What is the velocity of a wave if its peak-to-peark cycles each require .000058 second and its wavelength is .326 meters?
A single cycle, from peak to peak, is .326 meters long and passes in .000058 second. The wave therefore travels .326 meters in a time interval of .000058 second.
The wave must therefore be traveling at .326 meters / .000058 sec = 5620 meters/second.
A wave whose wavelength is `lambda, completing a period in time T, must move distance `lambda in time T. Its velocity must therefore be `lambda / T.
We note that the frequency of the wave is the reciprocal of T: frequency = f = 1 / T. Thus v = `lambda / T = f * `lambda, consistent with our knowledge that v = f `lambda.
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