In calm weather on the lake, a heavy anchor was dropped from the boat. Waves emanated from the place of throw.

In calm weather on the lake, a heavy anchor was dropped from the boat. Waves emanated from the place of throw. A man standing on the shore noticed that the wave reached him after 20 s, the distance between adjacent wave crests was 40 cm, and in 4 s there were 10 wave bursts on the shore. Calculate how far from the shore the boat was.

t = 20 s.

λ = 40 cm = 0.4 m.

t1 = 4 s.

N1 = 10.

S -?

To find the distance from the place where the anchor is dropped to the coast S, it is necessary to multiply the speed of propagation of the wave V by the time of its movement t: S = V * t.

The speed of propagation of the wave V is expressed by the formula: V = λ * v, where λ is the distance between two adjacent crests, which is called the wavelength, v is the frequency of wave oscillations.

v = N1 / t1.

S = λ * N1 * t / t1.

S = 0.4 m * 10 * 20 s / 4 s = 20 m.

Answer: the distance from the anchor to the shore is S = 20 m.