Talon Mcbride

2022-07-22

Rank in order, from largest to smallest, the wavelengths ${\lambda }_{1}$ to ${\lambda }_{3}$ fro sound waves having frequencies ${f}_{1}=100Hz,{f}_{2}=1000Hz$, and ${f}_{3}=10,000Hz$.

kamphundg4

Expert

The explanation here is:
The wavelength and frequency are given by the relationship:
$\lambda =v/f$
The equation above indicates that the wavelength varies inversely with the requency. Hence, the greater the frequency, the smaller the wavelength. Because ${f}_{1}<{f}_{2}<{f}_{3}$, it can be concluded that ${\lambda }_{1}>{\lambda }_{2}>{\lambda }_{3}$.

Do you have a similar question?