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Force and acceleration in a vibrating wire

  1. Sep 30, 2014 #1
    1. The problem statement, all variables and given/known data
    A wire of length 1[m] vibrates with the base frequency which is 200[Hz]. the specific mass is 8[gr/cm3].
    The maximal acceleration at the middle is 80,000[cm/sec2]. what is the amplitude

    2. Relevant equations
    Newton's second law: [itex]F=ma[/itex]
    The cosine sentence: [itex]A^2=B^2+C^2-2BC\cdot \cos \alpha[/itex]
    The speed of transverse waves in a wire: [itex]u=\sqrt{\frac{P}{\rho}}[/itex]
    P is the stress and [itex]\rho[/itex] is the specific mass.
    The speed as a function of wavelength and frequency: [itex]u=\lambda f[/itex]

    3. The attempt at a solution
    The tension is the stress times the area. and the force i take from newton's law:
    ##F=100[cm]\cdot 8[gr/cm^3]\cdot a \cdot 80,000=64E6\cdot A[dyn]##
    The velocity:
    ##u=\lambda f\rightarrow 200=\frac{u}{2\cdot 100}\rightarrow u=40,000[cm/s]##
    The stress:
    ##u=\sqrt{\frac{P}{\rho}}\rightarrow 40,000=\sqrt{\frac{P}{8}}\rightarrow P=200E6[dyn/cm^2]##
    I drew the forces in the stretched wire, i find the resultant force:
    ##F^2=2T^2-2T^2\cos 2\alpha##
    ##(64E6)^2\cdot A^2=(200E6)^2A^2(1-\cos 2\alpha)##
    ##\rightarrow \alpha=9.21^0 \rightarrow h=8.1[cm]##
    It's a little high, no? is my solution correct?
     

    Attached Files:

  2. jcsd
  3. Oct 1, 2014 #2

    ehild

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    The wire of length 1 m vibrates with its fundamental frequency. It means a standing wave. What is the wavelength?
    The frequency is given. What is the speed of the wave? If you know the speed and the specific mass, you can determine the tension.

    The problem text asks the amplitude at the middle. What function is the displacement of the time there? How do you get the acceleration from displacement?

    ehild
     
  4. Oct 2, 2014 #3
    The displacement function:
    ##y=A\sin\left(\frac{2\pi}{\lambda}x\right)##
    ##y''=-\frac{4\pi^2 A}{\lambda^2}\sin\left(\frac{2\pi}{\lambda}x\right)##
    ##-80,000=-\frac{4\pi^2 A}{200^2}##
    And A comes out huge
     
  5. Oct 2, 2014 #4

    ehild

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    You calculated the second derivative of a the spatial part of the wave, with respect to the position.
    How is acceleration defined?

    ehild
     
  6. Oct 2, 2014 #5
    ##y=2A\sin(2\pi f t)\sin\left(\frac{2\pi x}{\lambda}\right)##
    ##y=2A\sin(2\pi\cdot 200t)##
    ##y''=-32,000\pi^2A\sin(400\pi t)##
    ##-80,000=-32,000\pi^2A\rightarrow A=0.253[cm]##
    Is it correct and is the result logical?
     
  7. Oct 2, 2014 #6

    ehild

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    Why is that factor 2 there? y means the displacement from the middle position, and A is the maximum displacement.
    Why did you change the frequency to 400? And what is the maximum acceleration in parametric form?

    ehild
     
  8. Oct 2, 2014 #7
    The equation:
    ##y=2A\sin(2\pi f t)\sin\left(\frac{2\pi x}{\lambda}\right)##
    I took from 2 waves travelling in opposite directions. when they interfere positively the amplitude doubles. i could take here only A.
    I didn't change the frequency to 400, it's 400X2 that is in the brackets.
    The acceleration in parametric form:
    ##y'=4\pi f A\cos(2\pi ft)\sin\left(\frac{2\pi x}{\lambda}\right)##
    ##y''=-8\pi^2f^2A\sin(2\pi f t)\sin\left(\frac{2\pi x}{\lambda}\right)##
    ##80,000=8\pi^2\cdot 200^2\cdot A\rightarrow A=0.0253[cm]##
    And that's too small, no?
     
  9. Oct 2, 2014 #8

    ehild

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    The middle of the wire performs simple harmonic motion. The problem asks the amplitude of that vibration. It is simply A. The wire does not know that its motion can be written as the sum of two waves travelling in opposite directions.
    And why do you think it is too small? Do you think, the strings of a violin or guitar has to jump off? :D
     
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