Question

A surfer paddles out beyond the breaking surf to where waves are sinusoidal in shape, with crests 59.6 m apart. The surfer bobs a vertical distance of 4.28 m, taking 3.09 s to go from trough to crest. Describe the wave using equation y(x,t)=Acos(kx−ωt). Take the positive x-direction toward shore and x=0 at the location of a wave crest when t=0.

Answer 1

Answer: y(x, y) = 2.14cos(6.04x - 8.29t)

wave speed (v) =1.38m/s,

wavelength ( λ) =59.6m

Amplitude (A) =2.14m

Wave number (k) =6.04

Angular frequency (ω) = 8.29rad/s

Explanation:

The wavelength is simply defined as the distance between 2 successive crest or trough.

From the question it has been stated that the crest are 59.6m apart, thus (λ) =59.6m.

The wave travels a total vertical distance of 4.28m, this simply implies that along the vertical axis of the sine graph, the wave traveled from maximum displacement to the positive y axis (+A) through the center (o) and to the maximum displacement to the negative axis (-A).

Meaning total vertical distance =2A where A=Amplitude

This amplitude is 4.28/2 = 2.14m

Wave number (k) =2π/(λ)

Where π=180°

Thus, k=2*180/59.6

K=360/59.6

K=6.04

To get the angular frequency of the wave(ω), we use the formulae above

(ω)=kv

Where v is the wave speed

v=vertical distance/time taken

v= 4.28/3.09= 1.31m/s

Thus

(ω)=kv=6.04 * 1.31 =8.29rad/s.

Equation of wave is below

y(x,t)=Acos(kx−ωt)

By slotting in the parameters we have that

y(x, y) = 2.14cos(6.04x - 8.29t)

Above is the mathematical equation that describes the wave.