Question

A tsunami is a sort of pulse or “wave packet” consisting of several crests and troughs that become dramatically large as they enter shallow water at the shore. Suppose a tsunami of wavelength 235 km and velocity \(550\;{\rm{km/h}}\) travels across the Pacific Ocean. As it approaches Hawaii, people observe an unusual decrease of sea level in the harbors. Approximately how much time do they have to run to safety? (In the absence of knowledge and warning, people have died during tsunamis, some of them attracted to the shore to see stranded fishes and boats.)

Short answer

The people have approximately \(13\;\min \) to run to safety from tsunami.

Step-by-step solution

Given data

The wavelength of the tsunami is \(\lambda = \;235\;{\rm{km}}\).

The velocity of tsunami is \(v = \;550\;{\rm{km/h}}\).

Concepts

The unusual decrease in sea level is analogous to the wave trough, and the increase in the sea level is analogous to the wave crest.

The wavelength is the distance between a trough and a crest or two adjacent crests or two adjacent troughs.

Calculations

The unusual decrease in sea level (trough) during the tsunami is followed by a sudden sea level increase (crest). The people have to run to safety before the crest of the wave reaches the shore.

Now, the wavelength of the crest is given by,

\(\begin{aligned}{c}d\; &= \;\frac{\lambda }{2}\\ &= \;\frac{{235}}{2}\\ &= \;117.5\;{\rm{km}}\end{aligned}\)

The velocity is the rate of change of time. Let \(t\) be the time taken for the crest to travel a distance \(d\). Then,

\(\begin{aligned}{l}v\; = \;\frac{d}{t}\\t\; = \;\frac{d}{v}\end{aligned}\)

Substituting the values in the above expression, we get

\(\begin{aligned}{c}t\; &= \;\frac{{117.5\;{\rm{km}}}}{{550\;{\rm{km/h}}}}\\ &= \;0.214\;{\rm{h}}\\ &= 12.82\;\min \\ &\approx \;13\;\min \end{aligned}\)

Therefore, the people have approximately \(13\;\min \) to run to safety.