Question

A wave of frequency $500 Hz$ has a velocity of $360 m/s$. calculate the distance between the two points that are \[{{60}^{0}}\] out of phase.
A. $12 cm$
B. $18 cm$
C. $24 cm$
D. $32 cm$

Answer 1

Hint:A wave is nothing but a disturbance in the medium and there can be many ways of creating such disturbance. Frequency is the characteristic of the source and as the wave propagates through the medium, the frequency does not change. Also, with the change in the medium, the frequency remains the same.

Complete step by step answer:
Given the frequency, \[\nu =500Hz\], Velocity of the wave, \[s=360m/s\]
We know the relationship between the velocity, the frequency and the wavelength of a wave is given by: \[s=\nu \lambda \]
$s=\nu \lambda \\
\Rightarrow \lambda =\dfrac{s}{\nu } \\
\Rightarrow \lambda =\dfrac{360}{500} \\
\therefore \lambda =\dfrac{18}{25}m \\$
Also, the two nearest points are \[{{60}^{0}}\]\[=\dfrac{\pi }{3}\],out of phase, thus using the relationship between the path difference and the phase difference when waves interfere,
$\Delta x=\dfrac{\Delta \phi \times \lambda }{2\pi } \\
\Rightarrow \Delta x=\dfrac{\dfrac{\pi }{3}\times \dfrac{18}{25}}{2\pi } \\
\therefore \Delta x=0.12m $

That is 12 cm, so, the correct option is A.

Additional Information:
Frequency is the number of complete cycles per second for an oscillating body. It is the characteristic of the source and as the wave propagates through the medium it does not change. It is a constant. While other quantities like wavelength and velocity of the wave depends on the medium and changes.

Note: Here we have used the concept from the interference of light and in that we have used the relationship between the path difference which is usually measured in metres, and the phase difference which is measured in degree radians. That is the reason we had converted phase angle from degree to radians. We need to keep in mind that all the units are in standard SI units.