Question

Two boats are anchored in the harbor, separated by a horizontal distance of 30 m. The boats bob up and down as waves enter the harbor. When one boat is at its highest point, the other is at its lowest point and there is one wave crest between them. Each boat makes exactly 3 complete up-and-down cycles every 18.0 s. The vertical distance between a boat's highest and lowest position is 3.8 m.

Answer 1

Answer:

Explanation:

Given:

Horizontal distance = 30 m

Note that Wavelength, lambda is the distance between two consecutive crests/troughs. Since,

One boat is at trough, the other is at crest.

The distance between a crest and a trough next to it = lambda/2

Complete cycles = 3 cycles

Time taken for the 3 cycles = 15 s

Vertical distance = 3.8 m

Wavelength, lambda = 2 × horizontal distance

= 2 × 30

= 60 m

Amplitude = vertical distance from the extreme loint to the mid point

= y/2

= 3.8/2

A = 1.9 m

In one cycle = 18/3

= 6 s/cycle

frequency, f = 1/T

= 1/6 = 0.17 Hz

speed,v = lambda × frequency

= 60 × 0.17

= 10 m/s

Answer 2

The wavelength is the distance between two consecutive crests or troughs.  The speed of the wave is 1.2 m/s.

The speed of the wave:

$V = \lambda \times f$.......................1

Where,

$\lambda$ - wavelength

$f$ - frequency

The wavelength is the distance between two consecutive crests or troughs.

The boat is present in the crest and trough. And the horizontal distance between them is 30 m.

So the

$\lambda = 2\times 30 \\\lambda = 60 \rm \ m$

 

Wave complete 3 cycles in 15 sec, so the period of the wave,

$T = \dfrac {3}{15 }\\\\ T = 0.2 \rm\ s$

So the frequency of the wave,

$f = \dfrac 1{0.2}\\\\f = 0.02\rm \ Hz$

Now put the values in equation 1,

$V = 60 \times 0.02\\\\V = 1.2 \rm \ m/s$

Therefore, the speed of the wave is 1.2 m/s.

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