Question

A wave of frequency 500 Hz has a velocity of 360 m $s^{-1}$. Calculate the distance between two points that are ${60}^0$ out of phase.

• A. 12 cm
• B. 18 cm
• C. 24 cm
• D. 32 cm

Answer 1

The correct option is A 12 cm

frequency $\left(\nu \right)=500H_z$

velocity $\left(v\right)=360m/s$

$\therefore$ wavelength $=$ velocity/frequency

$=\frac{360}{500}=0.72m=72cm$

$\therefore \lambda =72cm$

${60}^{\circ }$ out of phase implies that the wavelength between two points is

$\frac{\lambda }{\frac{{360}^{\circ }}{{60}^{\circ }}}=\frac{\lambda }{6}$

$\therefore {360}^{\circ }=\lambda$

$1^{\circ }=\frac{\lambda }{360}$

${60}^{\circ }=\frac{\lambda \times 60}{360}=\frac{\lambda }{6}=\frac{72}{6}=12cm$

$\therefore$ The distance between point $A$ & $B$ is $12cm$ if they have a phase difference of ${60}^{\circ }$

$\therefore$ Option (A) is the correct answer.