Sound Waves Legacy Problem #19 Guided Solution

Perhaps you've seen this demonstration. It's quite fascinating. If you take a long aluminum rod, and you hold it exactly in the center, and get your fingers kind of sticky, maybe by getting a little rosin on them, and then you slightly pull your fingers across the aluminum rod, it begins to sing out, so to speak, with a very, very pure tone, a very, very high-pitched pure tone. And then, of course, you can hold it at varying locations and produce different harmonics by forcing different standing wave patterns up on that aluminum rod. Now, in this case, what we were given is the length of the rod, and we're told it's vibrating in such a way that there's one-half of a wavelength between the ends of the rod. The 2.13 meters is the length of the rod, and it's equal to one-half of the wavelength. We also know that the V equals 6320 meters per second. That's the speed of my vibrations traveling through aluminum. What we're asked to calculate is the frequency. So, obviously, I'm going to have to get to the wave equation here and do V equals f times lambda, solving for f. But before I can do that, I have to first get my wavelength, or my lambda. And so I use the relationship 2.13 meters is equal to one-half of lambda, or wavelength, and I solve for wavelength, and I get 4.26 meters. Now I can take the wave equation and use it to solve for frequency. I rearrange it first from V equals f times lambda to f equals V divided by lambda, where V is the speed and lambda is the wavelength. And so I do the 6320 meters per second divided by the 4.26 meters. I get 1,483.57 hertz, and I can round that to three significant digits, such that it becomes 1,480 hertz.