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anonymous

  • 4 years ago

A string of length 64cm and mass 3g is fixed at both ends. The string vibrates in its third harmonic with a maximum displacement of 1.9mm. The speed of transverse waves along the string is 462m/s. what is the wavelength and the frequency? also what is the amplitude?

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  1. anonymous
    • 4 years ago
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    hold up

  2. anonymous
    • 4 years ago
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    if we solve for fundemental frequency by using f=v/2L

  3. anonymous
    • 4 years ago
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    Amplitude; Wavelength; Frequency or light waves, when you change the frequency, you also change the wavelength. The term wavelength is also sometimes applied to modulated waves, and to the fixed wave speed, wavelength is inversely proportional to frequency: waves with as the wave slows down, the wavelength gets shorter and the amplitude .

  4. anonymous
    • 4 years ago
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    For a given wavelength, gravity waves in deeper water have a larger phase . Besides frequency dispersion, water waves also exhibit amplitude dispersion.

  5. anonymous
    • 4 years ago
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    do u get it?

  6. anonymous
    • 4 years ago
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    ok, i dont think it is hard to solve for wavelength or frequency we just do what i said previously, BUT for amplitude, we need to come up with an equation in the form of: y(x,t)=2Asin(kx)cos(wt) any ideas?

  7. anonymous
    • 4 years ago
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    your wavelength for the 3rd harmonic will be 2/3*L, finding L from your fundamental frequency equation. Then with your new lambda you can get the frequency for the 3rd harmonic using f = v/lambda...

  8. anonymous
    • 4 years ago
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    and Mythias can u help me

  9. anonymous
    • 4 years ago
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    u can get it throught the first way, the second thing is important because that f will calculate our period to solve omega. we can use that frequency should work as well. also we need to find k, which will need the second omega. ok i think we pretty much answered the question. just calculation time, i submit and then moment of truth :P

  10. anonymous
    • 4 years ago
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    k is just your 2pi/lambda

  11. anonymous
    • 4 years ago
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    good luck!

  12. anonymous
    • 4 years ago
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    lol thx man

  13. anonymous
    • 4 years ago
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    can someone help me

  14. anonymous
    • 4 years ago
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    don't see your question

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