Q. A wave of frequency $500\ Hz$ has velocity $360\ m/sec.$ The distance between two nearest points $60^{o}$ out of phase, is
A) $0.6\ cm$
B) $12\ cm$
C) $60\ cm$
D) $120\ cm$
Explanation
(b) The distance between two points $i.e.$ path difference
Between them $\mathrm{\Delta} = \frac{\lambda}{2\pi} \times \phi = \frac{\lambda}{2\pi} \times \frac{\pi}{3} = \frac{\lambda}{6} = \frac{v}{6n}$
$(\because v = n\lambda) \Rightarrow \Delta = \frac{360}{6 \times 500} = 0.12m = 12\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.