Question

A wave of frequency $500 \, Hz$ has a velocity of $350 \, m \, s^{- 1}$ . The distance between two nearest points on the wave which are $60^\circ $ out of phase with each other will be around

• A. $70 \, cm$
• B. $0.7 \, m$
• C. $12.0 \, cm$
• D. $120.0 \, cm$

Answer 1

Answer: (C)

Given, frequency $f=500Hz$ , velocity $v=350ms^{- 1}$ , phase $\phi = 6 0^{^\circ }$
By the formula, $\lambda =\frac{v}{f}=\frac{3 5 0}{5 0 0}=\text{0.7 m}$
Also we know that $\Delta \text{x} = \Delta \phi \frac{\lambda }{2 \pi }$
$= 6 0 \times \frac{\pi }{1 8 0} \times \frac{\text{0.7}}{2 \pi }$
$\approx \text{0.12 metre} \approx \text{12 cm}$