An unfingered guitar string is 0.68 m long and is tuned to pla...

Question

An unfingered guitar string is 0.68 m long and is tuned to play E above middle C (330 Hz). How far from the end of this string must a fret (and your finger, Fig. 16–8) be placed to play A above middle C (440 Hz)?

Accepted Answer

Hello, fellow physicists today, we're an assault the following practice from together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem. A cello string is typically a length of 0.70 m. A particular cello string is tuned to play the note A which is 220 Hertz when it is un fingered at what distance from the end of the string must one position their finger on the string in order to play the note C four which is 260 Hertz. So that is our angles. We're trying to figure out at what, where this musician has to position their finger in order to play the note C four that has a frequency of 260 Hertz. And that's our final answer. What is this distance value? That's our final answer we're ultimately trying to solve for what distance is this value? Awesome. So looking at our figure here, we have our cello here. So our first cello which is on the far left is un fingered. So it's just hanging out all by itself. And then our second figure or part of the figure to the right, our second cello, we have some length which is lowercase L is equal to question mark because we do not know the length or the distance where the musician has to position their hand in order to hit the C four or position their fingers to be exact to hit this note C four. So that what we're ultimately trying to solve for. So the pink little half of a hand that kind of looks like thing from Adam's family is just floating there indicating where the fingers will be at some length L on the cello itself. And that's what we're ultimately trying to solve for is where on this cello lengthwise, do they need to put their fingers in order to get the C four note? Awesome. Now that we have a visualization of what's going on in this particular problem, let's look at our multiple choice answers to see what our final answer might be. Let us note that it's L is equal to some value that's in units of meters. So all of our multiple choice answers are in units of meters. So let's read them off to see what our final answer might be. So A is 0.11 B is 0.32 C is 0.59 and D is 0.65. Awesome. So moving right along here. So first off let us recall and note that the speed of the string doesn't change if it is fretted. Now, we must recall and use the fundamental frequency equation which states that F is equal to V divided by two multiplied by capital L which to make, make sure we know what's, what, let's note that capital L is the length of the string that vibrates and V lowercase V is equal to the speed of the waves on the string. Let us also note that V is constant. Thus, we can write that F is directly. So F or frequency, so we can write that F or frequency is directly proportional to one divided by capital L or we can write that F multiplied by L is constant. So F multiplied by L is constant little arrow toward constant. So thus, we can recall and write that FC multiplied by LC is equal to fa multiplied by L A where the subscript C indicate in. So our subscript C indicates the note C four and of course, subscript capital A denotes our note A. So I've been saying our note C four. So that's what we're ultimately trying to solve for is at what distance does the musician need to position their fingers in order to play the C four note? Awesome. So now we need to take our equation and we need to rearrange it to isolate and solve for LC. So when we do that, using a little bit of algebra, we will determine that LC is equal to fa multiplied by L A all divided by FC, which now we could plug in our known variables and solve for LC. So we know that FA is equal to 220 Hertz. And we need to multiply this by L A which capital L A is equal to 0.70 meters. Because that's typically the length of a cello string, which is given to us in the prom itself divided by FC which FC is equal to 260 Hertz. And this is also given to us in the prom itself. So we will determine when we plug that into our calculator, that LC is equal to zero point 5 9 2 3 m. So now at this point, we need to solve for our final answer, which is the distance of the location where the musician places their finger in order to play the note C four. So lowercase L so lowercase L which is the distance that we're trying to solve for ultimately is equal to 0.70 m, which is the length of the cello stream minus our LC value, which is zero point 59. Because when you think about it, we have to take the length of the entire string minus the distance that we just determined together, which is 0.59 2 3 m. So when we plug that into our calculator, we will get 0.11 m will be round to two decimal places or to two significant figures. And that's it we've solved for our final answer. Hooray. So looking at our multiple choice answers, the correct answer has to be the letter A. Our length is equal to 0.11 m. Thank you so much for watching. Hopefully that helped and I can't wait to see in the next video. Bye.

An unfingered guitar string is 0.68 m long and is tuned to play E above middle C (330 Hz). How far from the end of this string must a fret (and your finger, Fig. 16–8) be placed to play A above middle C (440 Hz)?

Key Concepts

Fundamental Frequency and Harmonics

Frequency Ratio and Musical Intervals

String Length and Pitch Relationship

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