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Why does a wave travel with less frequency on a longer string than on a shorter one?

I don't need the math just the logical explain for why it happens.

https://primes.utm.edu/mersenne/LukeMirror/mirrors/eb/mers_02.htm

(P.S. my apologies for overlooking)

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  • $\begingroup$ Where does that come from? $\endgroup$
    – user45664
    Commented Apr 8, 2021 at 16:38
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    $\begingroup$ Quite simply, it doesn't. $\endgroup$ Commented Apr 8, 2021 at 16:48
  • $\begingroup$ Are you really talking about standing waves, or waves traveling in one direction only? $\endgroup$ Commented Apr 8, 2021 at 18:07
  • $\begingroup$ @DavidWhite about standing waves $\endgroup$
    – Scarlet
    Commented Apr 9, 2021 at 4:42

5 Answers 5

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The propagation speed does not depend on the length of a string, just on its tension $T$ and linear density of mass $\mu$, in the following way

$$v=\sqrt{\frac{T}{\mu}}.$$

However, if you have two strings of the same mass and different lengths, the speed of propagation will be higher in the longer one, since its linear density of mass will be lower.

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The above answer is correct. The speed is not dependent on the length. But it will take longer, because it is traveling a longer distance.

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The frequency needed to produce a resonant standing wave on a string fixed at the far end will depend on the length of the string. You will need an integer number of anti-nodes in the pattern.

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A standing wave doesn't necessarily travel with less frequency on a longer string, as the string tension and string density affect wave speed, as stated by AFG with the formula:

$v=\sqrt{\frac{T}{\mu}} \qquad \qquad(1)$

where $T$ is the tension in the string and $\mu$ is the linear density of the string. The velocity of waves on the string is also given by a second well known formula:

$v=f \lambda \qquad \qquad (2)$

For standing waves to exist on a string, the length of the string must be an integral multiple of half wavelengths, given by the formula:

$L=\frac{n \lambda}{2}\qquad \qquad (3)$

where $n$ is an integer greater than 0.

Solving for frequency with equations (1) and (2) yields:

$f=\frac{v}{\lambda}=\frac{1}{\lambda}\sqrt{\frac{T}{\mu}} \qquad \qquad (4)$

From equation (3),

$\lambda = \frac{2L}{n} \qquad \qquad (5)$

Substituting equation (5) into equation (4) yields a frequency equation that depends on string length, string tension, and string linear density:

$f=\frac{n}{2L}\sqrt{\frac{T}{\mu}} \qquad \qquad (6)$

Equation (6) indicates that the frequencies that can produce standing waves depend on which harmonic you want to see (n=0 is the fundamental frequency), the length of the string, the tension in the string, and the linear density of the string. If you specify $n$, keep the same string for several "experiments" (aka hold the linear density constant), and hold the tension in the string constant, it is immediately seen that the frequency that produces standing waves is proportional to the reciprocal of the string's length.

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Let say u have a USB cable and wish to connect from wall to your PC, but if the distance is short. Then you won't mind to untangle the cable, since it would connect anyhow. But if the same USB cable needs to connect a cell phone which is much further from wall, then u would think of stretching or minimizing coiling of cable, so that it could compensate to reach longer distance.

Since the cable length is limited, similar to limited energy provided to vibrate a string. So the string compensates energy by lowering frequency to reach longer distance.

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