5.For the arrangement shown in Figure 2, let the path length r1 = 1.20 m and the path length r2 = 0.80 m.

a)What is the highest frequency within the audible range (20 – 20000 Hz) that will result in a minimum at the receiver ? (Use v = 340 m/s).

Given answer is :19, 975 Hz

I don't know how to do.Can someone show or guide me the solution.I get the answer 50,233 Hz which is wrong.The correct answer is 19,975.I already attached the image for Figure 2.

Remember that for a minimum to occur, the two waves (one from each path) must reach the receiver with a phase difference of 180 degrees (or any odd-integer multiple thereof, i.e. +/- 180 degrees, +/- 540 degrees, +/- 900 degrees, etc.).

As a further shortcut, you can skip the part where you have to calculate the phase if you simply consider that a 180 phase difference is equivalent to a difference of a half-wavelength in path length. So really all you need to do is find how many wavelengths long each path is as a function of frequency. Then just find the frequencies where the difference is some odd multiple of a half (i.e. 0.5, 1.5, 2.5, etc.).

So what's the length of the bottom path, r1, in wavelengths for frequency f?

How about the top path, r2?

Now what's the difference between the two?

Now solve for f when the difference is equal to 0.5.

Too low? Nowhere even close to 20,000 Hz? Try 1.5.

Still too low? Maybe something a lot higher like 50.5?

Too high? Way over 20,000 Hz? Maybe try something in between like 20.5?

Keeping adjusting your value until you find the highest frequency that's not quite over 20,000 Hz.

One method, bringing you to an approximate frequency directly:

The path difference of the two waves is 1.2 - 0.8 = 0.4 m

As jcaron said, you get a minima when the path difference is half of a wavelength, 1.5 of a wavelength, etc, that is when you have (0.5 + n) of a wavelength as path difference, that is (0.5+n)\lambda

Now, this equals 0.4.

(0.5 + n)\lambda = 0.4

Then, recall the formula v = f\lambda

Take f = 20000 Hz

340 = 20000\lambda

Substitute lambda from what you got earlier to get:

340 = 20000\(\frac{0.4}{0.5+n}\)

Solve for n, the number of wavelengths to get an idea of how many n there are.

You'll get n = 23.029

Now, use n = 23 (n needs to be an integer) [You might also want to try one unit above and below to confirm, that is, try n = 22 and n = 24]

340 = f\(\frac{0.4}{0.5+23}\)

f = 19 975 Hz

Thanks Master

You're welcome. :)

I do hope though that you understood everything. It was an interesting problem I never really encountered before. I only applied the formulae I knew, and added a math trick to make it faster (that is using the values that you have, here 20,000 Hz to get an idea of the value of n)

So I answer three such questions for this guy and I don't even get an simple acknowledgment. Apparently he wanted the answer handed to him on a silver platter. :-)

How about we hand him another similar problem, with different figures (and some new tweaks)? ;)

Thanks jcaron2 for your answers for the 3 questions.It really helps me.Thanks.By the way that day I couldn't say thanks bcoz I slept off.Sorry.By the way thank you very much.

LOL! You're very welcome. Sorry to give you a hard time. Hopefully the explanations made sense. :-)