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AS physics. Waves

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Original post by isaac_forde
Lambda is one complete wavelength. The figure shows 3/2 wavelengths


How ? It's got 3 loops.
Original post by Seppuku
How ? It's got 3 loops.


2 ‘loops’ = 1 wavelength
Original post by isaac_forde
2 ‘loops’ = 1 wavelength


Oh okay. Damn. I didn't see that. Oh I got it
Original post by isaac_forde
2 ‘loops’ = 1 wavelength

But if Tension and length are constant, does speed and frequency change>
Original post by Seppuku
But if Tension and length are constant, does speed and frequency change>


You’re trying to work out the tension!!!! Frequency changes as harmonic changes
Original post by isaac_forde
You’re trying to work out the tension!!!! Frequency changes as harmonic changes


Oh. Okay.

But I'm saying, if you don't change any of the weights on the pulley, but do change frequency.
Original post by Seppuku
Oh. Okay.

But I'm saying, if you don't change any of the weights on the pulley, but do change frequency.


It’s a guitar not a pulley system
Original post by isaac_forde
It’s a guitar not a pulley system


Don't the same principles apply?

I feel cheated by the revision guides and textbooks and these sorts of gaps in my knowledge stress me out, honestly. Please help me.
@isaac_forde

Please explain why you used 0.66 as L here:


I thought L was the length of the whole string.
(edited 5 years ago)
Original post by Seppuku
@isaac_forde

Please explain why you used 0.66 as L here:


I thought L was the length of the whole string.


The distance between the two fixed point (x and y) is 0.66m so you use 0.66m as the distance for L
Original post by isaac_forde
The distance between the two fixed point (x and y) is 0.66m so you use 0.66m as the distance for L

So what is L defined as? Node distance?
Original post by Seppuku
So what is L defined as? Node distance?


As the equation only works for first harmonic then the distance of the string will be between the nodes as there’s two nodes and one antinode with the first harmonic
Original post by isaac_forde
As the equation only works for first harmonic then the distance of the string will be between the nodes as there’s two nodes and one antinode with the first harmonic


So that explains diagrams A and B. Is there any reason why this is tru? Thanks anyway.

This is another way of looking at it. It is probably an easier way of visualising it EE0E4109-3E8E-4A48-BDD4-126228DB8381.jpg.jpeg
Original post by Seppuku
So that explains diagrams A and B. Is there any reason why this is tru? Thanks anyway.



Why what is true?
Original post by isaac_forde
Why what is true?


Why do you have to use the fundamental frequency?
Original post by isaac_forde
Why what is true?


Alright I've understood it now. Thanks.
Original post by Seppuku
Why do you have to use the fundamental frequency?


You don’t. As long as you change the 1 to which ever frequency you’re using. I only just realised this when looking at it the formula for the second time. The third harmonic’s frequency is 3x the fundamental frequency
Original post by isaac_forde
You don’t. As long as you change the 1 to which ever frequency you’re using. I only just realised this when looking at it the formula for the second time. The third harmonic’s frequency is 3x the fundamental frequency


Yep, so in short:

nth frequency = number of nodes x fundamental frequency

So, fundamental frequency = nth frequency/number of nodes

Correct?
Original post by Seppuku
Yep, so in short:

nth frequency = number of nodes x fundamental frequency

So, fundamental frequency = nth frequency/number of nodes

Correct?


This is correct however the third frequency would have 4 nodes. Two at the ends and two in between. And the first harmonic would have 2 nodes. So saying number of nodes wouldn’t work. It would be the number of nodes minus 1

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