Question

A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 3.4s for the boat to travel from its highest point to its lowest, a total distance of 0.56 m. The fisherman sees that the wave crests are spaced 6.0m apart.
i. How fast are the waves traveling? in m/sii. What is the amplitude of each wave? in miii. 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 ? in m/siv. 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? in m

Answer 1

To solve this problem we will apply the concepts related to the Period. Speed as a division between wavelength and period. Finally, we will make the necessary considerations to solve the two points based on the information given.

PART A) The expression for the period motion in case of a wave moving up and down periodically is

$T = 2t$

Here

t= Time

Replacing,

$T = 2(3.4s)$

$T = 6.8s$

Now we know that speed is the change in wavelength per unit of time, so

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

Replacing,

$v = \frac{6m}{6.8s}$

$v = 0.8823m/s$

PART B) Amplitude is defined as the half of vertical distance, then

$A = \frac{d}{2}$

$A = \frac{0.56m}{2}$

$A = 0.280m$

PART C) The speed of the wave depends on the time period of the periodic motion and the distance between the two consecutive crest and troughs.

Therefore the change in the vertical distance traveled does not affect the speed of the wave and remains the same.

The speed of the wave is equal to 0.88m/s

PART D ) For the same concept about Amplitude we have that

$A = \frac{d}{2}$

$A = \frac{0.500m}{2}$

$A = 0.250m$

The amplitude of each wave is equal to 0.250m