Question

A fisherman notices that his boat is moving up and down periodically, owing to waves on the surface of the water. It takes 2.80 for the boat to travel from its highest point to its lowest, a total distance of 0.620 . The fisherman sees that the wave crests are spaced 6.40 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

Answer 1

Concepts and reason

The concepts required to solve the given question is periodic motion and the speed of propagation of the wave.

Initially, find the speed of the wave. Later, find the amplitude of the wave. Finally, find the amplitude of the wave when the total vertical distance travelled by the boat were 0.500.

Fundamentals

The expression for the period motion in case of a wave moving up and down periodically is as follows:

$T = 2t$

Here, t is the time taken by the boat to travel form the highest to the lowest point.

The expression for the speed of wave is as follows:

$v = f\lambda$

Here, f is the frequency and $\lambda$ is the wavelength.

The relation between the time period and frequency is as follows:

$f = \frac{1}{T}$

Here, T is the time of the periodic motion.

Therefore, expression for the speed of wave is as follows:

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

The expression for the amplitude of the wave is as follows:

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

Here, d is the total distance.

Substitute 2.80 s for t in the equation $T = 2t$.

$\begin{array}{c}\\T = 2\left( {2.80{\rm{ s}}} \right)\\\\ = 5.60{\rm{ s}}\\\end{array}$

Now, substitute 5.60 s for T and 6.40 m for $\lambda$ in the equation $v = \frac{\lambda }{T}$ and solve for T.

$\begin{array}{c}\\v = \frac{{6.40{\rm{ m}}}}{{5.60{\rm{ s}}}}\\\\ = 1.14{\rm{ m/s}}\\\end{array}$

Substitute 0.620 m for d in the equation $A = \frac{d}{2}$ and solve for A.

$\begin{array}{c}\\A = \frac{{0.620{\rm{ m}}}}{2}\\\\ = 0.310{\rm{ m}}\\\end{array}$

From the equation $v = \frac{\lambda }{T}$, it is clear that the speed of the wave depends on the time period of the periodic motion and the distance between the two consecutive crests and troughs.

Therefore, the change in the vertical distance travelled does not affect the speed of the wave.

Thus, the speed of the wave is as follows:

$v = 1.14{\rm{ m/s}}$

Substitute 0.500 m for d in the equation $A = \frac{d}{2}$ and solve for A.

$\begin{array}{c}\\A = \frac{{0.500{\rm{ m}}}}{2}\\\\ = 0.250{\rm{ m}}\\\end{array}$

Ans:

The speed of the wave is equal to 1.14 m/s.

The amplitude of each wave is equal to 0.310 m.

The speed of the wave is equal to 1.14 m/s.

The amplitude of each wave is equal to 0.250 m.