At a certain harbor, the tides cause the ocean surface to rise and fall a distance d(from highest level to lowest level) in simple harmonic motion, with a period of$\mathbf{12}\mathbf{.5}\mathbf{\text{\hspace{0.17em}h}}$. How long does it take for the water to fall a distance$\mathbf{0}\mathbf{.250}\mathbf{\text{\hspace{0.17em}d}}$from its highest level?

The time period of oscillation,$T=12.5h$.

The total amplitude is half the distance from the highest level to the lowest level. Using this relation and the relation between angular velocity and the period of the SHM, we can find the time for the water to fall at a distance offrom its highest level.

Formula:

The general expression for the velocity of motion

$x=x_m\text{cos}(\omega t+f)$ (i)

The angular frequency of a body in motion

$\omega =\frac{2\pi }{T}$ (ii)

As$x_m=0.5d,x=0.250d$

Using equation (ii) and the given values, we get the angular frequency to be

$\begin{array}{c}\omega =\frac{2\pi }{12.5h}\\ =0.503\text{\hspace{0.17em}rad}/\text{h}\end{array}$

And, the phase constant$f=0,$ because $x_0=x_m$using equation (i), we get

$\begin{array}{c}0.250d=0.5d\cos \left(0.503t\right)\\ 0.5=\cos \left(0.503t\right)\\ t=2.08h\end{array}$

Therefore, thetime for the water to fall a distance$0.250d$ from its highest level is$2.08h$ .