Question

Velocity of sound waves in air is \[330\;{\text{m/sec}}\]. For a particle sound wave in air, a path difference of 40 cm is equivalent to phase difference of \[1.6\pi \]. The frequency of this wave is?
A. 165 Hz
B. 150 Hz
C. 660 Hz
D. 330 Hz

Answer 1

Hint: A wave's frequency refers to how much the media particles vibrate as a wave is moving through the medium. For a wave, speed is the distance traveled in a given time span by a specific point on the wave (such as a crest). And while wave frequency refers to the amount of cycles per second that occur, wave velocity refers to the meters traveled by second.

Complete step-by-step answer:
Velocity of sound waves = \[330\;{\text{m/sec}}\]
\[\vartriangle \phi = K\vartriangle x\]
Here, \[\vartriangle \phi \] = Phase difference
\[K = \dfrac{{2\pi }}{\lambda }\]
and, \[\vartriangle x\] = Path difference
By putting the values given in the question to above equation we get,
\[1.6\pi = \dfrac{{2\pi }}{\lambda }\dfrac{{40}}{{100}}\]
\[\lambda = \dfrac{{2\pi }}{{1.6\pi }} \times \dfrac{{40 \times 10}}{{100}} = \dfrac{1}{2}m\]
Now,
\[V = \dfrac{f}{\lambda }\]
\[330 = f\dfrac{1}{2}\]
\[f = 660Hz\]
Therefore, the frequency of this wave is 660 Hz. Hence option C is the right answer.

Note: Sound velocity is a vector u whose magnitude is the velocity of sound u, and whose direction is natural to the constant phase surface. Sound speed is a property of the medium through which the sound travels and is therefore generally of greater interest than the velocity itself which depends on both u and the way the sound is produced.
The difference in phase is the difference of the two waves in phase angle. The difference in direction is the difference in the path the wave traverses. The relationship between difference in phase and difference in path is direct. They are proportional directly to each other.