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This question already has an answer here:

Consider a string with both ends fixed. If somewhere in middle of it be pulled and released, how would it oscillate and what is it's equation?

My solution assuming the result is a standing wave:

The point which is pulled will be an anti-node, so the length of string before this point is $(2n-1)\times\frac{\gamma}{4}$ and the length of string after this point will be $(2n^\prime-1)\times\frac{\gamma}{4}$. Considering the first length $l$ and the second length $l^\prime$, we will have: $$\frac{2n-1}{2n^\prime-1}=\frac{l}{l^\prime}$$ So the oscillator will be in it's $n+n^\prime -1$ harmonic. But what if that equation doesn't have natural answer, and what if it has answer so double of that is still an answer?

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marked as duplicate by ja72, Gert, Jon Custer, ZeroTheHero, John Rennie Feb 12 at 16:35

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Mathematically the shape of the string is the superposition of multiple standing waves. In general it has the form:

$$ y(x,t) = \sum_{i=1}^{\infty} \sin \left(i\frac{ \pi x}{\ell} \right) \left( A_i \sin \left( i\frac{ \pi c t}{\ell} \right)+ B_i \cos \left( i\frac{ \pi c t}{\ell} \right) \right) $$

where $\ell$ is the length from end to end, and $c$ is the speed of wave propagation along the string.

Now given an initial pluck of the string, of triangular shape you find the coefficients $A_i$ and $B_i$ using Fourier transform. The result is:

$$ \begin{align} A_i &= 0 \\ B_i &= Y \frac{2 \ell^2 \sin \left( \frac{i \pi x_p}{\ell} \right)}{\pi^2 i^2 x_p (\ell-x_p)} \end{align} $$

where $x_p$ is the position of the pluck point, $Y$ is the pluck amplitude, and $i=1 \ldots \infty$

To arrive at this use the pluck shape $y_0(x) = Y\,{\rm if}(x \leq x_p, \tfrac{x}{x_p}, 1-\tfrac{x-x_p}{\ell-x_p})$ and the known initial conditions $$ \begin{aligned} \lim \limits_{t\rightarrow 0} y(x,t) & = y_0(x) = {\rm triangle} \\ \lim \limits_{t\rightarrow 0} \frac{\partial}{\partial t} y(x,t) & = v_0(x) =0 \end{aligned}$$

Now pre-multiply with $\sin\left(i \frac{\pi x}{\ell} \right)$ and integrate over the length of the string

$$ \begin{aligned} \int \sin\left(i \frac{\pi x}{\ell} \right) y_0(x) {\rm d}x &= \int \sin\left(i \frac{\pi x}{\ell} \right) \lim_{t\rightarrow 0} y(x,t) {\rm d}x = B_i \frac{\ell}{2} \\ \int \sin\left(i \frac{\pi x}{\ell} \right) v_0(x) {\rm d}x &= \int \sin\left(i \frac{\pi x}{\ell} \right) \lim_{t\rightarrow 0} \frac{\partial}{\partial t} y(x,t) {\rm d}x = A_i \frac{i\,\pi\, c}{2} \end{aligned} $$

or

$$\begin{aligned} A_i &= \frac{2}{i\,\pi\,c} \int \sin\left(i \frac{\pi x}{\ell} \right) v_0(x) {\rm d}x = 0 \\ B_i & = \frac{2}{\ell} \int \sin\left(i \frac{\pi x}{\ell} \right) y_0(x) {\rm d}x = \frac{2 Y \ell^2}{\pi^2 i^2 x_P (\ell-x_p)} \sin\left(i \frac{\pi x_p}{\ell} \right)\end{aligned} $$

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    $\begingroup$ I have plotted this equation for i from 1 to 1000. It is just some lines. here's my plot: link $\endgroup$ – Pouya Esmaili Feb 14 at 8:55
  • $\begingroup$ What you see is the waveform. Essentially the initial shape of a plucked string is two lines. Then this shape travels down and up the string causing self-interference and that is why you see just lines. Each part of the string vibrates harmonically. Play the $t$ variable in Desmos very slowly (like $<0.1×$) and you can see that. Reduce also the sum points to $n=50$ or less to make the animation smoother. $\endgroup$ – ja72 Feb 14 at 13:19
  • $\begingroup$ I thought it must be a curve but it's always 2 to 5 lines. Is that what really happens? $\endgroup$ – Pouya Esmaili Feb 14 at 14:03
  • $\begingroup$ A single harmonic is a sine curve, but all of them combined result in this shape because of the initial conditions. In real life, though, damping will reduce the higher harmonics and the shape will degenerate down to just a few waves making the envelope look more like what you expected. $\endgroup$ – ja72 Feb 14 at 18:07

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