Question

Q 17.4: Two speakers both emit sound of frequency 320 Hz, and are in phase. A receiver sits 2.3 m from one speaker, and 2.9 m from the other. What is the phase difference between the two sounds detected by the receiver

Answer 1

Answer:

0.56*λ = 200º

Explanation:

• In order to know the phase difference between the two sounds detected by the receiver, we need first to know the wavelength of the sound.
• Assuming that the sound wave is a plane wave, there exists a fixed relationship between the speed of sound, the frequency and the wavelength, as follows:

        $v =\lambda * f$

• Assuming v= 343 m/s, and f = 320 Hz, we can find λ, as follows:

        $\lambda = \frac{v}{f} = \frac{343 m/s}{320 (1/s)} = 1.08 m$

• In order to know the phase difference, we need to know the path difference between both sounds, in units of wavelength:
• d = 2.9 m - 2.3 m = 0.6 m
• So, we can the fraction of the wavelength represented by the distance d, as follows:

       $\Delta\lambda = \frac{d}{\lambda} =\frac{0.6m}{1.08m} =0.56$

• As a difference of 1 λ, means that both sounds arrive in phase each other, a difference of 0.56*λ, in degrees, is as follows:

        $\Delta\theta = 0.56*360deg = 200 deg$