Question

Waves in a fish bowl jostled by the Thingamajigger move to the sides at an average velocity of 0.50 m/s. If they occur once every 0.25 seconds, what is the distance between crests of the waves? Show your work!

Answer 1

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:

$f=\frac{1}{T}=\frac{1}{0.25 s}=4 Hz$

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:

$\lambda=\frac{v}{f}$

Substituting the numerical values, we find

$\lambda=\frac{0.5 m/s}{4 Hz}=0.125 m$