# Video: Normal Vibrational Modes

In this video we learn what normal vibrational modes are and how to calculate their frequencies depending on system boundary conditions.

15:00

### Video Transcript

In this video, we’re going to learn about normal vibrational modes. We’ll see what these modes are, how they change based on boundary conditions, and how to describe them mathematically.

To get started, say that we have an elastic string. We take this string and we tie both its ends to fixed points. Then, we imagine further that we take hold of this string at the center and give it a good pluck. What will happen when we release this elastic string? One possibility is that a standing wave would be set up, with the string oscillating up and down as shown. The wave that moves along the string to create this string shape is a standing wave because it has two nodes, the points at either end. These are the locations where the string doesn’t move at all. And we know that in between nodes, halfway in between, are antinodes, the places on the string that experience maximum displacement. This standing wave is one option for the shape the string could take after we pluck it. But it’s not the only option. If we pull the string back even further than before and then released it, we might get a wave that looks like this blue curve.

Comparing these two possible standing waves with one another, the gold curve has twice as long a wavelength as the blue curve and one-half its frequency. And if we stretched our elastic string back even farther and then released it, then the string might take on the shape of this green curve in its standing wave form.

Considering these three curves, we can start to see a trend. For one thing, all the curves by necessity are fastened at their ends to the walls. If we look at the first standing wave we made, the gold one, that curve has one antinode, whereas the blue wave has two antinodes and the green wave has three. If we consider what parameters affect the shape of these three curves, it comes down to two factors. The first factor is the boundary conditions of this wave. The fact that the string is constrained to move between two fixed points has a lot to do with the shape that the string takes on. The second factor affecting the shape of these curves is what we can call the mode of vibration.

The first curve that we’ve drawn, the gold one, could be set to be in the 𝑛 is equal to one mode. That’s the first possible mode of vibration for this string under these boundary conditions. It’s at this mode that our string takes on the longest possible wavelength they can have under these boundary conditions. And it also takes on a frequency we call the fundamental frequency. And we symbolize it 𝑓 sub one. This symbol means that the string’s frequency is the frequency that corresponds to its 𝑛 equals one or first mode.

The blue curve that we’ve drawn in shows what the string would look like in the 𝑛 is equal to two mode. And the green curve, the one with three antinodes, is in the 𝑛 equals three mode. If we say that the original unstretched length of our string is capital 𝐿, then we could say that 𝜆 sub one, the wavelength that corresponds to the fundamental frequency of oscillation, is equal to two times that string length. But then we see that the wavelength of the second mode, 𝜆 sub two, is equal simply to the length of the string. And going further, 𝜆 sub three, the wavelength of the third mode, is equal to two-thirds times that length.

What we’re seeing in this diagram then is the first three normal vibrational modes of this string under these boundary conditions. Knowing that, in general, wave speed 𝑣 is equal to wavelength times frequency, we can say that the fundamental frequency of oscillation of this string, 𝑓 sub one, is equal to the speed of the wave on the string divided by 𝜆 sub one, the wavelength for the first mode. We’ve seen that that wavelength geometrically is equal to twice the length of the string, two times 𝐿. If we go on to calculate 𝑓 sub two, the frequency of the 𝑛 equals two mode, that’s equal to the wave speed over 𝜆 sub two. And since 𝜆 sub two is 𝐿, it’s equal to 𝑣 over 𝐿. And finally, 𝑓 sub three, the frequency of the 𝑛 equals three mode, is equal to wave speed, 𝑣, over 𝜆 sub three. And since 𝜆 sub three is two-thirds 𝐿, that means that 𝑓 sub three is equal to three times 𝑣 over two 𝐿.

If we look at all three of these frequencies, 𝑓 one, 𝑓 two, and 𝑓 three, we see they’re related. The frequency of the second mode, 𝑓 sub two, is equal to twice the frequency of the first mode, the fundamental frequency. And 𝑓 sub three is equal to three times 𝑓 sub one. We see that there’s a progression going on. And we can generalize that progression using this relationship. It says that the frequency of the 𝑛th mode of oscillation is equal to 𝑛, that mode, times the fundamental frequency, 𝑓 sub one, where 𝑛 can equal any positive integer. Even though we’ve only used the first three modes to arrive at this relationship, it turns out that this is accurate for any mode we might use for a string oscillating between two fixed ends. Let’s get a bit of practice solving for a few of these frequencies with a quick example.

Let’s say we have a string of length 4.0 meters long under a tension of 100 newtons with a linear mass density 𝜇 of 0.005 kilograms per meter. Let’s say we wanna solve for the fundamental frequency of oscillation of the string as well as the frequency of the second and third modes. We know that for a wave traveling along a string, the wave speed is equal to the square root of the string tension divided by its linear mass density, which in general is equal to the wavelength of the wave times its frequency. This means that the fundamental frequency, 𝑓 sub one, is equal to one over 𝜆 sub one, the wavelength of the 𝑛 equals one mode, multiplied by the square root of 𝐹 sub 𝑇 over 𝜇.

When we look at this string oscillating in its 𝑛 equals one mode, we see that one-half of a wavelength exists from one end of the string to the other. That tells us that 𝜆 sub one is equal to two times the string length 𝐿. So 𝑓 sub one is one over two 𝐿 times the square root of 𝐹 sub 𝑇 over 𝜇. Since we know 𝐿, 𝐹 sub 𝑇, and 𝜇, we’re ready to plug in and solve for 𝑓 sub one. When we do and we enter these values on our calculator, we find that to two significant figures the fundamental frequency of oscillation of this string is 18 hertz, 18 cycles every second.

Next, we wanna solve for the frequency of the 𝑛 equals two mode. And to do that, we’ll use our relationship 𝑓 sub 𝑛 is equal to 𝑛 times 𝑓 sub one. This means to solve for 𝑓 sub two, we can simply take the fundamental frequency and multiply it by two. And when we do using an exact value for 𝑓 sub one, we find that to two significant figures, it rounds to 35 hertz. And then since this relationship is true for any mode number 𝑛, we use the same relation to solve for 𝑓 sub three. To two significant figures, that comes out to 53 hertz. We see then that as the mode number goes up, we have a higher frequency and a shorter wavelength.

So far, we’ve thought of these modes of oscillation in the context of a physical string fixed at both ends and vibrating in a standing wave. It turns out though that the shape of these particular modes doesn’t depend on the fact that we have a string but rather, as we said, depends on the boundary conditions of our system. To show this, let’s imagine that instead of a string oscillating between two fixed points, we have simply an empty cylindrical tube. This is much like a musical instrument such as the pipe of an organ. Inside this tube, there’s nothing but air molecules. But when we fill the tube with a sound wave, that is, a longitudinal pressure wave, what we see is that, because of our boundary conditions having two closed ends to our tube, the shapes that the normal modes of oscillation can take of the sound waves in the tube are the same as the shapes of our vibrating string. That means our 𝑛 equals one mode has one-half wavelength that fits in the tube. The 𝑛 equals two mode has a full wavelength that fits inside the tube. And the 𝑛 equals three mode has one and a half wavelengths that fit.

One quick note about these wave shapes: when we worked with the string fixed between both ends, the waves that moved along the string were indeed moving with their peaks and troughs perpendicular to the direction of the wave motion. That is, the waves were transverse. Now that we’re working with sound waves though, we know that those waves are longitudinal rather than transverse, meaning the peaks and troughs move in the same direction as the wave motion. We’ll represent those waves here as transverse waves for clarity. But know that longitudinal waves will not have peaks and valleys perpendicular to the direction of wave motion. Rather, they’ll be in the same direction as the wave moves.

But coming back to these three modes of oscillation, we see that this physical system, an empty cylindrical tube, though it’s completely different than the physical system of a string fixed by its ends, produces the same-shaped standing waves, that is, the same-shaped normal vibrational modes. And this has to do with the boundary conditions of the two systems being the same. And the same mathematical relationship for frequencies of different modes, 𝑓 sub 𝑛 is equal to 𝑛 times the fundamental frequency, holds for this closed-ended tube. We can say that, in general, for closed-ended or fixed-ended systems, this mathematical relationship applies for the frequencies of various modes. And we noted that 𝑛, the index, can be any positive integer.

But a tube or a system with both ends closed isn’t the only kind of tube out there. For example, what if one end of our tube was open? In this case, this would be like a model of a clarinet, where one end of the instrument is closed and the other end open. The first thing we notice about this system is that the boundary conditions are different from our closed-ended tube. And those different boundary conditions will affect the shape of the normal vibrational modes that are allowed to exist inside this tube. Here’s why. When a sound wave is moving from left to right and reaches the open end of the tube, boundary condition continuity requires that the pressure of the sound wave at that point—at the end of the tube—matches the air pressure outside the tube. The air pressure outside the tube is at regular or standard atmospheric pressure. And therefore, the pressure of the sound waves just at the edge of this tube must match that.

This means that when it comes to the wave amplitude, the rightmost end of our tube, it either needs to be at a maximum or a minimum value. It’s under those conditions that the pressure of our wave is continuous with the pressure of the air outside the instrument. On the left side of our tube, the closed end, we again have a boundary condition where waves meet at the middle. This means that the first vibrational mode, the 𝑛 equals one mode, of standing waves in our system would look like this. If the length of our tube is 𝐿, then 𝜆 sub one is equal to four times that length, which means that the fundamental frequency of this wave would equal to the speed of the wave, the speed of sound, divided by four times 𝐿.

If we then try to draw the 𝑛 equals two mode for this tube, we see that actually this mode violates the boundary conditions of this tube. For this mode to be formed within the tube, at the right end, the open end, the wave would have to end at the middle of the tube. But that’s disallowed by our boundary conditions. This means that there is no such thing as an 𝑛 equals two mode for a tube with one end closed and one end open. We then move on to drawing the 𝑛 equals three mode. And we find that this mode is allowed by the boundary conditions of our system. The wavelength of this mode, 𝜆 three, is equal to four-thirds times 𝐿. So we find that 𝑓 sub three, the frequency of the third mode of oscillation, is equal to three times the wave speed over four times 𝐿 or three times the fundamental frequency.

If we move on from 𝑛 equals three to 𝑛 equals four and try to draw that curve, we find that, just like for 𝑛 equals two, this curve is disallowed by the boundary conditions of our tube. The 𝑛 equals four mode can’t exist under these conditions. Then, we move on to the 𝑛 equals five mode. And we find that this mode, the odd-numbered mode, is allowed. 𝜆 sub five is equal to four-fifths times 𝐿, which means that 𝑓 sub five is equal to five times the wave speed over four 𝐿 or five times the fundamental frequency. If we extrapolate the trend we’re seeing from the frequency of a given mode number to that mode number times the fundamental frequency, we find the same mathematical relationship as we did for a closed-ended system. But this time, the index 𝑛 needs to be an odd number. The even positive numbers are disallowed by the boundary conditions of our open end. We summarize this by saying that for a system that’s open at one end and closed at the other, 𝑓 sub 𝑛 is equal to 𝑛 times the fundamental frequency, where 𝑛 is a positive odd integer.

And lastly, let’s examine a system that has not one but both ends open to the atmosphere. In the world of musical instruments, this would correspond to a flute. Just like with our open-ended clarinet, our boundary conditions require that our waves terminate at a peak or a trough at either end of our tube. If we draw in the 𝑛 equals one mode, it would be a standing wave that looks like this, where, for a tube of length 𝐿, this wavelength, 𝜆 sub one, is equal to two times 𝐿, indicating that the fundamental frequency of a wave in this system is equal to the wave speed over two 𝐿. When we go to draw in the 𝑛 is equal to two mode of this system, we see that it’s possible. The boundary conditions allow this mode. 𝜆 sub two is equal to 𝐿, and this means that 𝑓 sub two, the frequency of the second mode, is equal to 𝑣 over 𝐿, or two times the fundamental frequency.

When we next draw in the 𝑛 is equal to three mode, we find that this fits as well under our boundary conditions and that 𝜆 sub three is equal to two-thirds times the length of the tube. This tells us that 𝑓 sub three is equal to three times the wave speed over two times the length of the tube or three times the fundamental frequency. We can write a general expression for the frequencies of the various modes of oscillation of this both ended tube. It’s the same mathematical relationship as we found before. This time though, the index 𝑛 can be any positive integer. This covers the various boundary conditions we might find within a wave-generating system: both ends closed, one end open and one end closed, and both ends open.

To summarize, we’ve developed mathematical relationships showing the frequencies of the different modes of oscillation for three types of systems. The relationship describing these frequencies is the same in each case. The only difference is the allowable modes within our different system. In two of the three cases, any mode number will do. And in one case, the system with one end open and one end closed, we need odd modes. These are the relationships that describe normal vibrational modes.