Find two lowest frequencies standing wave

[late347]

Homework Statement

A metal bar is attached into a vice in the middle. Bar is hit with a hammer creating a longitudinal standing wave
find the two lowest frequencies.
Bar's length l= 3m
wave velocity = 5100m/s

Homework Equations

wave equation $v = \lambda * f$

The Attempt at a Solution

Our teacher told us that the way to solve these kind of problems is to correctly affix antinode and node points, and these will indeed be the fixed boundary conditions. Using those conditions we will determine the lowest frequency first, for example. The boundary conditions should not be violated when finding the frequencies I seem to remember.

There must be antinodes at each end of 3m bar. And there's the node at 1.5m always in the center.

using this knowledge, the distance between an antinode and another antinode is $\lambda /2$

$\lambda / 2 = l$
$\lambda = 6m$
$\frac{v}{\lambda} =f$
f= 850 Hz

The second lowest frequency.

My question is essentially, does there exist a more analytical way of finding these upper frequencies? Or is it simply just supposed to be done by attempting different solutions with intuition and finding the next frequency step by step.

I looked at this source and it only gave a formula for the case of the string in a musical instrument (starting with the boundary conditions of node at each endpoint) http://www.physicsclassroom.com/class/sound/Lesson-4/Fundamental-Frequency-and-Harmonics

with some trial and error
Honestly I was a little bit confused about this second lowest frequency when doing this problem. I was initially confused about what is the valid "next" frequency and how it is suppsoed to be obtained.
For example, is this frequency supposed to have an affixed node also in the middle at about 1.5m of the metal bar, similar to the base frequency?

If this is true, why doesn't the other frequency example of a guitar string work in a similar fashion such that the guitar string has nodes at each end at the lowest frequency. But the guitar string also has the antinode in the center at this lowest frequency.

http://www.physicsclassroom.com/class/sound/Lesson-4/Fundamental-Frequency-and-Harmonics
So why doesn't the guitar string have the antinode affixed in the center in the next frequency (second harmonic) ?

I put the next frequency (second lowest)
such that you keep the node at 1.5m and antinodes at the ends.
But you add a node and an antinode on both sides of the central node. So there's A-N-A-N-A-N-A

And using this kind of setup we have three times the length between antinodes

$3* \lambda/2 = l$
$\lambda = \frac{2l}{3}$
$\lambda=2m$
$f = \frac{v}{\lambda} => f = 2550 Hz$

 

[TSny]

There must be antinodes at each end of 3m bar. And there's the node at 1.5m always in the center.

Yes. The vice clamps the rod at the center which prevents the rod from vibrating longitudinally at that point. So, there must be a node at the center for all of the possible standing waves. That's why you still need to assume a node in the middle when you are calculating the second frequency.

Your calculations for both the first and second frequencies look correct.

If you went on to do the next frequency, I'm sure you would see the pattern so that you could state all of the possible frequencies.

 

[late347]

Yes. The vice clamps the rod at the center which prevents the rod from vibrating longitudinally at that point. So, there must be a node at the center for all of the possible standing waves. That's why you still need to assume a node in the middle when you are calculating the second frequency.

Your calculations for both the first and second frequencies look correct.

If you went on to do the next frequency, I'm sure you would see the pattern so that you could state all of the possible frequencies.

For the guitar string case. The string is fixed at the ends, but when finding the 2nd harmonic from the base freq, you're still supposed to add the third node into the middle portion. And Then add correpondingly add another antinode.

But you're still saying that in the guitar string, it does seem indeed that the the buondary conditions are only really the nodes at the endpoints? This was a little bit confusing to find out exactly what must be followed, so that you know how many antinodes and nodes to add for the next frequency (I can usually get the basic case correct, the lowest frequency).

I'm not much of a musician, but I think youre supposed to put your finger on the guitar string to act as the central node in the 2nd harmonic frequency?!

 

[TSny]

Putting your finger on the guitar string helps to produce the second harmonic. But in principle, it is not necessary. If you have a rope that is stretched out and fixed at one end, and you hold the other end and wiggle it at the frequency of the 2nd harmonic, then you will set up the 2nd harmonic standing wave without needing any physical contact at the middle of the rope. The node in the middle will just naturally form.

 

[late347]

Putting your finger on the guitar string helps to produce the second harmonic. But in principle, it is not necessary. If you have a rope that is stretched out and fixed at one end, and you hold the other end and wiggle it at the frequency of the 2nd harmonic, then you will set up the 2nd harmonic standing wave without needing any physical contact at the middle of the rope. The node in the middle will just naturally form.

yea standing waves are easier to visualize when they are transverse standing wave. Isn't it just the sum wave? But this one in our example was supposed to be longitudinal standing wave. Do you know or have any good studying materials for these, like an animation or something?

 

[TSny]

I just found this one
http://www.acs.psu.edu/drussell/Demos/StandingWaves/StandingWaves.html

If you go down to the middle of the page you will see the animation of a particular longitudinal standing wave (not the fundamental). You can think of it as representing the left half of the clamped rod. So, you have an antinode on the left and a node on the right.