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The Physics Philes, lesson 100: Need for Wave Speed II

Oh my goodness. Has it been a week already? Let’s see, where were we? Oh that’s right. Last week we started talking about how to find the speed of a transverse wave on a string. The first method was fairly straight forward. Now it’s time to move on to the second method, which is a little more tricky.

Here’s what we can do to find the wave speed of a wave on a string. We can apply Newton’s second law of motion to a small segment of string that has the length Δx when its in its equilibrium position. The mass of that segment is μΔx. The free body diagram might look something like this:

Screen shot 2014-06-01 at 11.30.27 AM

OK, look. I never claimed to be an artist. Let me see if I can help make sense of this. Let’s start with trying to sort out the forces. The x-component of forces F1 and F2 is denoted as F. But since this wave is transverse, there is no movement in the x-direction, so those will add up to zero. We still need to figure out what F1y and F2y are. Hmm.

Ah ha! Remember that F1y/F is equal to the magnitude of the slope of the string at x and F2y/F is slope of the string at x + Δx. So, once we insert the appropriate signs, we get these equations:

Screen shot 2014-06-01 at 11.46.28 AM

Screen shot 2014-06-01 at 11.47.36 AM

Remember, we’re dealing with partial derivatives at points x and x + Δx. We can use the above equations to figure out the y-component of the force.

Screen shot 2014-06-01 at 11.58.03 AM

Now we can replace Fy with μΔx times the y-component to get the acceleration:

Screen shot 2014-06-01 at 12.01.31 PM

If we divide by FΔx, we get this monstrosity:

Screen shot 2014-06-01 at 12.05.30 PM

Ugh. Mindy, you must be thinking. This has gone too far. There must be a way to make this equation look less…blegh. Now, now. Just say with me. The payoff is coming. What we need to do now is take the limit as Δx goes to zero. That means that the left side of this equation becomes the second partial derivative of y with respect to x.

Screen shot 2014-06-01 at 12.08.55 PM

There. It’s looking better already, isn’t it? But wait…this looks familiar. It’s the same form as the wave equation we derived a while back. The two equations describe the same wave motion. For this to work, the wave speed must be

Screen shot 2014-06-01 at 12.14.06 PM

Ta da!

Notice that we didn’t make any assumptions about the shape of the wave as we derived this. The wave equation is valid for waves on a string of any shape.

What we’ve done so far gives us the wave speeds for only mechanical waves on a stretched string or rope. However, a lot of mechanical waves have the same general form: it’s the square root of the restoring force divided by the inertia resisting the return to equilibrium. Applying that to the situation we just got done discussing, the tension F in the string is the restoring force and the linear mass density is the inertia.

You may think that we’ve been talking about waves for a while, so we must be almost done. Hahaha! Nope! This chapter is the longest chapter ever and we’re only about halfway through! That means next week I’ll be back with more wave information.

(On a side note, I feel like I should at least mention that this is the 100th Physics Philes post. I honestly didn’t think I’d make it this far. I feel like I’ve learned a lot along the way and I hope I’ve managed to teach something, too. Here’s to 100 more!)

Featured image credit: Barbara Walsh 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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