The distance between two points differing in phase by $60^\circ$ on a wave having wave velocity $360\text{…
The distance between two points differing in phase by $60^\circ$ on a wave having wave velocity $360\text{ ms}^{-1}$ and frequency $500\text{ Hz}$ is
- $0.36\text{ m}$
- $0.18\text{ m}$
- $0.48\text{ m}$
- $0.12\text{ m}$
Solution
Phase difference, $\Delta\phi = \frac{2\pi}{\lambda} \cdot \Delta x$
The distance between two points,
$\Delta x = \frac{(\Delta\phi)(\lambda)}{2\pi} = \frac{(\Delta\phi)(v / f)}{2\pi} = \frac{(\pi / 3)(360 / 500)}{2\pi}$
$= 0.12\text{ m}$
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