A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 3.50 ss for the boat to travel from its highest point to its lowest, a total distance of 0.700 mm . The fisherman sees that the wave crests are spaced 6.00 mm apart.

How fast are the waves traveling? in m/s

What is the amplitude of each wave? in m

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/s

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

The correct values from the question are;

a total distance of 0.7m and wave crests are spaced 6.00 m apart

A) Speed = 0.8574 m/s

B) A = 0.35m

C) Velocity remains unchanged as 0.8574m/s and while A = 0.25m

Explanation:

A) We know that wavelength λ is the distance between two successive wave crests or troughs, so the wavelength here is 6m.

From the question, it takes 3.5 seconds to travel from its highest point to its lowest point which is half period. Thus, full period T is;

T = 2 x 3.5 = 7 seconds

Now, we know that frequency is given as;

f = 1/T.

Thus, f = 1/7 = 0.1429 Hz

We know that speed of wave is given by;

V = fλ

Thus V= 0.1429 x 6 = 0.8574 m/s

B) Amplitude(A) is the maximum magnitude of displacement from equilibrium. This implies that the distance between the boat's highest point to its lowest point is 2A. From the question, d = 0.7m

Thus, 2A = d = 0.7

Thus,A = 0.7/2 = 0.35m

C) Since the vertical distance is now 0.5m and the other values remain the same. Thus, the speed will remain the same because it doesn't depend on vertical displacement.

Thus, v remains 0.8574 m/s.

Amplitude will now be calculated;

A = 0.5/2 = 0.25m