Question

A sound wave has frequency 500 Hz and velocity 360 m/s , what is the distance between two particles having phase difference of ${60}^{\circ }$?

• A. 70 cm
• B. 120.0 cm
• C. 0.7 cm
• D. 12 cm

Answer 1

The correct option is D 12 cm

Given:

Frequency of wave, f= 500 Hz

Velocity of the wave, v = 360m/sec

Phase difference, $\Delta \phi =\frac{\pi }{3}$

Distance between the two particles is given by

$\Delta x=\frac{\lambda }{2\pi }\times \Delta \phi$

$\Rightarrow \Delta x=\frac{\left(v/f\right)}{2\pi }\times \Delta \phi$

Substituting the value in the above equation

$\Delta x=\frac{\left(360/500\right)}{2\pi }\times \frac{\pi }{3}$

$\Rightarrow \Delta x=\frac{0.72}{6}=0.12m=12cm$

Hence, the distance is 12 cm

Correct answer (D)