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So I have seen this exam question where water waves in a ripple tank are being used to demonstrate transverse waves.

enter image description here

The question says:

The frequency of the waves is increased. Describe what happens to the speed and the wavelength of the waves.

The answer to the question is that the speed would stay the same and the wavelength would decrease. However, I am struggling to see why this must be the case.

For example, could one not have the speed increase and the wavelength stay the same?

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The speed of waves such as this depends on the properties of the medium in which it is propagating, and not the source. The frequency, however, depends on the source producing the waves.

The statement $v = f \lambda$ is a "counting" formula (there's no new physics in it, simply the definition of speed, frequency, and wavelength) and since the medium hasn't changed, if $f$ increases, $\lambda$ must decrease.


EDIT: What I've written above is -- of course -- an approximation which is true to the spirit in which the question was asked. The real answer is more complicated: it depends very much on the type of waves that are formed on the water, whether they are "shallow" or "deep". It can be shown that the wave speed in a medium is given by

$$v^2 = \frac{g\lambda}{2\pi} + \frac{2\pi\gamma}{\lambda \rho},$$

where $g$ is the acceleration due to gravity, $\gamma$ the surface tension, $\rho$ the density of the water, and $\lambda$ the wavelength of the wave. This shows that in the general case, the wave's velocity will depend on the wavelength.

It turns out that if we are only dealing with shallow water waves that move only on the surface of the water, whose amplitude is much smaller than their wavelength, the above formula can be approximated to:

$$v_\text{shallow} = \sqrt{gh},$$

which is a constant. See here for more details.

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  • $\begingroup$ Perfect, thank you $\endgroup$
    – code noob
    Commented Aug 17, 2020 at 12:16
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    $\begingroup$ @codenoob You're welcome. The actual answer is a little more complicated, I'm making an edit right now ;) $\endgroup$
    – Philip
    Commented Aug 17, 2020 at 12:17

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