MCQ
In stationary waves, distance between a node and its nearest antinode is $20 \mathrm{~cm}$. The phase difference between two particles having a separation of $60 \mathrm{~cm}$ will be
  • A
    Zero
  • B
    $\pi / 2$
  • C
    $\pi$
  • $3 \pi / 2$

Answer

Correct option: D.
$3 \pi / 2$
$\frac{\lambda}{4}=20 \Rightarrow \lambda=80 \mathrm{~cm}, \text { also } \Delta \phi=\frac{\lambda}{2 \pi} \cdot \Delta x $
$\Rightarrow \Delta \phi=\frac{60}{80} \times 2 \pi=\frac{3 \pi}{2}$

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