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Question Number 140870 by jlewis last updated on 13/May/21

Ocean waves are observed to travel along the water surface during a developing storm. A Coast Guard weather station observes that there is a vertical distance from high point to low point of 4.6 meters and horizontal distance of 8.6 meters between adjacent crests.The waves splash into the station once every 6.2 seconds Determine the frequency and the speeed of these waves.

$${O}\mathrm{cean}\:\mathrm{waves}\:\mathrm{are}\:\mathrm{observed}\:\mathrm{to} \\ $$$$\mathrm{travel}\:\mathrm{along}\:\mathrm{the}\:\mathrm{water}\:\mathrm{surface} \\ $$$$\:\mathrm{during}\:\mathrm{a}\:\mathrm{developing}\:\mathrm{storm}. \\ $$$$\mathrm{A}\:\mathrm{Coast}\:\mathrm{Guard}\:\mathrm{weather}\:\mathrm{station} \\ $$$$\:\mathrm{observes}\:\mathrm{that}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{vertical}\: \\ $$$$\mathrm{distance}\:\mathrm{from}\:\mathrm{high}\:\mathrm{point}\:\mathrm{to}\:\mathrm{low} \\ $$$$\:\mathrm{point}\:\mathrm{of}\:\mathrm{4}.\mathrm{6}\:\mathrm{meters}\:\mathrm{and}\:\mathrm{horizontal} \\ $$$$\mathrm{distance}\:\mathrm{of}\:\mathrm{8}.\mathrm{6}\:\mathrm{meters}\:\mathrm{between}\: \\ $$$$\mathrm{adjacent}\:\mathrm{crests}.\mathrm{The}\:\mathrm{waves}\:\mathrm{splash}\: \\ $$$$\mathrm{into}\:\mathrm{the}\:\mathrm{station}\:\mathrm{once}\:\mathrm{every}\:\mathrm{6}.\mathrm{2}\:\mathrm{seconds} \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{frequency}\:\mathrm{and} \\ $$$$\:\mathrm{the}\:\mathrm{speeed}\:\mathrm{of}\:\mathrm{these}\:\mathrm{waves}. \\ $$

Answered by physicstutes last updated on 14/May/21

Amplitude of the wave = (1/2)(distance between high point and low point) ⇒ amplitude = (1/2)(4.6 m) = 2.3 m(well this won′t help anyway,just wanted you to know). waveleght = 8.6 m, since that is the distance between adjecent crest(points in phase). 1 cycle of the wave = 6.2 seconds frequency = ((#cycles)/(time)) = ((1 cycle)/(6.2 seconds)) = 0.16 Hz wave speed v = λf = (8.6 m)(0.16 Hz) = 1.4 m s^(−1)

$$\mathrm{Amplitude}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wave}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{distance}\:\mathrm{between}\:\mathrm{high}\:\mathrm{point}\:\mathrm{and}\:\mathrm{low}\:\mathrm{point}\right) \\ $$$$\Rightarrow\:\mathrm{amplitude}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}.\mathrm{6}\:\mathrm{m}\right)\:=\:\mathrm{2}.\mathrm{3}\:\mathrm{m}\left(\mathrm{well}\:\mathrm{this}\:\mathrm{won}'\mathrm{t}\:\mathrm{help}\:\mathrm{anyway},\mathrm{just}\:\mathrm{wanted}\right. \\ $$$$\left.\mathrm{you}\:\mathrm{to}\:\mathrm{know}\right). \\ $$$$\mathrm{waveleght}\:=\:\mathrm{8}.\mathrm{6}\:\mathrm{m},\:\mathrm{since}\:\mathrm{that}\:\mathrm{is}\:\mathrm{the}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{adjecent}\:\mathrm{crest}\left(\mathrm{points}\right. \\ $$$$\left.\mathrm{in}\:\mathrm{phase}\right). \\ $$$$\mathrm{1}\:\mathrm{cycle}\:\mathrm{of}\:\mathrm{the}\:\mathrm{wave}\:=\:\mathrm{6}.\mathrm{2}\:\mathrm{seconds} \\ $$$$\mathrm{frequency}\:=\:\frac{#\mathrm{cycles}}{\mathrm{time}}\:=\:\frac{\mathrm{1}\:\mathrm{cycle}}{\mathrm{6}.\mathrm{2}\:\mathrm{seconds}}\:=\:\mathrm{0}.\mathrm{16}\:\mathrm{Hz} \\ $$$$\mathrm{wave}\:\mathrm{speed}\:\:{v}\:=\:\lambda{f}\:=\:\left(\mathrm{8}.\mathrm{6}\:\mathrm{m}\right)\left(\mathrm{0}.\mathrm{16}\:\mathrm{Hz}\right)\:=\:\mathrm{1}.\mathrm{4}\:\mathrm{m}\:\mathrm{s}^{−\mathrm{1}} \\ $$

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