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

0.125 m

Step-by-step 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=(1)/(T)=(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=(v)/(f)`

Substituting the numerical values, we find

`\lambda=(0.5 m/s)/(4 Hz)=0.125 m`