
#1
Dec806, 06:04 PM

P: 32

Two loudspeakers emit sound waves of the same frequency along the xaxis. The amplitude of each wave is . The sound intensity is minimum when speaker 2 is 10 cm behind speaker 1. The intensity increases as speaker 2 is moved forward and first reaches maximum, with amplitude 2 , when it is 30 cm in front of speaker 1. What is the phase difference between the two speakers.
phase difference=2*pi*deltax/wavelength+initial phase difference I tried using this equation but I got the wrong answer. I said delta x was 30cm, wavelength is 80cm and initial phase difference is pi. Could anyone tell me where I am going wrong with this equation? 



#2
Dec806, 06:24 PM

Mentor
P: 39,597

I think you have the wavelength correct. How much of a wavelength is the initial offset of 10cm? And what is happening when they are 10cm apart? So the phase difference is not pi, it's pi with an offset of....




#3
Dec806, 07:16 PM

P: 32

I tried doing that but I must of misunderstood you because it still tells me Im wrong. I did 10cm/80cm * 2pi and this is still wrong.




#4
Dec806, 07:21 PM

Mentor
P: 39,597

Wave interference
Draw a diagram for yourself. When speaker 2 is 10cm behind speaker 1, you are acheiving what kind of interference? What does that say about the phase shift in the air between the two signals? So if speaker 2 was right next to speaker 1 and you had this kind of interference, what would the phase offset be? Now when you move speaker 2 10cm away from speaker 1, how much of a wavelength is that? So what do you have to offset the original phase difference by? Does it increase or decrease the original phase difference?
Just use the sketch and be careful about your signs. You'll get it. I gotta go for a while. Good luck! 


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