Solving the Volume Increase: Finding L with Two Speakers

[nemzy]

I got this wrong on a midterm, i have NO IDea absolutely how to solve this type of problem. I asked my TA too and he has no idea

-An observer hears the 343 Hz sound produced by two speakers. One is 5m from him, the other is at a distance L the first and can be moved. Taking the speed of sound to be v=343 m/s, for what values of L will the observer hear an increase in volume? (assume the speaker can only move to the right of the fixed one and that the amplitude of the waves does not decrease with distance)

the right answer is 0, 1m, 2m, 3m...

anyone have any idea why?

thanks

 

[krab]

The speed of sound is 343 m/s, so the wavelength at 343 Hz is 1 m. So 2 in-phase speakers will reinforce the signal at distances of n times 1 m apart. If you still don't get it, draw yourself a picture. P.S. Get another TA.

 

[charng]

The reason why the values for L are 0, 1m, 2m, 3m... is because of the concept of constructive interference. When two sound waves with the same frequency and amplitude overlap, they can either add up or cancel each other out depending on their relative positions. In this case, the two speakers are producing the same frequency (343 Hz) and are assumed to have the same amplitude. When the distance between the speakers is 0, the sound waves will overlap perfectly and the observer will hear a louder sound due to constructive interference. When the distance between the speakers is 1m, the waves will still overlap but not perfectly, resulting in a slightly lower volume increase. As the distance increases, the waves will overlap less and less, resulting in a decrease in the volume increase. Therefore, the values for L where the observer will hear an increase in volume are when the distance between the speakers is equal to a multiple of the wavelength of the sound wave (which is 343/343 = 1m). This is why the values for L are 0, 1m, 2m, 3m... and so on. I hope this helps clarify the concept for you.