in this video we're shown a standing wave on a string and this has been excited by some kind of mechanical oscillating device here at the ends in practice this is a speaker like device with a metal rod attached to it you hook it onto the string and we're told the length of the string is 1.8 meters and we notice that the standing wave we're looking at has four antinodes and we're told the frequency of oscillation that excited this standing wave is 30 hertz and then the string is tensioned by draping it over a pulley and hanging a mass of 350 grams off of it so in part a we're asked to find the fundamental frequency for the string and in a previous video we showed that all resonant frequencies on the string are multiples of a fundamental frequency the lowest one so that fundamental mode looks like this that's where the middle of the string wiggles up and down as the only anti-node that's our n equals 1 case and then if i examine the higher harmonics what i'm trying to do is figure out which harmonic i'm looking at here the n equals two cases when you have two antinodes and finally we get to the fourth harmonic which is the one that we're actually looking at in this problem so we can say that f4 the frequency of the fourth harmonic is equal to four times the frequency of the fundamental so the frequency of the fundamental is just the frequency of the fourth harmonic divided by four that's 30 hertz over four and i find that the frequency of the fundamental for this string is 7.5 hertz so what this means is if we tune our oscillator to seven and a half hertz we'll just see the center of the string flopping up and down as the only anti-node in part b we're asked to compute the linear density of the string and there are two parts coming together in this question first of all the wave speed is related to the frequency and the wavelength by v equals f lambda and i'll post a link real quick to the video where that was first introduced and second the wave speed is related to the tension in the string and the linear density of the string by square root t over mu where mu is the linear density and i'll post a link to the video where that was first introduced as well now linear density is kind of a strange quantity its units are kilograms per meter and this is just a way of talking about how heavy a string or a rope is how many kilograms does it have for every meter so you can imagine like a heavy chain might be two kilograms per meter of chain so that's the thing we're after in this problem for this string and i can go ahead and find the wave speed using frequency times wavelength and i have a couple choices here i can just go ahead and use the waves that i'm looking at in the original setup of the problem and what i notice is that an entire wavelength fits on half the string and that has a length of 0.9 meters so that's lambda for the fourth harmonic and then the frequency for the fourth harmonic was 30 hertz so i can take 30 hertz times 0.9 meters and i get a wave speed of 27 meters per second now that wave speed is going to be the same for every harmonic that we find and it's determined by how much tension we put on the string and how heavy the string is so i think i'll go ahead and get the tension on the string right now and this thing is tensioned by hanging a mass off of it and the mass is not accelerating and that means the force of gravity down on this thing is equal to the tension in the string and if i compute mg i need to put the mass in the right units so 0.350 kilograms times 9.8 meters per second squared and when i run the numbers on this i get 3.43 newtons okay now i have the wave speed and i have the tension the only unknown in that second equation is the linear density and i'm going to go ahead and solve for this symbolically i'm going to square both sides and i get v squared is t over mu multiply both sides by mu and divide by v squared i get that mu is t over v squared and we plug in our tension 3.43 newtons divide by the square of the wave speed so 27 meters per second all squared and when i crunch the numbers i get point zero zero four seven kilograms per meter it's probably more appropriate to state that in grams per meter so i get 4.71 grams per meter and that sounds pretty reasonable for like a guitar string for example if you find the physics content on zack's lab helpful click on the zaxlab logo on the right to browse playlists and subscribe to the channel i produce over 100 new videos per month and subscribing is the easiest way to find new content thanks for watching