Question

Two boats - Boat A and Boat B - are anchored a distance of 24 meters apart. The incoming water waves force the boats to oscillate up and down, making one complete cycle every 10 seconds. When Boat A is at its peak, Boat B is at its low point and there is a crest in between the two boats. The vertical distance between Boat A and Boat B at their extreme is 8 meters. The wavelength is ___ m, the period is ___ s, the frequency is ___ Hz, and the amplitude is ___ m.

Answer 1

The wavelength of the waves is 16.00 meters, the period is 10 seconds, the frequency is 1/10 Hz, and the amplitude is 4 meters.

Step-by-step explanation:

The wavelength of the waves is 16.00 meters, as given in the question. The period of the waves can be calculated by dividing the time for one complete cycle, which is 60 seconds (1 minute), by the number of crests that pass in that time. Since there are 6 crests in 1 minute, the period is 10 seconds. The frequency can be calculated by taking the reciprocal of the period, so the frequency is 1/10 Hz. The amplitude of the waves is half the vertical distance between Boat A and Boat B at their extreme, which is 8 meters/2 = 4 meters.

Answer 2

wavelength = 24 m

Period = 10 s

f = 0.1 Hz

Amplitude = 4 m

Step-by-step explanation:

Wavelength:

Since the boats are at crest and trough, respectively at the same time. Hence, the horizontal distance between them is the wavelength of the wave:

wavelength = 24 m

Period:

The period is given as:

`Period = (time)/(no.\ of\ cycles) \\\\Period = (10\ s)/(1)\\\\`

Period = 10 s

Frequency:

The frequency is given as:

`f = (1)/(time\ period)\\\\f = (1)/(10\ s)\\\\`

f = 0.1 Hz

Amplitude:

Amplitude will be half the distance between extreme points, that is, crest and trough:

Amplitude = 8 m/2

Amplitude = 4 m