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This question already has an answer here:

I am confused about simple harmonic motion. I understand that it follows a sine wave but is it possible to explain why it does?

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marked as duplicate by sammy gerbil, David Hammen, Kyle Kanos, Qmechanic Feb 23 '17 at 12:32

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Total kinetic energy remains constant, but is periodically changing between 2 forms in a continuous motion.

If you picture the unit circle, with x and y representing kinetic energy and potential energy respectively, you can visualize the situation. The radius of the circle (total magnitude of energy) doesn't change. Since the angle is replaced by time in the periodic motion, you can visualize time as going around the unit circle. It will gradually transition between maximums and minimums in x and y, while keeping the total magnitude the same. Sin and cos would be describing each type of energy as they change periodically.

I'm not sure if that helps at all, it didn't really explain why it is a sin wave, but it explains why a sin wave works. This Wikipedia link shows a good visual about what I mean.

The sin wave comes from the solution to the differential equation $F = -kx$ knowing $F = ma$ using the common notation $a = \ddot x$ since acceleration is the second derivative of position with respect to time, you get the equation $$m \ddot x = -kx$$ which has a sinusoidal solution for $x(t)$ characteristic of simple harmonic motion.

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A particle is said to be executing simple harmonic motion if it satisfies the following differential equation equation:

$$\frac{d^2\phi}{dt^2} = - \omega^2\phi \tag{1}$$

The $\sin$ wave is merely a solution to the above differential equation.

$$\phi = \phi_o \sin (\omega t + c) \tag{2}$$

You can verify it by substituting the function $(2)$ in equation $(1)$.

If a quantity is of the form $(2)$, then we say that the quantity is varying harmonically. However, this is not the only solution to the $(1)$ differential equation.

$$\phi = \phi_o \cos (\omega t + c)$$

$$\phi = \phi_{o1} \sin(\omega t) + \phi_{o2} \cos (\omega t)$$

are also the solutions for the $(1)$ differential equation.

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