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My physical intuition is rather poor so I would appreciate any help.

If a string is vibrating and is heavier towards one end and gradually gets lighter towards the other, I find after plotting the eigenfunctions that the standing waves have a smaller amplitude and smaller wavelength on the heavier end. I assume the reason the amplitude is lower is because more energy/force is required to move a greater mass, but I am not sure why the wavelength is smaller. I know the formula $v=\lambda f$, and I suppose by the same argument the speed of the wave will be lower when the mass is greater but it isn't obvious to me that the frequency should be constant throughout the whole string. What is the best way to think about this?

katemars
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The wave velocity on a string is $v =\sqrt{T/\varrho}$ where $T$ is the tension and $\varrho$ mass per unit lenngth. Tension is the same everywhere in the string. So in lighter parts wave speed is faster, wavelength shorter.

  • But why doesn't the wavelength stay the same and the frequency just decrease? ._. – katemars Jan 22 '18 at 17:54
  • The frequency is given by the driving force, the oscillator. By the modes of oscillation of the system (could be a superposition of different modes). –  Jan 22 '18 at 18:28
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You assumed that the frequency of oscillation is constant down the length of the string when you assumed a standing wave; combining that criteria with boundary conditions (e.g. fixed string ends) results in standing waves.

You could have different frequencies of oscillation in different parts of the string if you drove the string that way with an outside oscillator, but it won't be an eigenstate.

cms
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  • No, you cannot have different frequencies in different parts of the string. If you drive it with a sum of different frequencies, some ports of a structure could vibrate with a higher amplitude at that frequency, but the other frequencies would also be present.. –  Jan 22 '18 at 18:43