Chapter 16: Q85. (page 454)

A water wave is a shallow-water wave if the water depth d is less than ≈ l/10. It is shown in hydrodynamics that the speed of a shallow-water wave is v = 1gd, so waves slow down as they move into shallower water. Ocean waves, with wavelengths of typically 100 m, are shallow-water waves when the water depth is less than ≈ 10 m. Consider a beach where the depth increases linearly with distance from the shore until reaching a depth of 5.0 m at a distance of 100 m. How long does it take a wave to move the last 100 m to the shore? Assume that the waves are so small that they don’t break before reaching the shore

Short Answer

Expert verified

t = 28.571 sec

Step by step solution

Given Info:

$v = \sqrt{g d}$

Depth increases from 0 to 5m from distance 0 to 100m.

Slope of this surface = $\frac{5}{100} = 0.05$

If x is the distance from shore, depth of water can directly be found as d(x) = 0.05x

similarly velocity can be stated as a function of x as $v = \sqrt{g d (x)}$

Finding time

We know that time taken can be found as integral of inverse of velocity.

$\Rightarrow t = \int_{0}^{100} \frac{1}{v} d x$

Substituting values in above equation

$t = \int_{0}^{100} \frac{1}{\sqrt{g d (x)}} d x$

$\Rightarrow t = \int_{0}^{100} \frac{1}{\sqrt{g \times 0.05 x}} d x \Rightarrow t = \frac{1}{\sqrt{0.05 g}} \int_{0}^{100} \frac{1}{\sqrt{x}} d x \Rightarrow t = \frac{1}{\sqrt{0.05 g}} {[2 \sqrt{x}]}_{0}^{100} \Rightarrow t = \frac{1}{\sqrt{0.05 g}} \times 20 \Rightarrow t = 28.571 s e c$

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Finding time

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