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  • The distance between two points differing in phase by   on a wave having wave velocit

The distance between two points differing in phase by  60^{\circ} on a wave having wave velocity  360 \mathrm{~m} / \mathrm{s}^{-1} and frequency 500 Hz is 

 

Option: 1

0.36 m


Option: 2

0.18 m


Option: 3

0.43 m


Option: 4

0.12 m


Answers (1)

best_answer

Use, \,\,\Delta \phi=\frac{2 \pi}{\lambda} \cdot \Delta x \\

\therefore \Delta x=\frac{(\Delta \phi) \cdot \lambda}{2 \pi}=\frac{(\Delta \phi)(v / f)}{2 \pi} \\

\Delta x=\frac{(\pi / 3)(360 / 500)}{2 \pi} \\

\Delta x=0.12\, m

Posted by

Pankaj Sanodiya

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