### Video Transcript

A nylon guitar string is fixed between two lab posts 2.00 meters apart. The string has a linear mass density of 7.20 grams per meter and is placed under a tension of 160.0 newtons. The string resonates at the π equals three node when it is plucked. The string is placed next to a tube of length πΏ that is open at both ends, and the sound wave produced by the vibrating string resonates with the fundamental mode of the tube. Find πΏ, using a value of 343 meters per second for the speed of sound in the tube.

In this problem, weβll assume that the speed of sound in the tube is exactly 343 meters per second. Letβs highlight some of the vital information given to us in this statement. Weβre told that the length of the guitar string is 2.00 meters. And that the string has a linear mass density of 7.20 grams per meter, and that itβs placed under a tension of 160.0 newtons. The π equals three node of the string is equal to the fundamental mode of the tube its placed next to. And we want to solve for the length of the tube called πΏ.

Letβs start by drawing a diagram of the string resonating at its π equals three node. If we draw a diagram of the string vibrating at the π equals three mode, since weβre given πΏ sub π , we can see roughly how long the wavelength of that wave is. Weβre told that this vibrating string is placed next to an open-ended tube. The length of that tube, called capital πΏ, is not known. But what is known is that the π equals three mode of the vibrating string excites the fundamental mode of sound in the tube. We want to solve for the length of the tube πΏ.

In this problem, weβre given that πΏ sub π , the length of the string, is 2.00 meters. The mass per unit length of the string, π, is equal to 7.20 grams per meter. The string is plucked and held under a tension of π equals 160.0 newtons. And weβre told that the speed of sound in the tube is 343 meters per second. Weβll call that π£ sub π and treat it as an exact number. We want to figure out what the frequency of the π equals three mode is for the string. If we call that π sub three π , weβre told that that value is equal to the fundamental frequency of the open tube which weβll call π sub one π‘. If we can solve for the tubeβs fundamental frequency which we can use along with the speed of sound in the tube to solve for wavelength which will then let us solve for πΏ, the tube length, by analysis.

To begin, letβs recall a fact of standing wave modes. And that fact is that the frequency of the πth mode is equal to π times the fundamental frequency. As it relates to our problem, we can write that π sub three π is equal to three times π sub one π . Now letβs recall another way of relationship for the fundamental frequency of a string fixed at both ends. The fundamental frequency of a string fixed at both ends is equal to one divided by two times its length multiplied by the square root of the tension in the string divided by the mass per unit length π.

If we apply this relationship to our situation, we see that the fundamental frequency of the string is equal to one over two times its length multiplied by the square root of the string tension divided by its mass per unit length. Since πΏ sub π , π, and π are all given as part of the problem statement, we can insert those into this equation now. Before we enter these numbers into our calculator, letβs convert π into units of kilograms per meter, from its original given units of grams per meter. Since 1000 grams is equal to one kilogram, in units of kilograms per meter, π is 7.20 times 10 to the negative third kilograms per meter.

When we calculate out π sub one π , we find a value of 37.27 hertz. We can then use this value for π sub one π in our equation above to solve for the third mode of vibration in the string. Three times 37.27 hertz is equal to 111.80 hertz. Thatβs the frequency of the third mode of the string which weβre also told is equal to the fundamental frequency of the tube. So we can write π sub one π‘ is equal to 111.80 hertz.

At this point, letβs recall the relationship for the speed of a wave π£. Wave speed is equal to the frequency of the wave times its wavelength. Applying this relationship to the open tube, we see that π£ sub π , the speed of sound which is the speed of waves in the open tube, is equal to π sub one π‘ times the wavelength of the waves in the tube. If we divide both sides by π sub one π‘, that term cancels out at the right-hand side, and weβre left with an expression for the wavelength, π. Weβve solved for π sub one π‘ and weβre given π£ sub π , the speed of sound. So we can insert the values for those variables now.

When we calculate this value for wavelength, we find 3.07 meters is the wavelength of the wave in the tube. As we look at our diagram of the wave and the tube, if we continue drawing the wave past the open end of the tube on the right, then if we drew another half wavelength, altogether, that would give us one wavelength of the fundamental mode. We can tell by the shape of the wave in the tube that the tube is able to contain one-half π or half a wavelength. That means that πΏ, the tube length, is equal to π divided by two, or πΏ equals 3.07 meters divided by two.

That value, to three significant figures, is equal to 1.53 meters. Thatβs how long the open tube is.