Waves II. I figuring out the harmonics.

[afcwestwarrior]

Homework Statement

A steel rod of length 1.44 m is clamped at a point 1/4 of its length from one end. The ends of the rod are free to vibrate. If the velocity of longitudianl wave in steel is 5000 m/s, find the frequency of the first three allowed harmonics. Make diagrams to show the three standing waves.

Homework Equations

f = (nV) / 2L

because it has two open ends. Their not actually open but they both are free which means the same thing.

The Attempt at a Solution

I have the answers. I'm just clueless as to how you figure out n for each harmonic.

Shouldn't it just be 1, 2 and 3.

For some reason my classmates gor 2, 6 and 10.

And all you have to do is plug it into find each of the three frequencies.

 

[jbsteele]

First Harmonic: f = (2V) / 2L f = (2)(5000) / (2)(1.44) = 6944.44 Hz Second Harmonic: f = (6V) / 2L f = (6)(5000) / (2)(1.44) = 20833.3 Hz Third Harmonic: f = (10V) / 2L f = (10)(5000) / (2)(1.44) = 34583.3 Hz

 

[nlightner]

I can understand the confusion and frustration with figuring out the harmonics in this problem. The key to solving this is understanding the relationship between the length of the rod, the velocity of the wave, and the frequency. In this case, we are dealing with standing waves, which occur when two waves of the same frequency and amplitude travel in opposite directions and interfere with each other.

The equation f = (nV) / 2L is correct, but it is important to note that n represents the number of nodes in the standing wave. In other words, it is the number of points along the rod that are not moving. In the first harmonic, there is only one node, at the midpoint of the rod. In the second harmonic, there are two nodes, one at the midpoint and one at the end of the rod. In the third harmonic, there are three nodes, one at the midpoint, one at 1/3 of the length, and one at 2/3 of the length.

To find the frequency for each harmonic, we can plug in the values for n and solve for f. For the first harmonic, n = 1, so f = (1*5000)/2(1.44) = 1736.11 Hz. For the second harmonic, n = 2, so f = (2*5000)/2(1.44) = 3463.89 Hz. And for the third harmonic, n = 3, so f = (3*5000)/2(1.44) = 5190.56 Hz.

To make diagrams of the standing waves, we can plot the displacement of the rod at different points along its length for each harmonic. The first harmonic will have a single peak at the midpoint, the second harmonic will have two peaks at the midpoint and end, and the third harmonic will have three peaks at the midpoint, 1/3 of the length, and 2/3 of the length.

I hope this explanation helps you better understand the concept of harmonics and how to solve for them in this problem. Keep practicing and you will become more familiar with these types of problems. Good luck!