Why does the frequency of a wave increase as the wavelength decreases

1 answer

The relationship between the frequency and wavelength of a wave is described by the equation:

\[ v = f \cdot \lambda \]

Where:

  • \( v \) is the wave speed (constant for a given medium)
  • \( f \) is the frequency (number of wave cycles per second)
  • \( \lambda \) is the wavelength (distance between consecutive points of the same phase in the wave, such as crest to crest)

From this equation, we can rearrange it to express frequency in terms of wavelength:

\[ f = \frac{v}{\lambda} \]

This equation shows that frequency (\( f \)) is inversely proportional to wavelength (\( \lambda \)) when wave speed (\( v \)) is constant:

  1. If the wavelength (\( \lambda \)) decreases while the wave speed (\( v \)) remains constant, then the frequency (\( f \)) must increase to maintain the equality.
  2. Conversely, if the wavelength increases, the frequency must decrease.

This inverse relationship is why waves that have shorter wavelengths (like gamma rays or X-rays) correspond to higher frequencies, and waves with longer wavelengths (like radio waves) correspond to lower frequencies. Thus, as the wavelength decreases, the frequency increases.