MCQ
Two bodies performing $S.H.M.$ have same amplitude and frequency. Their phases at a certain instant are as shown in the figure. The phase difference between them is
- A$\frac {11}{6}\pi $
- ✓$\pi $
- C$\pi /3$
- D$\frac {3}{5}\pi $
✓
Answer
Correct option: B.
$\pi $
b
First particle is comming towards mean position
First particle is comming towards mean position
$\therefore \phi_{1}=\pi-\frac{\pi}{6}$
second particle is comming towards mean position but in negative side
$\therefore \phi_{2}=2 \pi-\frac{\pi}{6}$
$\therefore \phi_{2}-\phi_{1}=\left(2 \pi-\frac{\pi}{6}\right)-\left(\pi-\frac{\pi}{6}\right)=\pi$
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