A stone was thrown in the lake, and the wave created by it reached the shore in 10 seconds.

A stone was thrown in the lake, and the wave created by it reached the shore in 10 seconds. The distance between the two crests of the waves is 80 cm. Determine at what distance from the coast the stone was thrown. If in 2 seconds the wave hit the coast four times.

t1 = 10 s.

λ = 80 cm = 0.8 m.

t2 = 2 s.

N = 4.

S -?

The distance to the coast from the place of the fall of the stone S is expressed by the formula: S = V * t1, where V is the speed of wave propagation, t is the time during which the wave reached from the place where the stone fell to the coast.

The speed of propagation of the wave V is expressed by the formula: V = λ / T, where λ is the wavelength, T is the oscillation period.

A long wave is the distance between adjacent crests or troughs of a wave.

The oscillation period T is the time of one complete oscillation: T = t2 / N.

V = λ * N / t2.

The formula for determining the distance of the fall of the stone S will take the form: S = λ * N * t1 / t2.

S = 0.8 m * 4 * 10 s / 2 s = 16 m.

Answer: the distance from the place where the stone fell to the shore is S = 16 m.