Answer:
0.125 m
Explanation:
In this problem, we have:
v = 0.50 m/s is the average velocity of the wave
T = 0.25 s is the period of the wave
We can find the frequency of the wave, which is equal to the reciprocal of the period:
[tex]f=\frac{1}{T}=\frac{1}{0.25 s}=4 Hz[/tex]
The problem is asking us to find the distance between two crests of the wave: this is equivalent to the wavelength. The wavelength is related to the average velocity and the frequency by the formula:
[tex]\lambda=\frac{v}{f}[/tex]
Substituting the numerical values, we find
[tex]\lambda=\frac{0.5 m/s}{4 Hz}=0.125 m[/tex]
