mean position, the phase after time \( t \) is \( \phi = \omega t = \frac{2\pi}{T}\,t \).

Step 2: Phase of A
A completes one oscillation at \( t = T \), so \( \phi_A = \frac{2\pi}{T}\times T = 2\pi \).

Step 3: Phase of B
B has period \( \frac{3T}{2} \), so \( \phi_B = \frac{2\pi}{3T/2}\times T = \frac{4\pi}{3} \).

Step 4: Difference
\( \Delta\phi = 2\pi - \frac{4\pi}{3} = \frac{2\pi}{3} \), which is option (D).

Final Answer:
The phase difference is 2pi/3. \[ \boxed{\text{(D) }\left(\dfrac{2\pi}{3}\right)} \]