The motors that drive airplane propellers are, in some cases, tun...

Question

The motors that drive airplane propellers are, in some cases, tuned by using beats. The whirring motor produces a sound wave having the same frequency as the propeller. (a) If one single-bladed propeller is turning at 575 rpm and you hear 2.0-Hz beats when you run the second propeller, what are the two possible frequencies (in rpm) of the second propeller? (b) Suppose you increase the speed of the second propeller slightly and find that the beat frequency changes to 2.1 Hz. In part (a), which of the two answers was the correct one for the frequency of the second single-bladed propeller? How do you know?

Accepted Answer

Hey everyone in this problem we have a childhood game that involves tying light gauge polythene papers together and attaching a small mass on one end of the polythene paper string. Okay. The mass is going to be swung in a circle using that paper string emitting sound. Okay. So we're going to suppose that a child rotating a mass at RPMs When an observer here is a 0.5 hertz frequency due to the presence of another child rotating a second mass. Alright. We're asked to determine two possible rotational frequencies in rpm for the second mass. Okay. And then we're told that for part two the rotational speed of the second mass is decreased. Then the beat frequency decreases to 0.4 hertz. Okay. So we're asked which of the values in part one is the true rotational frequency of the second mass. Alright. So we're working with beat frequencies here. Okay. And let's recall that the beat frequency F. B. Okay. It's going to be equal to either F. A minus F. B. Or it's going to be equal to F. B minus F. A. Okay. We're gonna take whichever one is bigger and subtract the other. Huh? So when we're talking about frequencies, we're just talking about the difference in the two frequencies that are present. Okay. Alright. Now we can write this kind of neatly as the absolute value of F. A minus F. B. Good. All right, so we know that the beat frequency that the observer hears is 0.5 hertz. Okay. So R. F. Beat. It's going to be 0.5 hertz. Okay. No let's let F. A. Be the frequency of the first mass. Okay that that's the one that we know. Okay. And FB is gonna be the frequency of the second mass. Okay. And we're trying to determine some possible values for that one. Okay. Alright so if we look at the frequency F. A. Okay well FAA has given an rpm. Okay but we have our FV. And hurt. So let's convert our rpm to hurts. Okay and so we're gonna have 163 R. P. M. Okay and to convert to hurts we divide by and then we're going to subtract the frequency F. B. Yes we have 0.5 hertz Is equal to. Okay when we do this division we get 2. repeated hurts minus the frequency F. B. Okay and you'll see because we have this absolute value, that's where those two different possible frequencies are going to come into play. Okay when we wrote before these two different cases, okay you're going to get a different possible frequency for each and we're going to get the same with this absolute value. Okay. Alright so if we want to remove the absolute value we can consider having either positive or negative 0.5 hertz. Okay if 2.7166 hertz is bigger than F. B. The right hand side will be positive and we get the positive on the left hand side. If 2.7166 hertz is less than FB the right hand side is going to be negative and we're going to have to multiply by a negative in order to make it positive for absolute value. Okay so that's where that negative 0.5 hertz is going to come in to play. Alright so then on the right hand side we have 2.7166 hertz. Okay repeated minus F. F. B. So we get that the possible frequencies for FB we're gonna be 2.7166 hertz Plus or -0.5 Hz. Alright so we work these out. Okay the first one we're gonna take the smaller ones we're gonna subtract we get 2.2166 repeated hurts. Okay and for the larger value we add .5 and we get 3.2166 Hz. So these are our two possible rotational frequencies for that second mass. Okay. And we'll see if we go back up here. We're asked to find these in rpm. Okay? We've given them in hurt. So let's just convert to rpm now to go from hurts to rpm. We're going to multiply by 60. So we get 2.2166 repeated time 60 hertz or sorry rpm. And the second solution is 3.2166 repeated Times 60. And again that's going to be an rpm now and this is going to give us Rpm or rpm. Okay, so those are two possible frequencies for that second mass. Alright, so let's go back up for a second. And if we look at this, we see that we have 133 and 193 rpm. Okay, So we're looking at either option B or D. Okay. And now for part two, we need to figure out which of these is actually the true rotational frequency. Okay, given the information that if the rotational speed of the second mass is decreased, then the beat frequency decreases. Right? Alright, so for part two, we're told that the rotational speed omega decreases. Okay, And this causes the beat frequency. It's a decrease. Okay. And I'm gonna put omega b here because it's omega related to that second mass. All right, well, we know information about the relationship between the frequency F B and F B. Okay, so let's try to relate this omega b to the frequency. Okay, and let's recall that omega is going to be equal to two pi times the frequency F. So if omega decreases than our frequency F also decreases this is F B. Okay, so now we know that FB decreases causes F beat to decrease. Now we have two options. Okay, we either have F B. Is equal to F A minus F B. Where f beat is equal to F B minus F. A. So which one are we looking at here? Well if FB decreases in this left hand equation. Okay that means we're gonna be subtracting a smaller number which means that F beat will increase. Okay, so that is not the situation that we have. So we're not looking at that equation. Okay. And the equation on the right hand side, if FB decreases then F beat will also decrease, which is what we want. And so this is the equation we're looking at. Okay, if that's the case then F B. Has to be greater then F. A. Okay, we only use this equation when FB is bigger. Okay, we always take the bigger frequency. Subtract the smaller frequency. Okay. And remember that F. A. was 100 And 63 rpm. 163 Rpm. And so that means that our FB Must be the higher value. 193 rpm. Okay. Alright. So if we go back up to our answer choices, we see that we found possible rotational frequencies of 133 rpm in 100 and 93 rpm. And we found in part two that the true rotational frequency was rpm. And so we have answered D. Here, Thanks everyone for watching. I hope this video helped see you in the next one

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