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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?


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

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:

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

  • $\begingroup$ 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

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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