### Video Transcript

An organ pipe of length 3.00 metres is closed at both ends. The speed of sound is 343 metres per second. Compute the wavelength of the first mode of resonance. Compute the wavelength of the second mode of resonance. Compute the wavelength of the third mode of resonance. Compute the frequency of the first mode of resonance. Compute the frequency of the second mode of resonance. Compute the frequency of the third mode of resonance.

In this exercise, weβll treat 343 metres per second as an exact value for the speed of sound. And weβll use the fact that the length of the closed tube is 3.00 metres long. We want to solve for wavelengths and frequencies of various modes of resonance. Weβll call the wavelength of the first mode of resonance π sub one and so on through the third resonant mode. Weβll call the frequency of the first mode of resonance π sub one and so on down through the third resonant mode.

Letβs begin working towards a solution by drawing a diagram of this situation. We have a closed ended tube of length πΏ equals 3.00 metres long. Weβre studying the resonant modes of this tube. And because both ends of the tube are closed, that means standing waves resonant modes set up inside the tube must have nodes at either end of the tube.

The fundamental mode is the longest wavelength that could fit in the tube and meets these boundary conditions. That looks like this. And we call that the π equals one mode. The second mode of resonance, another standing wave that meets these boundary conditions, will have three nodes and two antinodes. That will look like this. And we call that the π equals two mode. And the π equals three mode will look like this.

Itβs for these three modes of oscillation that we want to examine the wavelength and the frequency. To do that, we can reference equations that tell us wavelength and frequency for progressive modes of resonance given these boundary conditions. For a closed-ended tube, like the one we have in our scenario, the wavelength of a given mode of resonance π sub π equals two times the length of the tube πΏ divided by the mode number π. And the frequency of each resonant mode π sub π is equal to the mode number π times π sub zero, the fundamental or most basic frequency allowable in that tube.

Letβs apply the wavelength equation to our particular scenario. Since π sub π is equal to two πΏ divided by π, for the first three modes: one, two, and three, π sub one equals two πΏ divided by one, π sub two equals two πΏ divided by two, and π sub three equals two πΏ divided by three. Weβve been given a value of πΏ equals 3.00 metres. So when we plug that in and then compute our three π values, we find that π sub one is 6.00 metres. This is the longest wavelength that can meet the boundary conditions of this closed ended tube. π sub two is 3.00 metres and π sub three is 2.00 metres. These are the wavelengths of the first three resonant modes of this tube.

Now that we know the wavelengths of these modes, letβs look at the frequencies. We can see from our equation for the frequencies within a closed-ended tube that it starts with something called the fundamental frequency, π sub zero. To find that frequency, we will rely on the relationship that wave speed π is equal to wave frequency π times wavelength π or equivalently frequency equals wave speed divided by wavelength. Applying this relationship to our scenario, the fundamental frequency π sub zero is equal to the speed of the wave in the tube divided by the fundamental wavelength π sub one.

We were told in the problem statement that the speed of standing waves in this tube, which are sound waves, is equal to exactly 343 metres per second. We will use that speed for π. We solved for π sub one earlier. Thatβs 6.00 metres. When we perform this division, we find that the fundamental frequency π sub zero is 57.2 hertz.

Based on that result, we can now solve π sub one, π sub two, and π sub three. According to our formula for the frequency of resonant modes, π sub one is one times π sub zero, π sub two is two times the fundamental frequency, and π sub three is three times π sub zero. When we multiply these values through, we can see that π sub one is equal to π sub zero, π sub two is twice that, 114 hertz, and π sub three to three significant figures is 172 hertz.

Based on our understanding of the boundary conditions of our problem, a tube with both ends closed, we use the relationships for π and frequency of progressive resonant modes under those conditions to answer the questions asked by the problem.