What is considered a wavelength? I am getting confused here. I keep seeing one wavelength is the distance when the wave repeats itself. So at the two highest points that's the wavelength. Now I'm seeing one wave cycle is when the wave comes back to the starting point. Which is it? enter image description here

enter image description here

Can i be missing something? Are these two different things and i am mistakenly thinking there the same?

  • 1
    $\begingroup$ Both of your images are correct examples to visualize wavelength. You could shift the upper images 'wavelength boundaries' one quarter of a wavelength to the left and get the same representation as in the lower image. $\endgroup$ – Communisty Dec 15 '17 at 13:51
  • $\begingroup$ In your post you have two images where one depicts a single oscillation [the second graph] and the other depicts two oscillations [the first]. The wavelength is just the distance between the same point travelling along a wave after a full period, and the point at $t=0$ $\endgroup$ – user177179 Dec 15 '17 at 15:16
  • $\begingroup$ Just as an exercise, figure out the wavelength of a 60Hz power line. $\endgroup$ – Howard Miller Dec 15 '17 at 17:51
  • $\begingroup$ In both figures, the horizontal axis is labeled "time". If that is time, then the segments indicated should be periods $T$ measured in seconds. That is not quite the same as the wavelength $\lambda$, is it? We would have: $$vT=\lambda$$ where $v$ is the speed of the wave. $\endgroup$ – Jeppe Stig Nielsen Dec 15 '17 at 20:00
  • $\begingroup$ "comes back to the starting point, going in the same direction". $\endgroup$ – Eric Towers Dec 16 '17 at 16:18

It is both as both lengths are the same. It doesn't matter which point you start measuring from - as long as you measure it to the same point in the next cycle. Obviously it's easier to choose starting points where you can easily tell where it is in the next cycle $($that is when $y=0$ or at the highest/lowest point in the wave$)$.

| cite | improve this answer | |
  • $\begingroup$ Ah got ya. This was seriously confusing me but there both the same. Just different ways to measure $\endgroup$ – user178750 Dec 15 '17 at 19:59
  • 5
    $\begingroup$ @user178750 Notice that in the second image--where the madding measuring point is don't at y=0--that the second measuring point is not the first time the wave has returned to 0, but the first time the wave had returned to zero with the same rate of change (rising towards 1). If you look at the first image, the same is true: the first time the wave has returned to both the starting value and the same rate of change. $\endgroup$ – Draco18s no longer trusts SE Dec 16 '17 at 2:20

A sine wave doesn't necessarily have an intrinsic "starting point", you usually can draw its curve starting at any phase and call the corresponding point the beginning of the cycle:

enter image description here enter image description here

Quoting Wikipedia:




| cite | improve this answer | |
  • $\begingroup$ Yeah i notice in some of the explanation videos i watch they draw the starting point anywhere. I thought there was a certain starting point but guess not lol $\endgroup$ – user178750 Dec 15 '17 at 20:00

A wave starts at zero, increases to a maximum in one direction, reverses direction and increases to a maximum in the other direction, then it reverses again; when it crosses zero is the length of a wave, a wavelength.

The number of waves per second is the frequency of oscillation.

The equation that relates wavelength and frequency for electromagnetic waves is: λν=c where λ is the wavelength, ν is the frequency and c is the speed of light. So, wavelength * frequency = speed of light.

A wave starts at the cycle start point in figure 2, it can't start at any value other than zero. You must wait until the cycle end point to measure the wavelength, how fast it gets there is the frequency.

It is possible to use a peak detector and save the location of the highest excursion of the waveform, hoping to measure the peaks in an equidistant manner. The waves in your two examples are sine waves, not all real waveforms look like that.

Ringing and Overshoot

It's better (more exact) to make one's wavelength measurement at the zero crossing. Your figures are 'demonstrative comparisons' the artist's depiction places leader lines, arrows and writing separated from each other for ease of viewing - not positioned in optimal locations or necessarily drawn to scale.

The greater the energy, the larger the frequency and the shorter (smaller) the wavelength. Given the relationship between wavelength and frequency — the higher the frequency, the shorter the wavelength — it follows that short wavelengths are more energetic than long wavelengths.

| cite | improve this answer | |
