PhysicsScience

The Physics Philes, lesson 108: The Speed of Sound

Last week we talked a bit about how our ears work and how they perceive sound. If we’re having a conversation, the sound seems to travel from my mouth to your ears in an instant. But what do you hear in a thunder storm? We know that thunder is the sound we hear as the air around a lightning bolt rapidly expands. We also probably know from experience that we see the lightning bolt before we hear the thunder. What gives?

In short, the sound travels less quickly than light. But it still moves really fast, and it will travel at different speeds depending on the medium. So this week and next we’ll discuss how to calculate the speed of sound.

OK, let’s start with the equation for the speed of a transverse wave on a string. Recall that it depends on the string tension F and the linear mass density μ:

Screen shot 2014-07-27 at 9.22.00 AM

Did you guess that there would be a corresponding equation for the speed of sound waves through a gas or liquid? If you did, you’d be right! Generally, the wave speed of a mechanical wave is found by taking the square root of the restoring force divided by the inertia that resists the restoring force. When a sound wave travels though a bulk fluid, it causes compressions and rarefactions (squeezing together to create greater pressure and pulling apart to create areas of less pressure). That means that the restoring force term must be related to how hard it is to compress the fluid. Luckily, that’s what the medium’s bulk modulus B tells us.

Now we need to figure out the inertia term. According to Newton’s second law, inertia is related to mass, which is in turn described by the medium’s density ρ. So the equation we’d expect to get for the speed of a sound wave is

Screen shot 2014-07-27 at 9.34.08 AM

This may be what we expect the equation to look like, but let’s derive it so we can know for sure. Let’s derive an equation for the speed of sound in a fluid.

Let’s think about a pipe of fluid with a piston that has a cross sectional area A. It looks something like this:

Screen shot 2014-07-27 at 9.45.59 AM

When the fluid is in equilibrium, the fluid is under uniform pressure p. Let’s also take the x-axis along the length of the pipe, which is also the direction of wave propagation.

Now, let’s move the piston to the right at time t = 0 at a constant speed v_y. This movement causes a wave motion that travels to the right. As the wave moves down the pipe, successive sections of fluid will be compressed at successively later times.

We’ve been pushing to piston to the right for t seconds. Now we have something that looks like this:

Screen shot 2014-07-27 at 9.53.38 AM

The fluid on the left of point P is moving with speed v_y. The fluid to the right is at rest. That boundary line in red is moving with a speed equal to the speed of propagation. At time t the piston has moved a distance that can be given by (v_y)t. The boundary line has moved a distance vt. We can use the impulse-momentum theorem to compute the speed of propagation.

The amount of fluid in motion at time t is the same amount that occupied a section of the tube with a length vt, cross sectional area A, and a volume of vtA. Multiply the volume by the density and you get the mass of the fluid, and if you multiply that result by v_y you get the longitudinal momentum:

Screen shot 2014-07-27 at 10.04.39 AM

Now we need to figure out what the increase in pressure in the moving fluid is. As stated above, the original volume is vtA. The volume has decreased by (v_y)tA. Recall that the definition of the bulk modulus is the negative of the change in pressure divided by the fractional change in volume. If we solve for the change in pressure, we get:

Screen shot 2014-07-27 at 10.09.29 AM

Screen shot 2014-07-27 at 10.10.53 AM

The pressure in the moving fluid is p + Δp. The force applied on the piston is (p + Δp)A. The net force of the moving fluid, then, is ΔpA, so the longitudinal impulse is:

Screen shot 2014-07-27 at 10.15.22 AM

Because we started at t = 0, the change in momentum up to time t is the same as the momentum at time t. If we apply the impulse-momentum theorem, we find that:

Screen shot 2014-07-27 at 10.17.20 AM

All we have to do now to find the speed of sound through a fluid is to solve for v:

Screen shot 2014-07-27 at 10.19.15 AM

Well, will you look at that?! That’s the same as what what we thought it would be at the beginning of this post! Nice! I guess you could say that we went through all that work for nothing, but can you really put a price on knowing for sure that you’re right? This equation doesn’t just hold for sound waves. It holds for any longitudinal wave in a bulk fluid, so it’s kind of important to know that we’re right.

This is only part of the story, though. Next week we’ll talk about the speed of sound through solids and gases.

Featured image credit: Stas Bukarev via Flickr

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Mindy

Mindy

Mindy is an attorney and Managing Editor of Teen Skepchick. She hates the law and loves stars. You can follow her on Twitter and on Google+.

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