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Tuning Fork - Simple Harmonic Motion

  1. Feb 2, 2010 #1
    This topic has proved itself to be a hard one in, in terms of looking it up online. I'm interested in simple harmonic motion, in specific that of a tuning fork vibrating between two electromagnetic devices, a microphone ad a detector.
    My main interest in it is to write a lab report about an experiment I have conducted on a tuning fork.

    I understand physically the notion of resonance (a frequency such that it pushes the fork always in just the right time to amplify it vibrations) but the mathematical side is somewhat difficult. So in terms of the electromagents in relation with the turning fork, can anyone please explain it to me in a bit more mathematical sense?

    Also the effect of damping by placing a magnet in between the forks. Thanks.
     
  2. jcsd
  3. Feb 2, 2010 #2
    In a tuning fork, just like a simple pendulum, energy is being exchanged between motion and stored energy. In the case of the pendulum, the stored energy is gravitational: mgh. In the case of the tuning fork, it is like a spring where F = kx. The stored energy is: ½kx2, where k is the spring constant and x is displacement from equilibrium point.

    Bob S
     
  4. Feb 2, 2010 #3
    That much I know about the tuning fork. My point is the ODE of the harmonic oscillator. In fact we have the ODE
    [tex] m\frac{d^2x}{dt^2} + b\frac{dx}{dt} + kx = F cos(wt) [/tex]

    Which leads eventually (with damping to)
    [tex] x = Ae^-^\gamma ^t cos(w_1 t + \theta) + \frac{Fsin(wt+ \theta _0)m}{[(w^2_0 - w^2)^2 +4\gamma^2 w^2]^1/2}[/tex]
     
    Last edited: Feb 2, 2010
  5. Feb 2, 2010 #4
    I don't know why the Latex failed but anyway it is quite a complicated equation with a damping effect and the part without the damping. My question was how can you get to the ODE in the first and keep going and second how does the damping come in.
     
  6. Feb 2, 2010 #5
    The tuning fork is not a forced oscillator, so the solution is

    [tex] x = Ae^-^\gamma ^t cos(w_1 t + \theta) [/tex]

    Bob S
     
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