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# Waves watch

1. 1st Harmonic - 20Hz - F0 -wavelength 2L
2nd Harmonic - 40Hz - 2F0 -wavelength L
3rd Harmonic - 60Hz - 3F0 - wavelength 2/3L

Can someone explain this how does the frequency relate to the wavelength as the harmonic increases?

If the Length of the string is L how does frequency effect this? How do you work out the wavelength?
2. ..
3. (Original post by Acrux)
1st Harmonic - 20Hz - F0 -wavelength 2L
2nd Harmonic - 40Hz - 2F0 -wavelength L
3rd Harmonic - 60Hz - 3F0 - wavelength 2/3L

Can someone explain this how does the frequency relate to the wavelength as the harmonic increases?

If the Length of the string is L how does frequency effect this? How do you work out the wavelength?
The equation to remember is:

wavelength / 2 = L

So if we have 3 fll wavlengths in the tube (standing waves), then it becomes:

3wavelengths / 2 = L
Rearrange to get wavlength:

Wavelength = 2L/3

HIgher harmonic means more wavelengths in the same tube/length, so the frequence will increase as there are more per unit time/length.

So: wavelength decreases as frequence increases.
4. (Original post by derpz)
The equation to remember is:

wavelength / 2 = L

So if we have 3 fll wavlengths in the tube (standing waves), then it becomes:

3wavelengths / 2 = L
Rearrange to get wavlength:

Wavelength = 2L/3

HIgher harmonic means more wavelengths in the same tube/length, so the frequence will increase as there are more per unit time/length.

So: wavelength decreases as frequence increases.
Is this the seperation between the adjacent nodes?

Secondly how do workout how many waves there are? So for the 3rd harmonic there is 3 wavelengths?
5. (Original post by Acrux)
Is this the seperation between the adjacent nodes?

Secondly how do workout how many waves there are? So for the 3rd harmonic there is 3 wavelengths?
The seperation is between adjacent nodes or antinodes.

For the second question, do you mean how many ful wavenlengths, because there is 1 standing wave, but 1.5 wavlengths. Im assuming you got 3 which is the number of full circles? Remember that one full circle is only half a wavelength, so 3 full circles = 3/2 wavelengths.
6. (Original post by derpz)
The seperation is between adjacent nodes or antinodes.

For the second question, do you mean how many ful wavenlengths, because there is 1 standing wave, but 1.5 wavlengths. Im assuming you got 3 which is the number of full circles? Remember that one full circle is only half a wavelength, so 3 full circles = 3/2 wavelengths.
Yes i mean full wavelength how are you getting 3 wavelengths? Yh i dont understand the working behind that
7. (Original post by Acrux)
Yes i mean full wavelength how are you getting 3 wavelengths? Yh i dont understand the working behind that
Think I've confused myself, which harmonic are we looking at?
8. (Original post by derpz)
Think I've confused myself, which harmonic are we looking at?
3rd harmonic has 4nodes and 3 Antinodes
9. (Original post by Acrux)
3rd harmonic has 4nodes and 3 Antinodes
The way I do it, I just look at the whole number of circles which in this case is 3 and divide it by 2 since each one it half a wavlength, sorry about the confusion
10. (Original post by derpz)
The way I do it, I just look at the whole number of circles which in this case is 3 and divide it by 2 since each one it half a wavlength, sorry about the confusion
3rd harmonic has 4nodes and 3 Antinodes
So one wavelength is 3/2 right

For the 4th harmonic one wavelength is 1/2

but for the 5th how many is one wavelength? How do you work this out
11. (Original post by Acrux)
3rd harmonic has 4nodes and 3 Antinodes
So one wavelength is 3/2 right

For the 4th harmonic one wavelength is 1/2

but for the 5th how many is one wavelength? How do you work this out
I thought you only need to know up to the 3rd harmonic. Also, I think it depends on if its an open ended or close ended tube
12. (Original post by derpz)
I thought you only need to know up to the 3rd harmonic. Also, I think it depends on if its an open ended or close ended tube

well its say the L is 2/5L
13. (Original post by Acrux)
well its say the L is 2/5L
So λ= 2/5L

Therefore
/2 = L
So the wavelength is 2.5

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