Chapter 21, Problem 1RQ

To determine

The wavelength of the periodic wave with high frequency.

Answer to Problem 1RQ

Shorter.

Explanation of Solution

Wavelength is the distance of one completed wave (crest to crest) and frequency is the inverse of the period, $\left(f=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$T$}\right.\right)$.

Wavelength is inversely proportional to the frequency of the periodic wave. In other words, if wavelength is high then the frequency is low and vice versa.

Writing mathematically,

$\lambda \propto \frac{1}{f}$

Therefore, the higher frequency periodic waves will have a shorter wavelength.

Conclusion:

For a given type of wave, periodic waves with higher frequency will have shorter wavelengths than those with lower frequencies.