Amplitude of Sound Waves from Two Sources at a Point

[e(ho0n3]

[SOLVED] Amplitude of Sound Waves from Two Sources at a Point

Problem. Two sources, A and B, emit sound waves, in phase, each of wavelength $\lambda$ and amplitude $D_M$. Consider a point P that is a distance $r_A$ from A and $r_B$ from B. Show that if $r_A$ and $r_B$ are nearly equal ($r_A - r_B \ll r_A$). then the amplitude varies approximately with position as

$$\frac{2D_M}{r_A} \, \cos \frac{\pi}{\lambda} (r_A - r_B)$$

Let D(x, t) be the function that describes the displacement of the sound waves at some time t and a distance x from the source. I figure that the displacement at point P must be $D(r_A, t) + D(r_B, t)$ right? One thing I'm noticing is that the expression for the amplitude given in the problem statement does not vary with time. What gives?

 

[e(ho0n3]

If I look at the situation when t = 0, I get that the displacement at P is

$$D_M \sin kr_A + D_M \sin kr_B = 2D_M \sin \frac{\pi}{\lambda} (r_A + r_B) \cos \frac{\pi}{\lambda} (r_A - r_B)$$

How in the world does the sine expression simplify to 1/r_A?

 

[andrevdh]

This problem describes the double slit interference pattern at a point near the central maximum. The signal might vary with time, but per definition the amplitude is the maximum (minimum) strength of the resulting signal.

 

[e(ho0n3]

Interesting. However, I'm still puzzled as why the sine term simplifies to 1/r_A.

 

[andrevdh]

I can get close :

The phase difference, $$\delta$$, between the two waves at the point P will be small under the stated conditions. Since the amplitude of the resultant at P is proportional to

$$\cos^2(\frac{\delta}{2})$$

which under the stated conditions reduces to

$$\cos(\frac{\delta}{2})$$

The phase difference at the point P is given by

$$\delta = k \Delta r$$

with the wave number $$k = \frac{2 \pi}{\lambda}$$. We therefore have that

$$\cos(\frac{\pi \Delta r}{\lambda})$$

for the above mentioned.

I am very suspicious about the $$\frac{1}{r}$$ term in the given solution, since the dimension is then incorrect.

 

[e(ho0n3]

Since the amplitude of the resultant at P is proportional to

$$\cos^2(\frac{\delta}{2})$$

How do you know that? Where did you get that expression from?

I am very suspicious about the $$\frac{1}{r}$$ term in the given solution, since the dimension is then incorrect.

You're right. Perhaps it is incorrect. I think we can safely ignore the 1/r_A term.

 

[e(ho0n3]

It just hit me. The problem wants the maximum displacement, i.e. the amplitude at point P. The displacement at P is given by:

$$2D_M \cos \frac{\pi}{\lambda} (r_A - r_B) \sin (\pi / \lambda (r_A + r_B) + \omega t)$$

and the maximum displacement or amplitude is just

$$2D_M \cos \frac{\pi}{\lambda} (r_A - r_B)$$

Duh!