Wave: Harmonics Question

[justadream]

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Wikipremed PF #220

How can you assume from the information provided that the current frequency is the 2nd harmonic?

 

[Cawolf]

With one fixed end you can fit a certain amount of standing waves on it.

n = 1 you get 1/4 wavelength

n = 3 you get 3/4 wavelength (this is what we see in the picture)

n = 5 you get 5/4 wavelength

I would just see in the picture that there is 3/4 of a wavelength, so n=3, the next one is n=5.

 

[sxld_cu]

Oh, this is an interesting question...
At first, I took at look at it and thought the same thing as you; I also thought that the frequency should be based off the formula f = nv/4L for strings.
It turns out there are actually many different cases for strings:

Case 1. 1D - 2 Fixed ends: Lambda = 2L/n and f = nv/2L
Case 2. 1D - 2 Free ends: Lambda = 2L/n and f = nv/2L
Case 3: 1D - 1 Fixed end and 1 Free end: Lambda = 4L/n and f = nv/4L

In this problem, it would be Case 3 since one end is attached to the peg. You can use the formula Lambda = 4L/n; since we know that the string only goes through 3/4 of the wavelength, you can multiply both sides by 3/4 to give (3/4)Lambda = 3L/n and you know the length L is fixed, (3/4)Lambda must still equal L (even though it only goes through 3/4 of the wavelength, it still uses the whole length L) so n = 3. Another way to think about it is to just rearrange the formula to give n=4L/Lambda which will give you the sequence that @Cawolf mentioned

Hope that makes sense! This was actually a great problem to think through and helped me to understand the situation better too 🙂 Thanks! Where did you find these questions?

 

[sxld_cu]

Also, just of curiosity, @Cawolf I saw in your post, that you are in a Post-Bac program. I was also thinking about Post-Bac programs, so that caught my interest. If I may ask, where are you attending and what do you think of it so far? 🙂

 

[sxld_cu]

For ref: http://physics.info/waves-standing/

 

[Cawolf]

@sxld_cu I will PM you.