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How can you check whether the given expression shows simple harmonic motion or not?And also how to calculate angular frequency of the given equation?

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marked as duplicate by Hritik Narayan, Kyle Kanos, Jon Custer, JMac, SRS Jul 10 '17 at 15:55

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  • $\begingroup$ Check if $\ddot{x}=-\omega^2 x$ for some constant $\omega$. $\endgroup$ – ZeroTheHero Jul 9 '17 at 14:44
  • $\begingroup$ You've got to be a bit more specific than that. $\endgroup$ – Bob Knighton Jul 9 '17 at 14:47
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If the equation describing either the displacement, velocity, or acceleration, contains just a single linear time-dependent sinusoidal term (perhaps with a phase offset - cosine is just sine with a phase shift) then it's simple harmonic. Examples of expressions of simple harmonic motion:

$$a(t) = -\omega^2 A e^{-i\omega t}\\ v(t) = B \cos(\omega t)\\ x(t) = C \cos(3t+5)+ D \sin(3t + 1) + 1.23$$

(That last one has two terms, but they both have the same frequency. A bit of manipulation will turn that into a single term, with a different amplitude and phase.)

Example of non-simple harmonic (periodic) motion:

$$x(t) = A\sin(\omega t) + B\sin(3\omega t)\\ x(t) = A\sin(\omega t^2)$$

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  • $\begingroup$ I would define periodic motion as motion that repeats after regular time intervals, called the period $p$, i.e. $f(x) = f(x + np)$ for all $x$, all (positive and negative) integers $n$, and fixed $p$. By that definition $x(t) = A \sin(\omega t^2)$ is not periodic. The fact that it happens to be repeat for some equally spaced values of $t$, like $t = 0, \pi/\omega, 2\pi/\omega,, 3\pi/\omega, \dots$ is not sufficient. $\endgroup$ – alephzero Jul 9 '17 at 15:57
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For a system if the acceleration $a$ is always proportional to the displacement from a fixed point $x$ and the acceleration is directed towards the fixed point, then the motion is simple harmonic.

In symbols this gives the relationship $a = - \omega^2 x$ where the constant $\omega^2$ is specially chosen to be a square so that it is positive.
$\omega$ is sometimes called the angular frequency and is related to the frequency $f$ and period $T$ of the motion.
$\omega = 2\pi f= \frac{2\pi}{T}$

The equation $a = \ddot x= - \omega^2 x$ is a second order differential equation and has solutions of the form $x = A \sin \omega t + B \cos \omega t,\, x = Ce^{i\omega t} + De^{-i\omega t}$, etc, where $A, \,B, \, C, \, D$ are constants which can be determined from the initial conditions.

Another clue that the a motion might be simple harmonic is that the frequency of the motion does not depend on the amplitude.

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