Question

A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 3.1 ss for the boat to travel from its highest point to its lowest, a total distance of 0.55 mm. The fisherman sees that the wave crests are spaced 4.2 mm 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

Answer:

Explanation:

It takes 3.1 s for the boat to travel from its highest point to its lowest, so the period of oscillation

T = 2 x 3.1 = 6.2 s

frequency of wave n = 1 / T = .1613 per sec

Amplitude of oscillation = .55/2 = .275 mm

The fisherman sees that the wave crests are spaced 4.2 mm apart. so wavelength of wave λ = 4.2 mm .

A ) velocity of wave v = n λ

.1613 x 4.2 = .677 mm /s

B ) Amplitude of wave = .275 mm

C ) The vertical distance determines only the amplitude which does not affect the velocity , so velocity will remain unchanged .

D ) Amplitude of wave depends only on the vertical displacement .

The amplitude will become .5 / 2 = .25 mm .