The hand holding the string would have to move such that its vertical displacement $y$ was proportional to the negative of its acceleration?

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 $$\psi(x, t) = A \cos \left( \omega t - kx - \dfrac{\pi}{2} \right)$$

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?

• Isn't that the definition of a harmonic wave? Sine or cosine satisfy the equation $\ddot{y}=-cy$. Dec 28 '20 at 7:44
• @kaylimekay I have never seen such a definition. Dec 28 '20 at 7:47

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.