What happens to wavelength in a ripple tank when frequency is doubled?

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When the frequency of a water wave in a ripple tank is doubled, the wavelength is indeed halved, assuming the depth of the water remains constant. In shallow water, the speed of the waves is determined solely by the water depth and gravitational acceleration. The relationship between wave speed, frequency, and wavelength is expressed as v = freq x wavelength. Therefore, if the frequency increases, the wavelength must decrease to maintain the equation's balance. This confirms that in shallow water, doubling the frequency results in halving the wavelength.
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If the frequency of a water wave is doubled in a ripple tank, is the wavelength halved? The depth of water is kept constant.
 
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Yes. In shallow water, the speed of the water waves is dependent only on the depth (and the gravitational acceleration g).

The general expression for the speed of a water wave is given by:
v = \sqrt{\frac{g\lambda}{2\pi}\,tanh(2\pi\frac{d}{\lambda}})

which simplifies in the shallow depth limit into
v = \sqrt{gd}
 
so,
v=freq x wavelength

Thus double the frequency the wavelength is halved in shallow water.
 
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