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 2.90s for the boat to travel from its highest point to its lowest, a total distance of 0.630m . The fisherman sees that the wave crests are spaced a horizontal distance of 5.60m apart.

Part A

How much is the wavelength?

express your answers in 3 sig. fig.

Part B

Find the period of the wave.

3 sig. fig.

Part C

How fast are the waves traveling?

Express the speed v in meters per second using three significant figures.

Part D

What is the amplitude A of each wave?

Express your answer in meters using three significant figures.

Answer 1

Concepts and reason

The concepts required to solve this problem is relation between wavelength, period, , and speed.

Initially, notice the motion of the boat then using the definition of wavelength, calculate the wavelength. Then, find the period using that it takes 2.90 s to travel from highest point to lowest point. Later, calculate the speed using the values of the wavelength and the period. Finally, using the distance between the highest point and the lowest point, find the amplitude of the wave.

Fundamentals

The wavelength is defined as the length of one cycle of a wave.

The period of a wave is defined as the duration of time to complete one cycle of wave.

The wave speed can be defined as the wavelength per unit time and it can be represented as:

$v = \frac{\lambda }{t}$

Here, v is the wave speed, $\lambda$ is the wavelength, and t is the period.

The amplitude $\left( A \right)$ of the wave is defined as the maximum height measured from the center of the wave.

(A)

The wave crests are spaced 5.60 m apart. This corresponds to the distance from one peak to the next peak of the waves. By definition, this is equal to the one wavelength.

The wavelength is,

$\lambda = 5.60{\rm{ m}}$

(B)

The time for a wave to travel from the highest point to the lowest point is 2.90 s. This time $\left( {{t_{1/2}} = 2.90{\rm{ s}}} \right)$ corresponds to half of one wavelength.

The time for one whole wavelength is,

$\begin{array}{c}\\T = 2{t_{1/2}}\\\\ = 2\left( {2.90{\rm{ s}}} \right)\\\\ = 5.80{\rm{ s}}\\\end{array}$

(C)

The speed of the wave is,

$v = \frac{\lambda }{T}$

Substitute 5.60 m for $\lambda$ and 5.80 s for T.

$\begin{array}{c}\\v = \frac{{5.60{\rm{ m}}}}{{5.80{\rm{ s}}}}\\\\ = 0.966{\rm{ m/s}}\\\end{array}$

(D)

The amplitude of the wave is the distance between the highest point of the wave to the lowest point of the wave. And here it is 0.630 m by assuming that the boat always sits on the wave without sinking.

The distance 0.630 m corresponds to the peak-peak vertical displacement of the wave. Therefore, the amplitude is half of this distance.

The amplitude is,

$\begin{array}{c}\\A = \frac{{0.630{\rm{ m}}}}{2}\\\\ = 0.315{\rm{ m}}\\\end{array}$

Ans: Part A

The wavelength is 5.60 m.

Part B

The period of the wave is 5.80 s.

Part C

The wave speed is $0.966{\rm{ m/s}}$.

Part D

The amplitude of each wave is 0.315 m.