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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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