MCQ
In the standing wave shown, particles at the positions $A$ and $B$ have a phase difference of
- A$0$
- B$\frac{\pi}{2}$
- C$\frac{5 \pi}{6}$
- ✓$\pi$
✓
Answer
Correct option: D.
$\pi$
d
(d)
(d)
Let, $y_A=A \sin \omega t$
$y_A=A \sin \omega t+\phi$
To convert $\sin \theta$ to $-\sin \theta, \phi$ must be $\pi$. Hence phase difference between is $\pi$.
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