Question

Answer in units of RADIANS Two loudspeakers are placed on a wall 2.21 m apart. A listener stands directly in front of one of the speakers, 3.21 m from the wall. The speakers are being driven by a single oscillator at a frequency of 254 Hz. The speed of the sound is 343 m/s. What is the phase difference between the two waves when they reach the observer? Answer in units of rad.

Answer 1

Answer:

phase difference = 3.21 rad  

Explanation:

given data  

wall apart =  2.21 m    

directly in front = 3.21 m

frequency f = 254 Hz  

speed of the sound v = 343 m/s

solution

we get  here first wavelength of sound wave that is  

wavelength = 343 ÷ 254  

wavelength = 1.35 m  

so  

path difference is  

path difference =   $\sqrt{2.21^2+3.21^2} - 3.21$

path difference = 0.687  

path difference = 0.69 m  

and  

phase difference will be  

phase difference =   $\frac{2\pi }{\lambda } \times \text{path\ difference}$  ............1  

phase difference = $\frac{2\pi }{1.35} \times 0.69$

phase difference = 3.21 rad