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AS physics. Waves

Could someone please explain part c of the following question. I don't understand why they used the frequency of the first harmonic in the calculation to find the tension in the string for the third harmonic. Or is the tension the same regardless of the harmonic and this equation does only work for first harmonic (as I previously thought)? Thank you

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Have you got a picture of the diagram it’s referring to?
EC33407D-D96E-4508-8BFB-2D465AFA9546.jpg.jpeg
Reply 4
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Original post by isaac_forde
Have you got a picture of the diagram it’s referring to?

True.

I tried to answer it and got 195.1 N :



=〖((2 × 0.66 ×330)/(0.91 ×2))〗^2 × (3.1 X 0.01)/0.91

IS this expression correct then?

T= f ^ 2 x m/L

so
m/L x (v/lambda)^2
(edited 5 years ago)
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Original post by MissMathsxo
EC33407D-D96E-4508-8BFB-2D465AFA9546.jpg.jpeg


Got it!
Original post by isaac_forde
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Yeah this is what the mark scheme had. I was just wondering why they used the first harmonic. Am I right to think this equation only works for the first harmonic and in this situation, the tension is the same for both the harmonics?
Original post by MissMathsxo
Yeah this is what the mark scheme had. I was just wondering why they used the first harmonic. Am I right to think this equation only works for the first harmonic and in this situation, the tension is the same for both the harmonics?


Yeah the equation can only be used for the first harmonic. That’s why you had to divide the thirds harmonics frequency by 3. The tension will stay the same for both harmonics, it’s only the frequency which changes
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here
Capture.PNG
Original post by MissMathsxo
Yeah this is what the mark scheme had. I was just wondering why they used the first harmonic. Am I right to think this equation only works for the first harmonic and in this situation, the tension is the same for both the harmonics?




You could try this method and see if it works.
Original post by isaac_forde
.The tension will stay the same for both harmonics, it’s only the frequency which changes


I'm confused; is this because the length changes?

Also, is the phase difference 2 pi rad?
Original post by Seppuku
I'm confused; is this because the length changes?

Also, is the phase difference 2 pi rad?


Remember frequency of a wave is defined by ‘the number of crests of a wave past a point per second.’ The phase difference between A and B is pi rad as they’re exactly 1/2 wavelength apart. 1 wavelength = 2pi
This may help regarding the frequency and length. The length doesn’t change.
(edited 5 years ago)
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Original post by isaac_forde
This may help regarding the frequency and length. The length doesn’t change.

So if T doesn't change and length doesn't change what does?
Original post by isaac_forde
This may help regarding the frequency and length. The length doesn’t change.


Why is lambda 0.44, I though it was 0.66m

Damn it I hate this topic.
Original post by Seppuku
So if T doesn't change and length doesn't change what does?


You’re trying to show that T is a certain value. You are given the frequency of the 3rd harmonic and have to change the frequency to the 1st harmonic then rearrange the equation f = 1/2L (T/μ)^(.5) to make T the subject. Then solve for T and show it is approx. 70N
Original post by isaac_forde
This may help regarding the frequency and length. The length doesn’t change.


Oh so the formula is fnew = fundamental f/loops?
Original post by Seppuku
Why is lambda 0.44, I though it was 0.66m

Damn it I hate this topic.


Lambda is one complete wavelength. The figure shows 3/2 wavelengths

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