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Homework Help: Increase in frequency of tuning forks

  1. May 29, 2006 #1
    1)When one of the prongs of a tuning fork is cut, why does its frequency increase? Is it because the damping forces exerted by the prongs on each other decreases?
    2)When an incident wave and its reflected wave superimpose, it is said that both standing waves and progressive waves are present in the medium. Is it true?
  2. jcsd
  3. May 29, 2006 #2
    I'd like to believe that waves are set up between the two prongs which are sustained by the vibrations of either prong as they lose energy to the surrounding medium. Damping does play a role yes, but if you cut one prog, you effectively reduce one source of waves. But if OP have a better explanation, then you should trash this one.

    Yes in general the incident wave and reflected wave differ in amplitude (this happens because of the difference in densities of the two media) and so the wave generated by their superposition has a traveling wave component in addition to a standing wave component. So it is not a "true" standing wave in the sense that nodes no longer have zero amplitude. You can work this out taking the two wavefunctions

    [tex]y_{I}(x,t) = A_{I}sin(k_{1}x-\omega t)[/tex]
    [tex]y_{R}(x,t) = A_{R}sin(k_{1}x+\omega t)[/tex]

    and adding them (don't worry how the amplitudes are related for now....also don't worry about the phase...just add them).
  4. May 29, 2006 #3
    1) Hmm..Not sure about this but I'd say decreasing the length of the prongs would increase the frequency, the same way as decreasing the length of a simple pendulum would increase its frequency .
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