Question

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.10 s for the boat to travel from its highest point to its lowest, a total distance of 0.700 m. The fisherman sees that the wave crests are spaced a horizontal distance of 6.10 m apart. A. How fast are the waves traveling?B. What is the amplitude of each wave? C. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, how fast are the waves traveling? D. If the total vertical distance traveled by the boat were 0.500 , but the other data remained the same, what is the amplitude of each wave?

Answer 1

a) v = 2.9 m / s, b) A = 0.350 m, c) v = 2.9 m / s, d) A = 1.00 m

Step-by-step explanation:

The oscillatory motion is described by the expression

x = A cos (wt + Ф)

the wavelength which is the distance for the wave to repeat and the frequency which is the number of times a wave oscillates per unit of time

a) In this part they ask us for the speed of the wave.

Let's use the relationship between speed, wavelength and frequency

v = λ f

For the wavelength they indicate that the distance between two crest is 6.1 m

λ / 2 = 6.10

λ = 12.20 m

They give us the period of the wave is the time it takes to return to the same point, in this case they give half a period

A / 2 = 2.10

A = 4.20 me

f = 1 / t

f = ¼, 2

f = 0.238 Hz

let's calculate

v = 12.20 0.238

v = 2.9 m / s

b) the amplitude of the wave, is the distance from zero to some maximum

2A = 0.700

A = 0.350 m

c) the speed of the wave is not function of the amplitude, so the speed is the same

v = 2.9 m / s

d) the amplitude is

2A = 0.50

A = 1.00 m