An explosion occurs at the end of a pier. the sound reaches theother end of the pier by traveling through three media: air, freshwater, and a slender metal handrail. the speeds of sound in air, water, and the handrail are 343, 1482, and 5040 m/s, respectively. the sound travels a distance of 135 min each medium.
(a) through which medium does the sound arrivefirst, second, and third?
watersteel handrailair (first)
handrailairfresh water (second)
watersteel handrailair (third)

(b) after the first sound arrives, how much later do the second andthird sounds arrive?
s (second sound)
s (third sound)

a) The sound will arrive first through the slender metal handrail, second through fresh water, and third through air.

b) 0.064 s  (2nd sound)  0.36 s (third sound)

Explanation:

a) Assuming that the sound travels along a straight line, at a constant speed, we can apply the definition of velocity, in order to obtain the time needed for the sound to reach to the other end of the pier.

We will call t₁ to the time needed to travel along the handrail, t₂ to the time for fresh water propagation, and finally t₃ for the time through air.

⇒ t₁ = Δx / v = 135 m / 5040 m/s = 0.027 s

⇒ t₂ = Δx / v  = 135 m / 1482 m/s = 0.091 s

⇒ t₃ = Δx / v = 135 m / 343 m/s = 0.39 s

We can see that the order in which the sound will arrive will be as follows:

1) handrail

2) fresh water

3) air

b) After  the first sound arrives, the second will arrive at a time equal to the difference between the time for the first sound to arrive and the second one, as follows:

Δt₁ = t₂ - t₁ = 0.091 s - 0.027 s = 0.064 s

The third sound will arrive at a time equal to the difference between the time for  the second sound to arrive and the third one (regarding  the time for the second time to arrive) , as follows:

Δt₂₃ = t₃ - t₂ = 0.39 s - 0.091 s = 0.3 s

Now, counting from the instant that the first sound arrived, the third one arrived after a time that is equal to the sum of the difference between the first and the second one, and the difference between the second sound and the third one, as follows:

Δt₂ = Δt₁ + Δt₂₃ = 0.064 s + 0.3 s =0.36 s

wavelength divides in half and speed stays the same.