Wave interference, "beats" problem

[Brian_D]

Homework Statement

You're in an airplane whose two engines are running at 560 rpm and 570 rpm. How often do you hear the sound intensity increased as a result of wave interference?

Relevant Equations

$f\! \left(t\right)=A \cos\! \left(\omega_{1} t\right)+A \cos\! \left(\omega_{2} t\right)$

$f(t) = 2Acos((\omega_1 - \omega_2)/2t) + cos((\omega_1+\omega_2)/(2)t)$

The first equation represents the superposition of two waves, each representing one of the engines. The second equation is a transformation of the first equation using the "cos alpha + cos beta" trigonometric identity. I would like to know how to solve this problem analytically, but don't know how to set it up. In the meantime, I attempted to graph the function and see if I could find the period by visual inspection. Inserting the given values for omega1 and omega2 into the second equation, we get f(t)=cos(5t)*cos(565t). Putting this equation into my TI-84 Plus graphing calculator and using a window for t of zero to pi (in radians), I got the display that I have attached. This appears to be a complete cycle of the function, so if the period is pi, the answer should be that you hear a beat every pi seconds. But the book answer key says a beat every 6 seconds. Also, when I graph the same function in Maple, it looks completely different from what the calculator shows. I would appreciate help solving this problem analytically (algebraically) as well as advice about why I'm getting a different answer than the book answer key as well as feedback about whether function looks like the one I attached when you graph it on your calculator.

 

[Brian_D]

For some reason, my second equation got cut off. The last part is supposed to read "omega1 + omega2 divided by two times t."

 

[berkeman]

For some reason, my second equation got cut off. The last part is supposed to read "omega1 + omega2 divided by two times t."

Yeah, that is weird. It may be a limitation on the length of equations in the Relevant Equations section of a post (which is weird considering it's a place to put equations). I'll try to edit the equations to shorten them up. BTW, there is a Greek symbol for Omega available in LaTeX, so you don't need to spell that out.

 

[berkeman]

I tried to fix up your post, but please check it and let me know if there are additional fixes needed. BTW, you can use \frac to post fractions in LaTeX. See the "LaTeX Guide" link below the Edit window.

 

[TSny]

Homework Statement: You're in an airplane whose two engines are running at 560 rpm and 570 rpm. How often do you hear the sound intensity increased as a result of wave interference?
Relevant Equations: $f\! \left(t\right)=A \cos\! \left(\omega_{1} t\right)+A \cos\! \left(\omega_{2} t\right)$

$f(t) = 2Acos((\omega_1 - \omega_2)/2t) + cos((\omega_1+\omega_2)/(2)t)$

Keep in mind that $\omega$ in your formulas represents angular frequency and needs to be in radians per unit time (not revolutions per unit time).

 

[haruspex]

560 rpm

Which is not 560 radians per second.

This appears to be a complete cycle of the function, so if the period is pi, the answer should be that you hear a beat every pi seconds.

You hear a beat at every peak amplitude. How many of those are there in a cycle?

 

[Brian_D]

Yeah, that is weird. It may be a limitation on the length of equations in the Relevant Equations section of a post (which is weird considering it's a place to put equations). I'll try to edit the equations to shorten them up. BTW, there is a Greek symbol for Omega available in LaTeX, so you don't need to spell that out.

I tried to fix up your post, but please check it and let me know if there are additional fixes needed. BTW, you can use \frac to post fractions in LaTeX. See the "LaTeX Guide" link below the Edit window.

Thank you, Berkeman. I did use the Greek symbol for Omega, but I was getting my LaTex code by entering the expression into Maple and then saving it as LaTex code using Maple; for some reason, Maple saved it as the English word. I'm trying to focus on the physics and I'm juggling a lot of other balls, so was letting Maple produce the LaTex code. But this is not working very well, so I'm going to have to study LaTeX.

In the meantime, thanks for the edit. However, I wrote the second equation incorrectly, and now I don't see how to fix it. The right side of the equation is supposed to be a product and not a sum and the "t" is supposed to be in the numerator, not the denominator.

 

[Brian_D]

Which is not 560 radians per second.

You hear a beat at every peak amplitude. How many of those are there in a cycle?

Thank you, TSny and haruspex. Very helpful. So I converted revolutions per second to radians per second. Then Omega1=.525 rad/s and Omega2=59.165 rad/s. Putting the function with these values into my graphing calculator (with an x window of 0 to 30 seconds), there are peaks at at 0 seconds and 30 seconds, so the period should be 30 seconds.

Now, to answer haruspex's question, if we count both peaks as part of the cycle, then the beats would occur every 15 seconds. But it doesn't make sense to count both beats as part of the cycle, because we had to wait 30 seconds to hear the second beat, not 15 seconds. So my answer is that we hear a beat every 30 seconds, while the book answer key says every 6 seconds.

 

[haruspex]

using a window for t of zero to pi (in radians),

Why? For a complete cycle, you want $\cos(((\omega_1 - \omega_2)/2)t)$ to go from 1 through -1 and back to 1.
But graphing it is overkill. Just solve $((\omega_1 - \omega_2)/2)t=2\pi$.

it doesn't make sense to count both beats as part of the cycle, because we had to wait 30 seconds to hear the second beat, not 15 seconds

No, you will hear peak amplitude both when $\cos(((\omega_1 - \omega_2)/2)t)=1$ and when $\cos(((\omega_1 - \omega_2)/2)t)=-1$. That's twice per cycle.
Remember that this acts as a multiplier on the much faster $\cos(((\omega_1 +\omega_2)/2)t)$ term, which oscillates between -1 and +1 many times both while $\cos(((\omega_1 - \omega_2)/2)t)$ is close to +1 and when it is close to -1. As a result, the listener cannot tell the difference between those two periods.

 

[Brian_D]

Why? For a complete cycle, you want $\cos(((\omega_1 - \omega_2)/2)t)$ to go from 1 through -1 and back to 1.
But graphing it is overkill. Just solve $((\omega_1 - \omega_2)/2)t=2\pi$.


No, you will hear peak amplitude both when $\cos(((\omega_1 - \omega_2)/2)t)=1$ and when $\cos(((\omega_1 - \omega_2)/2)t)=-1$. That's twice per cycle.
Remember that this acts as a multiplier on the much faster $\cos(((\omega_1 +\omega_2)/2)t)$ term, which oscillates between -1 and +1 many times both while $\cos(((\omega_1 - \omega_2)/2)t)$ is close to +1 and when it is close to -1. As a result, the listener cannot tell the difference between those two periods.

OK, except that the cos(.525t) part of the function equals 1 or -1 fives times in 30 seconds, not twice. This is consistent with the book's answer because 30 seconds divided by 5 equals 6 seconds. However, I am confused by the graph. If I display the whole function and the cos(.525t) part on the same graph, the extrema only coincide twice (at t=0 and t=24) and appear to be only approximately correlated the other three times.

 

[Brian_D]

Also, haruspex said previously that you could find the period just by solving cos(.525t)=2pi. But that gives us 4.8 i, while the period as seen in a graphing calculator is 12 seconds. Why the discrepancy?

 

[haruspex]

If I display the whole function and the cos(.525t) part on the same graph, the extrema only coincide twice (at t=0 and t=24) and appear to be only approximately correlated the other three times.

The brain interprets the sound as a pure tone (the cos(average) part) with a varying amplitude (the cos(half difference) part). Coinciding peaks of the two are not relevant.

Also, haruspex said previously that you could find the period just by solving cos(.525t)=2pi.

That is not what I wrote. Take another look.

 

[olivermsun]

Also, haruspex said previously that you could find the period just by solving cos(.525t)=2pi. But that gives us 4.8 i, while the period as seen in a graphing calculator is 12 seconds. Why the discrepancy?

Look carefully at $\cos(t)$. What's the period?

Now what if I have $\cos(2t)$. Now what's the period?

 

[Brian_D]

Thank you, olivermsun. So increasing the coefficient of t reduces the period, and vice versa. But I still don't understand haruspex's suggestion for finding the period algebraically. That method gives us 4.8 i, while the graphing calculator shows a period of 12 by visual inspection.

 

[Brian_D]

The brain interprets the sound as a pure tone (the cos(average) part) with a varying amplitude (the cos(half difference) part). Coinciding peaks of the two are not relevant.

That is not what I wrote. Take another look.

Thank you. I understand the first part of your message; another mystery solved. :-) But as for the second part, you said, "For a complete cycle, you want to go from 1 through -1 and back to 1." For the cosine, isn't that the same as the period? Solving cos(.525t)=2pi algebraically gives us 4.8 i. If not the period, what does 4.8 i represent?

.

 

[olivermsun]

Thank you, olivermsun. So increasing the coefficient of t reduces the period, and vice versa. But I still don't understand haruspex's suggestion for finding the period algebraically. That method gives us 4.8 i, while the graphing calculator shows a period of 12 by visual inspection.

Be as specific as you can, and I think you will see what’s going on.

What exactly is the period of $\cos t$?

What do you know about the value of $\cos t$ at the beginning and end of a period?

What about $t$ at the beginning and end of a period?

 

[renormalize]

However, I am confused by the graph.

Perhaps the low-resolution of the calculator graph is confusing you? Here's a more detailed plot:

 

[haruspex]

Solving cos(.525t)=2pi algebraically gives us 4.8 i.

You are still misreading it. I wrote

Just solve $((\omega_1 - \omega_2)/2)t=2\pi$.

Not "solve $\cos(((\omega_1 - \omega_2)/2)t)=2\pi$".

 

[Brian_D]

OK, thank you.

 

[Brian_D]

Perhaps the low-resolution of the calculator graph is confusing you? Here's a more detailed plot:
View attachment 359573

Yes, the resolution is what threw me off. Thank you.