Question

A nearly plane water wave with a wavelength of 2.5 m passes through two narrow openings within a concrete pier. The openings are 6.6 m apart. Calculate the angle of the third nodal line.

Answer 1

ANSWER:

71.3°

STEP-BY-STEP EXPLANATION:

Given:

Wavelength (λ) = 2.5 m

Source separation (d) = 6.6 m

Nodal line number (n) = 3

We can calculate the angle using the following formula:

$\sin\theta=\left(n-\frac{1}{2}\right)\frac{\lambda}{d}$

We substitute each value and calculate the angle like this:

$\begin{gathered} \sinθ=\left(3-\frac{1}{2}\right)\frac{2.5}{6.6} \\ \\ \sinθ=0.94696 \\ \\ θ=\sin^{-1}(0.94696) \\ \\ θ=71.3\degree \end{gathered}$

The angle of the third nodal line. es 71.3°