The lowest note on a grand piano has a frequency of 27.5 Hz. The entire string is 2.00 m long and has a mass of 400 g. The vibrating section of the string is 1.90 m long. What tension is needed to tune this string properly?

Video transcript

Hey everyone in this problem, a student builds a tunable stringed musical instrument using a fishing line of mass 6g and length 60 cm. The fishing lied slides through a tuning peg that allows a student to adjust attention. The developed instrument was tested in a room where the speed of sound is 344 m/s. The student adjusts the tension so that when it vibrates in its second overtone it produces sound with a wavelength of 0.63 m. And were asked to do two things. First, were asked to determine the tension in the fishing line in order to vibrate in the second overtone. Okay. And then were asked to determine the frequency of the sound produced by the line in its fundamental mode of vibration. Alright, so let's start with part one. Okay, we're asked to find the tension. And let's recall that we have the following equation which involves tension. We have V. The speed is equal to the square root of the tension T divided by mu. The linear density. Alright, so this equation allows us to find T. Okay, let's go ahead and isolate for T. So if we square both sides, we get V squared is equal to tension T. Over the linear density mu Which gives us attention. T is equal to V squared times meal. Alright, so now we have an equation for our attention. Okay, and in order to solve attention, we need the speed. V. And the linear density mu. Okay, let's start with the linear density mu mu. The linear density is going to be equal to the mass divided by the length. Okay, recall the only of linear density. Its mass divided by length. Now, what is our mass? Our mass is six g And our length is 60 cm. Now we want this to be in our fundamental units. So we want this to be kilograms per meter. So if we multiply We have one kg per 1000 gramps. The unit of grams will divide and were left at the unit of kilogram in the numerator. And similarly for the 60 centimeters we have centimeters per one m. Okay, And when we multiply the unit of centimeter will cancel. And we're left with the unit of meter in the denominator. Okay, so we get six g times one kg times 100 centimeters divided by 60 centimeters times 1000 g times one m. This is going to give us 0. kilograms per meter. So, that's our linear density mu that we're gonna use in our attention equation. Now, for attention equation, we also need to figure out V. Okay, all right, And we want to figure out V We're talking about tension in the second overtone. So, we want to figure out V in the second overtone. Okay, now, when we're talking about the second overtone, Okay, this tells us that N is equal to three And the second overtone is equal to the third harmonic and is equal to three Alright, and how would we calculate speed? The speed V. We're gonna say the speed V. Of the string, It's going to be equal to the frequency F three Times λ three. The wave wave recall this relationship between the velocity of the frequency and the wavelength. And so we need to find f. three and λ three. Alright well f. three. The frequency is going to be equal to V. Over land three. Now we're given the speed of sound in the room is 344 m/s. So we get 344 meters per second Divided by λ three is going to be the wavelength of the sound. Okay this is the frequency of the sound. We get 0.63 m. We divide the two. This is gonna give us a frequency of .032 Hz. So we have F. Three. Now we need to find lambda three. And when we're talking about lambda three here we're talking about the wave length of the string. Not the wavelength of the sound. Okay so in this frequency calculation in the problem we have the wavelength of the sound is 0.63 m. But we want to calculate the wavelength of the string. Okay And so we're gonna have to do that using the properties of that string. Okay so let's recall that the wavelength λ three. And let me write it as lambda in It's gonna be equal to two L. Divided by. Alright, so in our case we want λ three. This is gonna be two times L the length of that string. And again we have 60 centimeters okay, 60 cm. And we want to convert this to meters. So we're gonna multiply 51 m. Her 100 cm. Okay? And then we're dividing by our end value of three. And this gives us a wave length of the string of 0.4 m. Okay? So we have the wavelength of the sound and the wave length of the string. All right, so now we have F. Three. We have lamb 23 and we can calculate R. V. String. Okay, the speed string is going to be the frequency 546.32 hertz Times 0.4 m. Okay. And recall that it hurts is one over second. So this gives us units of meters per second. Okay, we get the velocity or the speed is .412, m/s. Alright, so back to what we're trying to find. Okay, remember that we're trying to find the tension team and in order to do that we needed the speed V squared. Sorry, the speed V. Okay. And the linear density mu Hey, that's gonna find us the tension in the second overturn. So now we have V. And we have mu we can get back to this tension equation. And so our attention T. is going to be .412, m/s times the linear density we found above, which was 0.1 kg per meter. Okay, let me just do a dotted line here. So we don't get confused because it's getting a little squishy. Alright. And this is going to give us attention t Of approximately 477 newtons. Okay, So we've found our attention t. Okay. And if we look at our answer choices, we see at this point we've narrowed it down to either C. Or D. Because we have our attention of 477 newtons. Okay, So let's work out part two to get our final answer. So part two. And let's go up and remind ourselves of what part two is asking her two is asking us to find the frequency of the sound in the fundamental mode. Alright, so when we're talking about the fundamental mode, this means that we have N. Is equal to one. And when we're talking about the second overtone, that's equivalent to the third harmonic. We have N equals three. Now we're in the fundamental mode and we have N. Is equal to one. Now we're trying to find the frequency and we call that the frequency F one is going to be equal to the speed V. Divided by the Wavelength λ one. Now the speed we have we just found the speed And this is going to be 218. eight m/s. But what about the wavelength lambda one? Okay well recall. We just calculated the wavelength lambda three. The wavelength lambda N. Is equal to two L. Divided by N. Okay so lambda one is going to be two times the length just 60 centimeters. And again converting two m times one m per centimeters, all divided by N. Which is just one. This gives us a wavelength λ one of 1.2 m. So in our frequency calculation now we have the speed v. m/s divided by that wavelength 1.2 m. Which gives us a frequency of 182 hertz approximately. Okay. Alright so back up to our answer choices. We found that the tension in the fishing line and the second Overtone was 477 newtons. And the frequency of sound produced by the line in its fundamental mode was 182 hertz. So we have answer C. Thanks everyone for watching. I hope this video helped see you in the next one