## Chapter 4

## Q. 4.9

Nonlocal forcing results in an oscillation of frequency ω_{0} whose amplitude at the mouth of a channel is Q. If the channel is of constant depth H and of length A, determine a general expression for the amplitude of the oscillation at the head of the channel. Calculate specific values for the case when H = 20 m, ω_{0} = 0.001 radians/sec, Q = 0.5 m and A = 20 km and for A = 21 km.

## Step-by-Step

## Verified Solution

The governing equation is

\frac{\partial²\eta }{\partial t^{2} } =c²\frac{\partial²\eta }{\partial x^{2} }where c = √(gH). The solution that satisfies the condition of no flow across the closed end of the channel has the form of a standing wave (see also Section 2.3.10) or a superposition of them. Choosing the origin, x = 0, to be at the closed end, the solution is

\eta =\eta _{0} \cos (kx) \cos (w_{0}t ),

where ω_{0} =kc. Now, at x = A, η = Q. Substituting these values into the above solution gives the amplitude of the oscillation at the head of the channel as η_{0} = Q sec(kA) = Q sec(A ω_{0} /√(gH)).

Substituting the numerical values gives η_{0} = 3.5 m for A = 20 km and η_{0} = 7.0 m for A = 21 km. The increase in amplitude is very large (resonance occurs) when the frequency of the forcing is close to one of the natural frequencies of oscillation. The remarkable tidal ranges found in the Bristol Channel and Bay of Fundy can be attributed to the near resonant response. Resonant responses can also occur in ports and harbours, sometimes exacerbated by the neat geometrical outlines adopted in such cases.