# Wave Interference and wavelength

• satchmo05
In summary, the conversation discusses a problem involving sound interference in a tube with a straight portion and a half-circle. The goal is to find the smallest radius that results in an intensity minimum at the detector. The relevant equations to consider are I = P/(4pi*r^(2)), where I is intensity, P is power, and r is radius, and P = .5(4*pi^(2))*frequency^(2)*A^(2)*v, where A = 4*pi*r^(2) and v is the velocity of sound. After attempting to solve the problem using these equations, it is suggested to simplify the equation to r^(2) = .5*pi*velocity, but this solution is incorrect.

## Homework Statement

In Fig. 17-35, sound with a 15.6 cm wavelength travels rightward from a source and through a tube that consists of a straight portion and a half-circle. Part of the sound wave travels through the half-circle and then rejoins the rest of the wave, which goes directly through the straight portion. This rejoining results in interference. What is the smallest radius r (cm) that results in an intensity minimum at the detector? (Take the speed of sound to be 343 m/s.)

## Homework Equations

If it wants the minimum intensity, we have to minimize the following equation: I = P/(4pi*r^(2)), where I is intensity, P equals power, and r equals radius. But then again, I also think of derivative when I hear maximum/minimum values...

## The Attempt at a Solution

343/15.6 = 21.99 Hz = frequency
2pi * frequency = angular velocity = 138.15
k = wave number = 0.403

So, these are the values we know, but I do not know how to manipulate them into the formula above...maybe I'm using the wrong formula?! Please help! I appreciate all the help. Thanks a bunch!

Oh, another piece of info. that I know, (but after I enter in the problem, it tells me I'm wrong) is that P = .5(4*pi^(2))*frequency^(2)*A^(2)*v, and A = 4*pi*r^(2). They also say on this one website that I = f^(2)*A^(2). Therefore, the equation simplifies down to:

r^(2) = .5*pi*velocity --> r = 23.2cm = WRONG!