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If the displacement of a string follows

$$ y(x,t) = A \sin(kx - \omega t + \phi_0) $$

how can I show that the "hand" generating the wave must be moving vertically in a simple harmonic motion?

Simple harmonic motion means that the restoring force is proportional to the displacement, which basically is Hooke's law:

$$ \vec{F} = -k\vec{x} $$

where $\vec{x}$ is the displacement. I am not sure how to tackle this exercise.

Hooke's law in differential form is

$$ m\cdot \frac{\partial^2 y(x,t)}{\partial t^2} = -k x $$

$$ \frac{\partial^2 y(x,t)}{\partial t^2} = \omega^2 A \sin (kx -\omega t + \phi _0) $$

I'm not sure how this shows that $y(x,t)$ is a solution to the differential Hooke's law

$$ -kx = m \omega^2 A \sin (kx -\omega t + \phi _0) $$

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    $\begingroup$ Write Hooke's law as a differential equation and show that your sinusoidal wave-form is a solution. $\endgroup$ – lemon Sep 24 '16 at 10:16
  • $\begingroup$ I tried, but I don't see how the last equation holds. $\endgroup$ – Yoda Sep 24 '16 at 10:28
  • $\begingroup$ All you have to do is to show that $\frac{\partial^2 y(x,t)}{\partial t^2}= - \omega ^2 y(x,t)$ which you have done here $\frac{\partial^2 y(x,t)}{\partial t^2} = \omega^2 A \sin (kx -\omega t + \phi _0)$ provided that you remember to put in the missing minus sign. $\endgroup$ – Farcher Sep 24 '16 at 11:24
  • $\begingroup$ Why is that all I have to show? I don't understand. $\endgroup$ – Yoda Sep 24 '16 at 11:37
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    $\begingroup$ In your equation $x$ and $y$ are denoting the same thing, so you should use one symbol for both. $x$ inside the $\sin()$ term implies that there is a wave in $x$-direction which is not what you intend. $\endgroup$ – Deep Sep 27 '16 at 5:27
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Consider the following proof: The wave equation is given by

$$ \frac{\partial ^2 y}{\partial x^2} = c^2 \nabla ^2 y. $$

Suppose, i consider a rope with on end attached to a wall and the other end free to move. Clearly, we can find the boundary condition for this system at any time t : y(L,t) = 0 and y(0,t) = $Asin(\phi_0 - \omega t)$. (note this is already stated in the fact that the rope undergoes shm). Also at any given x, the laplacian is defined. Now,we want to find out how the source moves. It is not difficult to guess such a scenario, where the rope and source both move together in shm. Now for the fun part. Uniqueness theorem gurantees that this is the only solution to the system given the boundary conditions for any x. Hence the source must be undergoing shm too. You can read up on uniqueness theorem online. I may be wrong in the proof, so any doubts are welcome.

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