Find the periods of free oscillations in the following cases:

1. A long narrow lake of uniform depth 15 m and length 5 km.

2. An open-ended channel of depth 20 m and length 10 km.

Step-by-Step

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In the first case we may use the 1-dimensional version of Equation 4.55

T=\frac{2}{\sqrt{g h} \sqrt{\left(m^2 / A^2\right)+\left(n^2 / B^2\right)}} \quad(m=0,1, \ldots ; \quad n=0,1, \ldots) (4.55)

as the lake is narrow. The periods of free oscillations are given by

T=\frac{2}{\sqrt{gh} }\frac{A}{m} \left(m=0,1,…\right)\\ =\frac{10,000}{m\times 12.1} \\=826.4,413.2,275.5,206.6,…\left(sec\right).In the second case as the channel is open-ended we use Equation 4.56, giving the periods of free oscillation as

T\frac{4A}{\sqrt{gh}\left(2m+ 1\right) } \left(m=0,1,…\right) (4.56)\\=\frac{40,000}{\left(2m+ 1\right)\times 14.0 }\\ =2857,952.4,571.4,408.2,…\left(sec\right).Question: 4.1

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