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In the string which has both end fixed then the end point have to be $n (\lambda/2)$ from the beginning point in order to have standing waves. I know it has to start with a node and end with a node, but why?

Is it because the end point is fixed, so no matter which phase the sending wave end, the reflex waves at this point always start from the beginning phase (meaning $A \cos(\varphi) = 0)$, so the sending wave must end at $\cos (\varphi)=0$ in order to make standing wave?

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  • $\begingroup$ please accept my answer if it answers the question. $\endgroup$ – SamD97 Dec 6 '18 at 9:07
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Standing wave or Stationary wave is a result of two waves (incident wave and reflected wave) propagating in opposite direction with same amplitude and same frequency. This phenomena is a result of interference of wave where energies of two waves add or cancel out depending on the phase. (energies add up in case of complete in-phase and cancel out in case of complete out-phase)

In a case of a string fixed at both of the ends, the following equation has to be satisfied. $$ \lambda = \frac {2l}{n} $$ where
$\lambda$ - wavelength
$l$ - length of the string
$n$ - number of loops or order of harmonic

You are correct it has to start and end with a node because of the boundry conditions. (i.e displacement of string from equilibrium position at both the ends should be zero)

Now, suppose you create a wave of wavelength $2l$ at left end, after a half time period it will reach the other end and it will undergo inversion because of the fixed end. If you imagine right end of the string at this instant it is under effect of a upward wave going right and a downward wave going left. This particular superposition will have zero displacement at the end all the time because (node) of inversion (out of phase interference). After a full time period you will see that the wave has twice downward displacements at the center of the string because of in-phase superposition, and this will continue to happen at all times (anti-node). You can think similarly for a wave with a wavelentgh of $l$ or $\frac l2$ and so on.

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One may describe the waves in terms of $\Delta x$, the deviation of the position of strings' atoms from the equilibrium locations. Because they're attached, $\Delta x(\sigma)=0$ for $\sigma$, the coordinate along the string, equal to any of the end point values, $\sigma=0$ and $\sigma=L$.

But $\Delta x$ obeys a wave equation, so the eigenstates of the frequency have to depend on $\sigma$ as sines and cosines: $$\Delta x(\sigma,0)=A\sin(k\sigma)+B\cos(k\sigma)$$ I wrote the solution at $t=0$, a moment when $\Delta x $ is nonzero (or maximized).

The condition $\Delta x=0$ at $\sigma=0$ says $B=0$, so we only have signs, and $\Delta x =0$ at $\sigma=L$ implies $kL=n\pi$ because the sine vanishes when the argument is a multiple of $\pi$ i.e. $L=n\lambda /2$ because $k=2\pi/\lambda $. The calculation works both for longitudinal and transverse oscillations $\Delta x$ – but they are usually transverse. (In string theory where this exact same calculation is important for the basics as well, the longitudinal oscillations are absolutely unphysical.)

I reduced the spatial dependence on a sine. The sine is a standing wave – much like a cosine, just a shifted sine – and it may also be understood as an equal mixture of the right-going and left-going (signs) complex wave $\exp(\pm ik\sigma)$.

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