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I am currently studying the textbook Optics, fifth edition, by Hecht. Chapter 2.3 Phase and Phase Velocity says the following:

Examine any one of the harmonic wave functions, such as $$\psi(x, t) = A \sin(kx - \omega t) \tag{2.26}$$ The entire argument of the sine is the phase $\varphi$ of the wave, where $$\varphi = (kx - \omega t) \tag{2.27}$$ At $t = x = 0$, $$ \psi(x,t)|_{\begin{subarray}{l}x=0\\t=0\end{subarray}}=\psi(0,0)=0 $$ which is certainly a special case. More generally, we can write $$\psi(x, t) = A \sin(kx - \omega t + \epsilon) \tag{2.28}$$ where $\epsilon$ is the initial phase. To get a sense of the physical meaning of $\epsilon$, imagine that we wish to produce a progressive harmonic wave on a stretched string, as in Fig. 2.12. In order to generate harmonic waves, the hand holding the string would have to move such that its vertical displacement $y$ was proportional to the negative of its acceleration, that is, in simple harmonic motion (see Problem 2.27). But at $t = 0$ and $x = 0$, the hand certainly need not be on the $x$-axis about to move down-ward, as in Fig. 2.12. It could, of course, begin its motion on an upward swing, in which case $\epsilon = \pi$, as in Fig. 2.13. In this latter case, $$\psi(x, t) = y(x, t) = A \sin(kx - \omega t + \pi)$$ which is equivalent to $$\psi(x, t) = A \sin(\omega t - kx) \tag{2.29}$$ or $$\psi(x, t) = A \cos \left( \omega t - kx - \dfrac{\pi}{2} \right)$$ enter image description here

I'm confused by this:

In order to generate harmonic waves, the hand holding the string would have to move such that its vertical displacement $y$ was proportional to the negative of its acceleration, that is, in simple harmonic motion (see Problem 2.27).

Why is this required to generate harmonic waves?

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  • $\begingroup$ Isn't that the definition of a harmonic wave? Sine or cosine satisfy the equation $\ddot{y}=-cy$. $\endgroup$
    – kaylimekay
    Dec 28 '20 at 7:44
  • $\begingroup$ @kaylimekay I have never seen such a definition. $\endgroup$ Dec 28 '20 at 7:47
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Thanks for the question. Understanding certain concepts related to the phase of cyclic processes may require subtle thought. Here, the author is simply saying that if $y = A \sin(\omega t - k x)$ with $0 < A, \omega, k$, then $\ddot{y} = - A \omega^2 \sin(\omega t - k x) \propto - y$ with positive proportionality constant. More simply, the position of the hand corresponds to $x = 0$ so that $y = A \sin(\omega t)$ and $\ddot{y} = - A \omega^2 \sin(\omega t) \propto - y$ or $y \propto -\ddot{y}$, which put in words means that, in general, the position of the hand ($y$) and the acceleration ($\ddot{y}$) have opposite signs.

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