Question

Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when speaker 2 is at the origin and speaker 1 is at x = 0.50 m. If speaker 1 is slowly moved forward, the sound intensity decreases and then increases, reaching another maximum when speaker 1 is at x = 0.90 m.

a. What is the frequency of the sound? Assume vsound = 340 m/s.

b. What is the phase difference between the speakers?

Short answer

a) The frequency of the sound is 850 Hz

b) The difference between the speakers are$\mathrm{\pi }/2$

Step-by-step solution

Understanding the frequency of the sound 

The frequency of the sound is the number of occurrences of repertaing events per unit.

Understanding the frequency of the sound 

The two consecutive maxima are given with two consecutive positions in which the phase difference is 0. On the other hand, based on the provided information it is$∆\varnothing =2\mathrm{\pi }\frac{∆\mathrm{x}}{\mathrm{\lambda }}$

Thus, the distance between the consecutive maxima is similar to the wavelength, thus, the wavelength is, $\underline{\lambda =0.9-0.5=0.4m}$

Thus, the frequency is $f=\frac{v}{\lambda }=\frac{340}{0.4}=\boxed{850Hz}$

Since the maxima are at x=0.5 then the phase difference will be 0. The phase difference is coming from the position of the signal phase difference. In numerically,

$∆\varnothing =2\mathrm{\pi }\frac{∆\mathrm{x}}{\mathrm{\lambda }}+∆{\varnothing }_0=0$

Thus the difference is$∆\varnothing =-2\mathrm{\pi }\frac{0.5}{0.4}=-\frac{5\mathrm{\pi }}{2}=-\frac{\mathrm{\pi }}{2}$