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I seem to be puzzled by the importance of using angular frequency $\omega$ as the frequency scalar to the mathematical model of Simple Harmonic Motion rather than merely using $f$, and by $f$ I mean $f$ such that $f = \frac{1}{T}$, $T$ being the time/distance (time in this case) for a complete cycle. Rather, I can accept both of them as being plausible scalars for $t$ at the same time, and I'll elaborate on this.

$\omega$ is defined as $2\pi f$, or the number of cycles/oscillations/revolutions per second. This sounds fine to me, but I'm confused as to why $f$ is in there other than from knowing that $\omega \ T = 2\pi$ Due to the deduction that $cos(\omega (t + T)) = cos(\omega t + 2\pi)$ and thus it would make sense that $\omega \ T = 2\pi$ by equating arguments. However, I don't see how merely saying $f$ as the time scalar is going to fail to portray simple harmonic motion other than the fact that $ft$ do not resolve to radians. So, I can see why it's a poor choice semantically but not intuitively.

Say my equation for simple harmonic motion is $x(t)=A\cos(ft + \phi)$. If my object undergoes two cycles from $0$ to $2\pi$, I would merely say $f = 2$ and it can model that appropriately. The only thing I find odd here is "having two cycles from $0$ to $2\pi$" doesn't really make any physical sense, but I've failed to notice how making $\omega$ the scalar for $t$ is going to resolve this. If this made absolutely no sense please let me know and I'll try to elaborate what I don't get further.

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  • $\begingroup$ You don't think it's handier in using phasors, to feature omega? Your question, no offence, is not as clear to me as the person that upvoted it though. $\endgroup$ – user171879 Oct 14 '17 at 15:02
  • $\begingroup$ Not sure what phasors are, but I'm trying to wrap my head around the important of using $\omega$ when describing harmonic motion rather than using some arbitrary scalar $f$. Like, why it's necessary to make our frequency scalar $\omega = 2\pi f$ rather than $f$. $\endgroup$ – sangstar Oct 14 '17 at 15:09
  • $\begingroup$ Phasors are phase vectors, we use them to determine driven SHM, so we can work out electrical circuits properly, why badly designed buildings fall down in earthquakes, and in, I dunno, 50 percent of engineering problems. Have a look at en.wikipedia.org/wiki/Phasor if you want diagrams. $\endgroup$ – user171879 Oct 14 '17 at 17:34
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Do you know that sines and cosines are defined in terms of projections on to a diameter of a point on the circumference of a circle of unit radius? [This enables us to understand how we can have sines and cosines of angle greater than 90°.] Thus if the frequency of SHM is f, then $\omega=2 \pi f$ represents the angular velocity of a point moving at frequency f around a circle of radius A, $\omega t$ represents the angle swept out by the point in time t (with the point at one end of a horizontal diameter at t = 0), and $x=A\cos(\omega t)$ represents the projection of the point on to a horizontal diameter, and indeed represents the SHM itself (if x is at its maximum at t = 0).

I realise that without diagrams this must be hard to follow, but it is a standard approach to shm and will be found, explained properly, in decent textbooks. I've given this potted version just to bring out the significance of $\omega$ as angular velocity.

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  • $\begingroup$ Philip, I edited in a diagram, (waiting approval or not), but obviously please remove it if you don't like it, thanks $\endgroup$ – user171879 Oct 14 '17 at 17:40
  • $\begingroup$ I am aware of this idea for sines and cosines, which is why it's totally understandable as to why its a (likely empirical) model to SHM. However, my question sort of boils down to -- why does $f$ itself not represent the angular velocity moving around a circle of radius $A$? $\endgroup$ – sangstar Oct 15 '17 at 13:01
  • $\begingroup$ Angular velocity is best represented by a vector, but in this case, as you say yourself, frequency is a scalar. So the vector addition we need to relate various aspects of shm, would not be suitable using scalar quantity. math.stackexchange.com/questions/1400421/… is related, slightly, to this, but if you are learning about SHM, this is a later part of the course $\endgroup$ – user171879 Oct 15 '17 at 16:14
  • $\begingroup$ Because $f$ is the number of cycles per second, which, in terms of the point going round the circle, is the number of revolutions per second. But for each revolution, the point sweeps out an angle of $2 \pi$ radians, so the angle swept out per second is $2 \pi f$ and the angle swept out in time $t$ is $2 \pi f t$, and that's what you take the sine or cosine of! $\endgroup$ – Philip Wood Oct 15 '17 at 16:16

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