A violinist places her finger so that the vibrating section of a 1.0 g/m string has a length of 30 cm, then she draws her bow across it. A listener nearby in a 20°C room hears a note with a wavelength of 40 cm. What is the tension in the string?

Video transcript

Hey everyone. So this problem is dealing with sound waves. Let's see what it's asking us. A physicist uses a 40 centimeter plastic line with a linear density of 0.3 g per meter to create sound by making the line vibrate the physicists test this instrument in a lecture ball by making the line vibrate in its fundamental mode. The temperature in the hall was 20 degrees Celsius. The students in the hall catch a sound with a wavelength of 50 centimeters, calculate the tension in the line. Our multiple choice answers here are a 25 newtons b 48 newtons C 90 newtons, four D 112 newtons. OK. So first we're going to go through the givens from this problem. So we have the length of line L is centimeters. We're gonna rewrite that as 0.4 m to keep everything in standard units similar to this linear density linear density is um represented by mu by the character mu. So 0.3 g per meter can be rewritten as three times 10 to the negative four kg per meter. His kilogram is our standard unit there. Our fundamental mode fundamental tells us that N equals one as far as our harmonic number is and the wavelength that they caught the of the sound is 50 centimeters. So again, I'm going to rewrite that as 500.5 m. And we're asked to calculate the tension in the line. So we can recall that this, that speed in terms of tension is given by the equation B equals the square root of T divided by mu. So when we rearrange that for tension, we get V squared U where T is the tension in the line V is the speed of the line, that speed of the line vibrating and U is the linear density of the line. So we are solving for tea, we were given me now, we need to find V we can recall that frequency is given by V divided by lambda and the frequency of sound, whether it's the frequency of the line or the frequency of the air, that's going to be the same. So we can take the frequency in of the line and the frequency and air and set them equal to each other. So that looks like the air divided by lambda. Air is equal to V line divided by lambda lying. Now, we don't have lambda for the line, but we can recall at lambda of the lambda, the wavelength of a string is given by the equation two pi or sorry, two L divided by and, and so we have that we have that link of the line and we have N which is just equal to one and its problem. So LAMBDA is going to be two L for the line. So now we're going to solve for the, of the line and that's going to equal speed of sound and air. So speed of sound and air at 20 degrees Celsius, that's, we can recall that that is 343 m per second. We'll multiply that by two, multiplied by the length of the string, which is 20.4 m and we'll divide all of that by that wavelength that they caught in air of 0.5 m. And that gives us a speed of a line of 548.8 m per second. And now we can go back into our tension equation and solve for that. So 548.8 m per second squared multiplied by mu or three times 10 to the negative four kilograms per meter plug that in and get detention of 90 mutants. And that's it. That's all we have for this problem. We go back up to our multiple choice answers. Not a lack of answer choice. B All right. We'll see you in the next one.