A piano tuner stretches a steel piano wire with a tension of 800 ...

Question

A piano tuner stretches a steel piano wire with a tension of 800 N. The steel wire is 0.400 m long and has a mass of 3.00 g. (a) What is the frequency of its fundamental mode of vibration?

Accepted Answer

Hey everyone, welcome back in this problem. We have a 0.8 m strength. Okay? With a mass of 1.6 g, it's going to be attached to a tuning peg on one end and a wooden board on the other. It's turned clockwise and it's going to obtain a tension of 80 newtons in the strength. And we are asked to calculate the frequency of the fundamental mode of oscillation. Alright, so let's just write out some of the information. And so we're told the mass Of the string is 1. g. Okay, we're gonna go ahead and put this into kg. This is gonna be 0.0016 kg. We're told the length of the string is 0.8 m. And that the tension in the string which we'll call f T. K. The force to detention Is equal to Newtons. Okay? And we're asked to find the frequency of the fundamental mode case of the fundamental frequency. Now, let's recall, we have the following formula. We have B. The velocity is equal to the square root of ft. Okay, The force due to tension over mu and mu is a mass per unit length. Okay, So this allows us to relate the tension in the string to the motion of the string. Okay, so we have um you again, mass per unit length. So let's go ahead and find mew for the string that we have. So mu is going to be mass per unit length. M divided by l In this case we have 0.0016 kg divided by 0.8 m. And we get a mu value of 0.002 kilograms per meter. Okay. Alright, great. So we have mu we have F. T. Okay, so we can figure out the right side of this. On the left hand side that we have the velocity. We're not looking for the velocity. We're looking for the frequency of that fundamental mode. Okay, well, recall that velocity can be written as lambda. The wavelength times F. The frequency. Okay, So we can find that frequency using this ft over meal. Okay. Alright, so again, we're looking for F. We know F. T. We know mu were not given any information about the wavelength. Okay, this is where we have to think about all of the information. They're telling us. They're telling us we're looking for the fundamental frequency. We know the fundamental frequency will have a wavelength that's two times the length of the string. Okay, So we know what the wavelength will be based on the fact that we are looking for the fundamental frequency. Okay, so we know that the wavelength lambda is going to be two times the length of the string when we're at the fundamental frequency. Because we now have two L times the frequency F. Is going to equal the square root of the tension divided by new. Okay, so this here we can use to find this fundamental frequency F. We know L. We know F T. We know mu and we can just solve for F. Okay, so let's use this equation. We're just gonna move up here so we can still see all of the values we have. Okay, substituting in what we have, we have to Time 0.8 m times F is going to equal the square root Of 80 Newtons divided by 0.002. Okay. And let's remember that we can write newton as kilogram meter per second squared. Okay. And our unit for 0.0 to the mass per unit area is kilogram per meter. Okay, So on the next line when we simplify this, we're going to assume that the newton is written as kilogram meter per second squared. That's going to allow us to simplify our units as well. Okay, so on the left hand side we're gonna have 1.6 m times of frequency F. It's going to be equal to the square root of 40,000. And the units will we have kilogram meter per second squared divided by kilogram per meter? Okay, So the unit of kilogram will cancel and we're gonna be left with meters squared per second squared. Alright. So dividing by 1.6 m we have the frequency is equal to the square root of four or 40,000. Sorry, is going to be 200. Okay, When we take the square root of meters squared per second squared? We're gonna be left with meters per second. Okay. And then we're going to divide by 1.6 m. And that will leave us with a frequency of 125 Okay, inverse seconds. The unit of meters cancel. And we know that we can write this as 125 hertz. Okay, so that is the fundamental frequency we were looking for. That's gonna correspond with answer. A thanks everyone for watching. See you in the next video.

A piano tuner stretches a steel piano wire with a tension of 800 N. The steel wire is 0.400 m long and has a mass of 3.00 g. (a) What is the frequency of its fundamental mode of vibration?

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