Question:
Two particles A and B execute S.H.M.s of periods T and \(\frac{5T}{2}\). If they start from the mean position, then the phase difference between them when the particle A completes two oscillations will be (in radian)
Two particles A and B execute S.H.M.s of periods T and \(\frac{5T}{2}\). If they start from the mean position, then the phase difference between them when the particle A completes two oscillations will be (in radian)
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Find the phase of each particle at the instant A finishes two oscillations, using phase equal to 2 pi times time divided by period, and subtract.
Updated On: Nov 12, 2026
- \(\frac{\pi }{5}\)
- \(\frac{3\pi }{5}\)
- \(\frac{6\pi }{5}\)
- \(\frac{12\pi }{5}\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
Both start at the mean position, so the phase after time \(t\) is \(\phi=\dfrac{2\pi}{T}t\).
Step 2: Phase of each
A completes two oscillations at \(t=2T\): \(\phi_{A}=\dfrac{2\pi}{T}(2T)=4\pi\).
For B with period \(\frac{5T}{2}\): \(\phi_{B}=\dfrac{2\pi\cdot2T}{5T/2}=\dfrac{8\pi}{5}\).
Step 3: Difference
\(\phi_{A}-\phi_{B}=4\pi-\dfrac{8\pi}{5}=\dfrac{12\pi}{5}\ \text{rad}\)
Final Answer:
The phase difference is \(\frac{12\pi}{5}\) radian. \[ \boxed{\text{(D) }\dfrac{12\pi}{5}} \]
Both start at the mean position, so the phase after time \(t\) is \(\phi=\dfrac{2\pi}{T}t\).
Step 2: Phase of each
A completes two oscillations at \(t=2T\): \(\phi_{A}=\dfrac{2\pi}{T}(2T)=4\pi\).
For B with period \(\frac{5T}{2}\): \(\phi_{B}=\dfrac{2\pi\cdot2T}{5T/2}=\dfrac{8\pi}{5}\).
Step 3: Difference
\(\phi_{A}-\phi_{B}=4\pi-\dfrac{8\pi}{5}=\dfrac{12\pi}{5}\ \text{rad}\)
Final Answer:
The phase difference is \(\frac{12\pi}{5}\) radian. \[ \boxed{\text{(D) }\dfrac{12\pi}{5}} \]
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