Turn on thread page Beta
    • Thread Starter
    Offline

    12
    ReputationRep:
    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
    Posted on the TSR App. Download from Apple or Google Play
    • Thread Starter
    Offline

    12
    ReputationRep:
    Name:  E68DBCD1-8DB9-489D-AFA9-816760E3EF8F.jpg.jpeg
Views: 4
Size:  15.2 KB
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    10
    ReputationRep:
    Have you got a picture of the diagram it’s referring to?
    Posted on the TSR App. Download from Apple or Google Play
    • Thread Starter
    Offline

    12
    ReputationRep:
    Name:  EC33407D-D96E-4508-8BFB-2D465AFA9546.jpg.jpeg
Views: 7
Size:  36.7 KB
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    12
    ReputationRep:
    (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
    Offline

    10
    ReputationRep:
    Name:  9ECC6E03-CE05-4461-A258-44B9F90A1694.jpg.jpeg
Views: 7
Size:  44.6 KB
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    12
    ReputationRep:
    (Original post by MissMathsxo)
    Name:  EC33407D-D96E-4508-8BFB-2D465AFA9546.jpg.jpeg
Views: 7
Size:  36.7 KB
    Got it!
    • Thread Starter
    Offline

    12
    ReputationRep:
    (Original post by isaac_forde)
    Name:  9ECC6E03-CE05-4461-A258-44B9F90A1694.jpg.jpeg
Views: 7
Size:  44.6 KB
    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?
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    10
    ReputationRep:
    (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
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    12
    ReputationRep:
    here
    Offline

    12
    ReputationRep:
    Name:  Capture.PNG
Views: 4
Size:  5.9 KB
    Offline

    12
    ReputationRep:
    (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.
    Offline

    12
    ReputationRep:
    (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?
    Offline

    10
    ReputationRep:
    (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
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    10
    ReputationRep:
    This may help regarding the frequency and length. The length doesn’t change.
    Attached Images
      
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    12
    ReputationRep:
    (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?
    Offline

    12
    ReputationRep:
    (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.
    Offline

    10
    ReputationRep:
    (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
    Posted on the TSR App. Download from Apple or Google Play
    Offline

    12
    ReputationRep:
    (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?
    Offline

    10
    ReputationRep:
    (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
    Posted on the TSR App. Download from Apple or Google Play
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: May 25, 2018

2,744

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.