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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 2.60 s for the boat to travel from its highest point to its lowest, a total distance of 0.630 m . The fisherman sees that the wave crests are spaced a horizontal distance of 5.70 m apart.

Required:
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 0.30 m but the other data remained the same, how would the answers to parts (a) and (b) be affected?

Sagot :

cjrodriguez89

Answer:

a) v = 1.1 m/s

b) A = 0.315 m

c) v = 1.1 m/s  A= 0.15 m

Explanation:

a)

  • In any travelling wave, there exists a fixed relationship between the propagation speed, the wavelength and the frequency, as follows:

        [tex]v = \lambda * f (1)[/tex]

  • If the wave crests are spaced a horizontal distance of 5.7 m apart, this means that the wavelength of the wave is just the same, i.e., 5.70 m.
  • Regarding the frequency, we know that the frequency is just the inverse of the period, i.e., the time needed to complete one oscillation.
  • If it takes a time of 2.60 s to go from the highest point to the lowest, the time needed to complete an oscillation (the period T) will be just double of this time:
  • ⇒ T = 2.60 s * 2 = 5.20 s (2)
  • Since we have now T, we can find the frequency f as follows:

       [tex]f = \frac{1}{T} = \frac{1}{5.20s} = 0.19 Hz (3)[/tex]

  • Replacing f and λ in (1) we get:

      [tex]v = \lambda * f = 5.70 m * 0.19 Hz = 1.10 m/s (4)[/tex]

b)

  • The amplitude of the wave is just the amount that the water aparts from its equilibrium level, which is just the half of the distance between its highest point and the lowest one, as follows:

      [tex]A = \frac{0.630m}{2} = 0.315 m (5)[/tex]

c)  

  • Part a) will not be affected by the new amplitude, because we have showed that the speed is independent of the amplitude, so v can be written as follows:

       v = 1.10 m/s (6)

  • Part b) will change , due to the amplitude changes. If the total vertical distance traveled by the boat is 0.30 m, by the same token as explained in b), the new amplitude will be just half of this, as follows:

       [tex]A = \frac{0.30m}{2} = 0.15 m (7)[/tex]

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