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Trouble with waves 1D

  1. Mar 26, 2013 #1
    1. A given string length L, connected at one end while the other end is free. Assume that the string moves in one dimensional, ie, the amplitude can be described by a function of the shape of his y (x, t).
    What is the average kinetic energy (as a string), assuming it moves with the lowest frequency possible?
    Given that when the string is perfectly horizontal its tension = To and amplitude motion is A.
    T0 = 2.76 [gram*cm/sec2]
    L = 5.25 [cm]
    A = 5.86 [cm]


    2. <sin2(at)>=<cos2(at)>=1/2. (average)

    3. With boundary conditions find the frequency of the wave vector, and then calculate the kinetic energy
    1. The problem statement, all variables and given/known data
     
  2. jcsd
  3. Mar 26, 2013 #2

    Simon Bridge

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    Welcome to PF;
    I see three questions there - how have you attempted to answer them?
     
  4. Mar 26, 2013 #3
    hey

    its only one question not three... I have tried to find the frequency using: y(0,t)=0
    y=A*cos(kx+wt)
    but then I dont know how to find the σ to use: Ek= σ*(A^2)*(ω^2)/2T * ∫cos^2(kx+wt)

    thnx
     
  5. Mar 26, 2013 #4

    Simon Bridge

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    Oh I see, the numbers refer to the sections in the standard question template... so "3" is your attempt.

    The question refers to a standing wave on a string length L, with only one end fixed.
    The lowest frequency standing wave has a special name - what is it called?
    You should be able to work out the shape of that wave and, thus, the wavelength, and from that and the wave speed, you get the frequency.
     
  6. Mar 26, 2013 #5
    I think it called the fundamental.
    the shape of the wave is y=Acos(kx-wt) (asuming that the left edge of the string is stable). then I know that y(0,t)=0. the wave length is 2*L ? but how do I use the tension...
     
  7. Mar 26, 2013 #6

    Simon Bridge

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    By that equation, y(0,t)=Acos(wt) ... i.e. the x=0 end is not fixed, totally contradicting what you write below. Also according to that equation, the amplitude is the same for any value of x - does that make sense?

    Wavelength would be 2L if the string were fixed at both ends. Is it?

    The tension and the linear mass density are used to calculate the wave speed.
    That's the next step after getting the wavelength.
    You should have notes about this.
     
  8. Mar 26, 2013 #7
    I think I should use: y(x,t)= A*e^i(kx+wt) + B*e^-i(kx+wt)
    nd then y(0,t)=0 should I also try dy/dt =0 ?
    its all missed up Im tring to solve this question more than two days.. the problem is I dont have material to stdy about this type of questions... do u have some examples or some good material that I should read befor trying solving this?
    thnx alot!
     
  9. Mar 26, 2013 #8
    1. The problem statement, all variables and given/known data
    A given string length L, connected at one end while the other end is free. Assume that the string moves in one dimensional, ie, the amplitude can be described by a function of the shape of his y (x, t).
    What is the average kinetic energy (as a string), assuming it moves with the lowest frequency possible?
    Given that when the string is perfectly horizontal its tension = To and amplitude motion is A.
    T0 = 2.76 [gram*cm/sec2]
    L = 5.25 [cm]
    A = 5.86 [cm]

    2. Relevant equations
    <sin2(at)>=<cos2(at)>=1/2. (average)

    3. The attempt at a solution
    With boundary conditions I should find the frequency of the wave vector, and then calculate the kinetic energy.. right? but how! "I should be able to work out the shape of that wave and, thus, the wavelength, and from that and the wave speed, you get the frequency."
     
  10. Mar 26, 2013 #9

    Simon Bridge

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    Good grief, whatever for?!

    I guess you need a primer for standing waves ... putting standing waves into a search engine should get you any number.

    A string fixed at both ends, the waveform has a node at each end and an antinode in the middle. ##\lambda=2L##
    http://hyperphysics.phy-astr.gsu.edu/hbase/waves/string.html

    A string fixed only at one end has a node at that end and an antinode at the other end.
    The math is the same as the link above, but the wavelength is longer. You should be able to see by sketching it out.
     
  11. Mar 26, 2013 #10
    I think you need the linear density of the string to get the wave speed.
     
  12. Mar 26, 2013 #11
    oh so in my case wavelength= 4*L !! right?!!
     
  13. Mar 26, 2013 #12

    Simon Bridge

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  14. Mar 26, 2013 #13

    Simon Bridge

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    Well done!
    The same webpage tells you how to get the wave speed too.
    From the wave speed and the wave length you can get the frequency, period, and angular frequency.
    Average kinetic energy is a bit trickier.
     
  15. Mar 26, 2013 #14

    berkeman

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    (Two threads merged into one)
     
  16. Mar 26, 2013 #15
    ok now I have a problem, I think the wave speed is the same as the two sides fixed string... if Im right, then I need the density to work it out but I dont have it!!
     
  17. Mar 26, 2013 #16

    Simon Bridge

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    Yeah - no density ... but you have a better picture of what is going on than before.
    The equation would be: ##y(x,t)=A\sin(\pi x/2L)\cos(\omega t)## - but you don't know ##\omega##.
    You need extra information to get it - check that what you have supplied in post #1 is everything.
     
  18. Mar 26, 2013 #17
    I think I should use this: "Given that when the string is perfectly horizontal its tension = To and amplitude motion is A." ??
     
  19. Mar 27, 2013 #18
    The prof. says: "Mu is is the mass per unit volume , if you look at what;s given in the question you can see that you can make it up from the given data as well"
    can u give me a hint?
     
  20. Mar 27, 2013 #19

    Simon Bridge

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    OK "mass per unit volume" is the density.
    Do you know the volume of the string? (You have the length - you need the diameter or the radius or the cross-sectional area.)
    (note: this is stuff that is not in post #1 - please make sure that all the information you are supplied is present.)

    If you can get the area, then you have the mass per unit length.
    From there you can find the wave-speed ... and then the frequency.
     
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