Video Transcript
In this video, we’re going to learn about energy and power of waves on a string. We’ll see how to derive the energy of a wave on a string and how that relates to the power it delivers. To get started, imagine that you and a friend are racing cars. But you’re racing them in an unusual way. The cars are two smaller-than-normal-size model cars, which run on electrical power supplied by wires connected to a converter box, where the power is supplied by mechanical motion. And that mechanical motion is supplied by you and your friend as you move ropes that are connected to these converter boxes. In order to send the most power possible through your rope into the converter box and then into your car so that you win, you want to know how you should move the rope optimally.
To figure this out, it will be helpful to know about energy and power of waves on a string. Say that we have a wavelength pulse which is moving down a physical string from left to right. As the wave moves along down the string, the leading edge of the wave moves the string off of its equilibrium location and stretches it. We know that since the string is being moved off of its equilibrium location, there must be energy in the string creating that motion. And since this movement happens over some period of time, there must also be power that this wave can deliver.
To figure out the energy that this wave transmits, we can focus in on a very small segment of the string. The energy in that segment is equal to the kinetic energy of the segment plus its potential energy. In this case, the potential energy would be due to an elastic effect rather than a gravitational effect. Let’s solve for the total energy transmitted by this wave. And we’ll start by considering the potential energy of the wave. Say that we consider this tiny segment of our string before the wave reaches it. In that case, this segment of string is lying along the horizontal or 𝑥-axis. And we can label this very tiny length of string Δ𝑥. When the incoming wave reaches this segment of string, it will lift the string up. And it will stretch it a bit so that it has a new length we’ve called 𝑙.
The length 𝑙 is not the same as Δ𝑥. It’s more than Δ𝑥. And that’s because the string is being elastically stretched when the wave passes through it. We can create a right triangle with Δ𝑥 and 𝑙 by raising a vertical line. And that will be the 𝑦-displacement of the string Δ𝑦. Here is the basic idea when it comes to the potential energy of this string segment. Because the string has been stretched from Δ𝑥 to a length 𝑙, this new length segment of string has elastic potential energy. It wants to contract back to its original length Δ𝑥.
In order to stretch the string to this new longer length, we needed to do work on it. And that’s equal to the force we applied multiplied by the stretch of the string. The force doing that work is the tension force that runs through the string. And the stretch of the string is equal to 𝑙 minus Δ𝑥. With the string stretched like this and wanting to return to its equilibrium length, we know that it has potential energy. In fact, its potential energy is equal to the work done to stretch the string. Looking at our expression for potential energy, we can expand on this expression by using the fact that 𝑙 is the hypotenuse of a right triangle. We can, therefore, rewrite 𝑙 as the square root of Δ𝑥 squared plus Δ𝑦 squared.
And if we factor out Δ𝑥 from under the square root, we can rewrite the expression under the square root as one plus Δ𝑦 over Δ𝑥 quantity squared. Let’s consider this quantity Δ𝑦 over Δ𝑥, looking back at our right triangle. If we consider the angle of this triangle that faces the side Δ𝑦 to be small, which is a reasonable assumption, when we’re speaking of a string being stretched, then that means that Δ𝑦, what we could call the rise, over Δ𝑥, what we could call the run, is a small value. Then looking back at our square root, if Δ𝑦 over Δ𝑥 is small, then Δ𝑦 over Δ𝑥 quantity squared is quite small.
On that basis, we’re going to use an approximation. If 𝑥, which represents any number, is a small number, then the square root of one plus 𝑥 is approximately equal to one plus 𝑥 over two. If we let Δ𝑦 over Δ𝑥 quantity squared be represented by 𝑥 here, then we can replace our square root with one plus one-half the quantity Δ𝑦 over Δ𝑥 squared. With this substitution, notice that we have a positive Δ𝑥 and a negative Δ𝑥. So those terms cancel out, leaving us with the expression the stretch in the string is equal to one-half Δ𝑦 over Δ𝑥 quantity squared multiplied by Δ𝑥.
Now, well we’ll remember that Δ𝑥 and Δ𝑦 refer to a very very small segment of string. If we let that segment become infinitesimally small, that means we can replace Δ𝑥 with 𝑑𝑥 and Δ𝑦 with 𝑑𝑦. When we make that substitution for our expression for the stretch of the string, we leave the 𝑑s inside the parentheses as script 𝑑s because this represents a partial derivative of the function 𝑦 with respect to 𝑥. The derivative is partial because 𝑦 is also a function of time 𝑡.
We recall now that this expression, which is equal to 𝑙 minus Δ𝑥, is just part of the equation for the potential energy of this segment of string. To make it complete, we multiply by the tension force, capital 𝑇. And then because capital 𝑇 can also represent the period of a wave, to avoid confusion, we’ll replace capital 𝑇 for tension force with 𝐹 sub 𝑇. We’ve now solved for the potential energy of this small segment of string as it’s stretched as the wave passes through it. To make our solution more complete though, we’d like to insert the actual function describing the wave’s position 𝑦.
For a transverse wave, 𝑦, which is a function of position and time, is equal to wave amplitude times the sine of wavenumber times position minus wave angular frequency 𝜔 times time. This means that if we take the partial derivative of 𝑦 with respect to 𝑥, we get the result 𝑘 times 𝐴 times the cosine of 𝑘𝑥 minus 𝜔𝑡. When we substitute in this expression for 𝑑𝑦 𝑑𝑥 and square it, we now have a complete expression for the potential energy of this string segment. Keep in mind though, this is the potential energy of an infinitesimally small segment of the string.
If we wanted instead to calculate the potential energy of a meter length of the string, then we would integrate with respect to 𝑑𝑥 from zero to one meter. When we integrate cosine squared of some value over one wavelength, we know that value would be zero because the cosine is positive just as often and as much as it’s negative. But when we integrate cosine squared, that’s a different story. The average value of a cosine squared function over one wavelength is one-half. We can say then that the potential energy of our string per one meter of string length is equal to one-half 𝐹 sub 𝑇, the tension, times 𝑘 squared 𝐴 squared times one-half, or simply one-quarter 𝐹 sub 𝑇 𝑘 squared 𝐴 squared.
Before we finalize this expression, let’s connect it with another expression we know about waves on a string. The speed of a wave 𝑣 is equal to the square root of the tension force over the linear mass density 𝜇. In other words, 𝐹 sub 𝑇 can be replaced by 𝜇𝑣 squared. With that substitution done, we can also recall another relationship. 𝜔, the angular speed of our wave, is equal to the linear speed 𝑣 multiplied by the wavenumber 𝑘. And looking at our expression for potential energy, we see two factors of 𝜔 are there.
With this substitution made, we now have our finalized expression for the potential energy of a wave on a string of a meter length. With this expression, we recall that the total energy of the string segment, which is what we’re seeking, is equal to the sum of potential plus kinetic energy. Now that we have the potential energy solved, we’ll work on the kinetic energy part. Considering the kinetic energy of an infinitesimally small segment of string, we know it’s equal to one-half the mass of that segment times its speed in the 𝑦-direction squared. If we consider the fact that this string has a linear mass density 𝜇 which is equal to the mass of the string over its length, when we focus in on a very small segment of the string, that means that our length, capital 𝐿, becomes infinitesimally small.
This means that if we multiply both sides of our equation by 𝑑𝑥, an infinitesimally small displacement in the horizontal direction, then the 𝑑𝑥 and the 𝐿 effectively cancel one another out. That means we can replace the 𝑚 in our equation for kinetic energy with 𝑑𝑥 times 𝜇. And then as we consider the speed of our string in the 𝑦-direction, that’s equal to the derivative of our string function, 𝑦 as a function of 𝑥 and 𝑡, with respect to time 𝑡. 𝑑𝑦 𝑑𝑡 is equal to negative 𝜔𝐴 times the cosine of 𝑘𝑥 minus 𝜔𝑡. And when we substitute that in for 𝑑𝑦 𝑑𝑡 and then square the expression, we now have a complete expression for the kinetic energy of our infinitesimally small segment of string.
Just like before, if we wanna solve for the kinetic energy not just of an infinitesimal segment but a larger stretch of our string, then we’ll integrate this expression over 𝑥, the horizontal displacement. And we’ll once again integrate for one meter. Just like with potential energy, the integral of cosine squared over one-meter stretch has an average value of one-half. We can say then that the kinetic energy of a string with a wave moving through it one meter long is equal to one-half 𝜇 𝜔 squared 𝐴 squared times one-half, or one-quarter 𝜇𝜔 squared 𝐴 squared.
Now that we have both the potential and the kinetic energy per meter of our string with a wave moving through it, we’re ready to solve for the total energy of that string. Since the potential energy term and the kinetic energy term are identical, their sum is just one-half 𝜇𝜔 squared 𝐴 squared. That’s the total energy in one-meter length of moving string. Here, 𝜇 is the linear mass density of our string, 𝜔 is the angular frequency of the wave, and 𝐴 is its amplitude. That’s the energy of a wave on a string. Now, how about the power?
Imagine that our wave pulse was moving along with a speed we could call 𝑣 and units of meters per second. If we imagine some surface that this wave could run into and perfectly transmit its energy to, then that would mean that, every 𝑣 seconds, where 𝑣 is again a speed in meters per second, 𝑣 times the amount of energy held in the string per meter, that would equal the amount of power that’s delivered. For example, if the wave was moving at two meters per second, then every second, two times 𝐸 per one meter of energy would hit the wall. And this means that the power in our wave is equal simply to the energy times the speed of the wave 𝑣. Written out as an equation, the power in one meter of the wave is equal to one-half 𝜇 times 𝜔 squared 𝐴 squared times the wave speed 𝑣.
With these results for the energy and power of a wave on a string, let’s review what we’ve learned so far. In this segment, we’ve seen that the function 𝑦 of 𝑥 and 𝑡 equals 𝐴 sin 𝑘𝑥 minus 𝜔𝑡 describes transverse waves that are moving on a string. We also saw that the total energy per meter of a wave on a string, 𝐸 per one meter, is equal to one-half the string’s linear mass density multiplied by its angular frequency squared times its amplitude squared. And finally, we saw that the total power per meter of a wave on a string is equal to the energy per meter, all multiplied by the wave speed 𝑣.
