# Homework Help: Variable frequency sound waves

1. Feb 1, 2013

### Von Neumann

Problem:

The problem asks to find which frequencies a listener, whom is receiving sound waves in the same direction from two variable-frequency speakers 50ft and 55ft away, will not hear anything at all.

Solution (so far):

My approach is to find an expression for the wave coming from each speaker, add them, and find when the sum equals 0. Doing so I get,

y_1=y_m*sin(k(x-x_0)-wt) [for speaker #1]

y_2=y_m*sin(k(x+x_0)-wt) [for speaker #2]

(where x_0=2.5 ft., y_m is the amplitude of the waves, k is the wave number, x is the position of the observer, w is the angular frequency, and t is the time)

For the next part, I know that the sum is

y_1+y_2=2*y_m*cos(k*x_0)sin(kx-wt)

However I am unable to get this answer using the my expressions for y_1 & y_2. Any suggestions?

Last edited: Feb 1, 2013
2. Feb 1, 2013

### Von Neumann

Ah just an algebraic error! To finish the problem one must set cos(kx_0)=0.

So then kx_0=pi/2, 3pi/2, ... & so on.

Then,

f_1=(v/4)*(1/x_0)=113 Hz
f_2=3*f_1=339 Hz
f_3=5*f_1=565 Hz
etc...

3. Feb 2, 2013

4. Feb 2, 2013

### Von Neumann

Hey rude man,

Why must the expressions for y_1 and y_2 contain + x_0 and - x_0, respectively? I did so by following a procedure from a related problem. However the related problem was asking for the location between two speakers where the sound would be the loudest. So it intuitively makes sense that x_0 for y_1 and y_2 would contain opposite signs since they are traveling in opposite direction However, in this problem the sound is coming from one direction only; mustn't the sign be the same in this case? Obviously this logic is wrong since I obtained the correct answer, but I'm trying to develop a sound intuition (pun not intended!).

Now that I'm typing out my thought process I'm beginning to think that the significance of x_0 is to alter the wave number slightly for each wave (subtracting 2.5 from 55 and adding 2.5 to 50) so that situation can be viewed as a superposition of two waves of equal wavelength. Is this correct?

Last edited: Feb 2, 2013
5. Feb 2, 2013

### rude man

OK, Dr. Neumann , here's my reasoning which I think may be more fundamentally based than what you did:

y1 = cos(kx - ωt), therefore
-y1 = cos[kx - ωt + (2n + 1)π], n = 0, 1, ...
y2 = cos[k(x + x0) - ωt] where x0 = 5ft.

We arbitrarily set x = t = 0 WLOG ("Without Loss Of Generality"),
since y1 + y2 = 0 → -y1 = y2 we get
cos[(2n + 1)π] = cos(kx0) or
(2n + 1)π = kx0

But k = 2π/λ = 2πf/v so we finally get
(2n + 1) = 2fx0/v and f = (2n + 1)v/2x0.

6. Feb 2, 2013

### rude man

No, the wave number is k = 2pi/lambda and lambda = v/f so there is only one wave number per solution frequency f.

Your equations were OK but I did not applaud using x0/2 and -x0/2. I like to think of one wave as the fundamental wave equation cos(kx - wt) and then contrast that to the second, interfering wave which then becomes cos[k(x + x0) - wt]. Since we want destructive interference the first wave can be made negative by the simple expedient of adding (2n+1)pi to its argument. You could just as well have put the (2n+1)pi into y2 but then you would have wound up with negative n's which is OK mathematically of course. The main thing is to put odd integer multiples of pi into either y1 or y2 so that they subtract at the same x and t.

7. Feb 2, 2013

### Von Neumann

I do very much agree with this alternate solution. Thank you for your time! My professor went over this particular solution in class, and stressed heavily that it is key to use +/-x_0/2. However, I side with you regarding this treatment being much simpler, as well as more fundamental and intuitive.

Also, did you set x=t=0 from the fact that they end up canceling out anyway (much as in my solution)? Or, is there some other reason that permitted you to know it would not affect the end result?

Last edited: Feb 2, 2013
8. Feb 8, 2013

### xatu

Von Neumann,

Your approach is fine, do what is intuitive to you and what makes sense to you, not rude man.

9. Feb 8, 2013

### rude man

x and t were arbitrary so I made them zero. Remember we were interested in differences.