MCQ
In stationary waves, distance between a node and its nearest antinode is $20 cm$. The phase difference between two particles having a separation of $60 cm$ will be
- AZero
- B$\pi /2$
- C$\pi$
- ✓$3 \pi /2$
✓
Answer
Correct option: D.
$3 \pi /2$
d
(d) $\frac{\lambda }{4} = 20 \Rightarrow \lambda = \,80\,cm$,
(d) $\frac{\lambda }{4} = 20 \Rightarrow \lambda = \,80\,cm$,
also $\Delta \phi = \frac{\lambda }{{2\pi }}.\,\Delta x$
==> $\Delta \phi = $ $\frac{{60}}{{80}} \times 2\pi = \frac{{3\pi }}{2}$
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