Answer:
6.88 s
Explanation:
First of all, we can find the frequency of the wave, given by:
[tex]f=\frac{v}{\lambda}=\frac{450 m/s}{0.50 m}=900 Hz[/tex]
where v is the wave speed and [tex]\lambda[/tex] the wavelength.
Then we can find the period of the wave, T, which is the reciprocal of the frequency:
[tex]T=\frac{1}{f}=\frac{1}{900 Hz}=0.0011 s[/tex]
The amplitude of the wave is
[tex]A=20.0 cm=0.20 m[/tex]
and this corresponds to the distance between the maximum displacement of the wave and the equilibrium position. Therefore, the wave covers a total distance of 4A (4 times the amplitude) during one period T:
[tex]d=4A=4(0.20 m)=0.80 m[/tex]
and this distance is covered in T = 0.0011 s.
Therefore, the total time needed for the wave to cover a total distance of
d' = 5.0 km = 5000 m
is given by:
[tex]t= \frac{d'}{d}T=\frac{5000 m}{0.80 m}(0.0011 s)=6.88 s[/tex]
