I’m currently going through harmonics, and I do not at all understand the fundamental frequency. I understand that it is the simplest vibration of a string, but I don't understand how can it have frequency at all if it is only half a wavelength. Isn’t frequency how many cycles are completed per second, and isn’t the fundamental frequency only half a cycle if it is half a wavelength? How can there be frequency of (say) 162 cycles per second if one cycle doesn’t even complete in the medium of the string? Is it measuring the frequency of the half wavelength as a full cycle? Is there frequency measured as a whole cycle from one half because it is a resultant wave of two waves making the half wavelength? If so, why do we have to multiply by two to get the wavelength of the cycle from the string length?

  • 1
    $\begingroup$ Get a slinky or something and watch its oscillations. $\endgroup$
    – user137289
    Jun 23, 2018 at 20:37
  • $\begingroup$ Yes but how is half a wavelength one cycle? How is there a frequency made from that half a cycle in the medium? I understand one wavelength=one cycle $\endgroup$ Jun 23, 2018 at 20:38
  • $\begingroup$ Or play with phet.colorado.edu/sims/html/wave-on-a-string/latest/… . (But nothing is as enlightening as a Slinky.) $\endgroup$
    – user137289
    Jun 23, 2018 at 20:40
  • $\begingroup$ Thank you for the simulation, but that still helps me very little with understanding the why with standing waves frequency. $\endgroup$ Jun 23, 2018 at 20:44
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    $\begingroup$ I think there's terminology confusion here. The simplest vibration of a string fixed at both ends is the fundamental mode. Yes, the spatial variation along the length of the string, e.g., $\phi(x) = \sin(\pi x/L)$ is a 'half-cycle' but the fundamental frequency refers to time dependent amplitude, e.g., $A(t) = A_0\cos(\omega_0 t)$ where $\omega_0 = \frac{\pi}{L}\sqrt{T/\rho}$. So, the amplitude of the mode completes $\omega_0/2\pi$ cycles per second. See the animated gif at this answer $\endgroup$ Jun 23, 2018 at 21:21

3 Answers 3


Isn’t frequency how many cycles are completed per second, and isn’t the fundamental frequency only half a cycle

When a string, fixed a both ends, vibrates in the fundamental mode, the perpendicular displacement $\phi_1(x,t)$ of a point located at $x$ along the length of the string is given by

$$\phi_1(x,t) = A_1(t)\phi_1(x) = A_1\cos(2\pi f_1t + \varphi)\sin\left(\frac{\pi}{L}x\right)$$

Now, it is true that the spatial variation of the fundamental mode is a 'half-cycle' since the argument of the $\sin$ ranges from $0$ to $\pi$.

However, the fundamental frequency refers to the time dependent amplitude $A_1(t)$. Note that $A_1(t)$ executes $f_1$ cycles per second. Take a look at this animated gif of the fundamental mode and the first three harmonics:

enter image description here

Animated gif credit

See that although the fundamental mode has a 'half-cycle' spatial variation, the time dependent amplitude goes from a maximum, through zero, a minimum, back through zero back to the maximum in a time $T_1 = \frac{1}{f_1}$ where $f_1$ is the fundamental frequency.

Also note that the frequency of the 2nd harmonic is twice the frequency of the fundamental, the frequency of the 3rd harmonic is thrice the frequency of the fundamental and so on.

  • $\begingroup$ Ohh, okay, so, the period of this wave is not the spatial variation, half a wavelength, but depending on the amplitude going up and down as it were, and that is the period, and therefore the cycle, and so the amount of up and down amplitude movements that equals a cycle would make up a frequency of say 162 cycles per second, but in terms of proclaiming the wavelength of such we treat it as a sine wave, as though we were folding it out. As in the gif, the first would have a lower frequency, the second, frequency is doubled as the cycles are being completed by twice as many in a second, so on. $\endgroup$ Jun 23, 2018 at 22:21

A standing wave on a string can be thought of as a traveling wave that is bouncing back & forth in one dimension. Every time it reaches one of the ends, it is reflected, either inverted or upright depending on what conditions you have at the end. Assuming you have the same boundary conditions at both ends, this means that by the time the traveling wave makes a full cycle, it will be upright: either it got inverted twice, or it never got inverted.

However, by the time the wave has made a round trip, the "first peak" of the wave may not be in phase with the later peaks any more. If it is a little bit out of phase, then after it makes another round trip, it will be even more out of phase; and after many round trips, it will be even more out of phase (and so will the second peak, and the third, and....) All of these out-of-phase waves will exhibit destructive interference, and cancel each other out. So you can't have a standing wave if this is the case.

But there's a way around this: suppose that the time for the wave to make one round-trip is exactly equal to the period of the wave. In this case, the "first peak" lines up with the second peak, and you get constructive interference between the first peak and the second peak. Thus, you get a standing wave.

But notice that you needed the round-trip time to be equal to the period. This means that the distance traveled by the first peak in one cycle must be twice the length of the string. Since the distance traveled by a peak of a wave in one cycle is equal to the wavelength, this means that the wavelength of the corresponding traveling wave is double the length of the string. In other words, one cycle of the wave has to fit into the "round-trip" distance, not the length of the string.

You can generalize this, by the way: the higher harmonics can be thought of as traveling waves that make a round trip in $n$ cycles of the wave, so that the first peak lines up with the $(n+1)$th peak after it makes a round trip.

  • $\begingroup$ Thank you for the very detailed answer! So is a standing wave, half a wavelength, technically a “round trip” of a wave and therefore one cycle, but the standing waves corresponding wavelength would be twice that of the string for the fundamental frequency. And then the wavelength decreases by making that round trip happen once again, second harmonic and one loop is still one round trip so there are now two cycles in second harmonic and the wavelength is equal to the length of the two cycles loops? $\endgroup$ Jun 23, 2018 at 21:25

From the many questions you make, it seems that you don't understand the difference between frequency and wavelength. Frequency is inversely proportional to wavelength. This is expressed by the equation $f = \frac{v}{\lambda}$. (v = speed of sound)
Also, for a string, $\lambda$ = 2l. (l = length of string)

Some examples might be useful:

For sound, v =343 m/s, so for a string with a wavelength of 1 m, f would be 343 cps. However, since for a string $l = \frac{\lambda}{2}$, the length (l) of the string must be 0.5 m long!

If the length of the string is 1 m, then $\lambda$ = 2l = 2 m, and the frequency would be f = 343/2 = 171.5 cps.

Keep in mind that whatever the string length is, it only represents 1/2 of the wavelength. So, if you want the length of a string that would vibrate at 162 cps, the answer would be $l = \frac{\lambda}{2} = \frac{v}{2f}$ = 343/2x162 = 1.058 m.


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