Another Question on Natural Frequencies

[bleedblue1234]

Homework Statement

A thin, flexible metal rod has a length of 0.606 meters. One end of the rod is clamped to a table, and the other end can vibrate freely. What are the first three natural frequencies of the rod if the wave speed in the rod is 27.4 meters per second? Hint: Draw pictures of what the rod will look like when it is vibrating at the first three harmonics.

The Attempt at a Solution

How would the open end affect the motion? I am confused here...

 

[bleedblue1234]

Nevermind i figured this one out also

 

[kerimek]

I would like to clarify that the open end would not affect the motion of the rod. The vibration of the rod is determined by its length and the wave speed, not by the clamped end or open end. The open end simply allows for the free vibration of the rod, meaning that it can vibrate with larger amplitudes compared to a closed end.

To find the first three natural frequencies, we can use the formula f = n(v/2L), where f is the frequency, n is the harmonic number, v is the wave speed, and L is the length of the rod.

For the first harmonic (n=1), the rod will have one half-wavelength, with one node at the clamped end and one antinode at the open end. This gives us a frequency of f1 = (1)(27.4/2(0.606)) = 22.7 Hz.

For the second harmonic (n=2), the rod will have one full wavelength, with two nodes and one antinode. This gives us a frequency of f2 = (2)(27.4/2(0.606)) = 45.5 Hz.

For the third harmonic (n=3), the rod will have one and a half wavelengths, with three nodes and two antinodes. This gives us a frequency of f3 = (3)(27.4/2(0.606)) = 68.2 Hz.

These are the first three natural frequencies of the rod, and they represent the fundamental mode and the first two overtones. I hope this helps clarify your understanding of natural frequencies and their relationship to the length and wave speed of a vibrating object.