0

Why so the oscillations which are not circular also have angular frequency which is a quantity related to the circular motion? I have referred many articles related to simple harmonic motion where they directly include the term angular frequency. Is every periodic oscillation correlated to circular motion such that it has the angular frequency term?

Qmechanic
  • 201,751
Karthikeyan s
  • 1
  • 1

2 Answers2

1

I think the root of this is Euler's relationship in complex numbers.

A complex number can be represented as $R\exp(i\omega t)$, where $R$ is an amplitude and $t$ is time and $i^2=-1$. This complex number can be drawn as a coordinate in the complex plane (real numbers vs imaginary numbers) and will trace out a circular path with angular frequency $\omega$.

A linear oscillation (e.g. SHM) can then easily be represented as the projection of this circular motion onto the real or imaginary axis through Euler's relation.

$$R\exp(i\omega t) = R\cos \omega t + iR\sin \omega t$$

I.e. the real part is a (co)sinusoidal oscillation, whilst the imaginary part is an oscillation with the same amplitude and angular frequency, but 90 degrees out of phase.

Possibly this explanation doesn't help if you've never come across complex numbers, but even if you haven't, you should be able to see that a linear sinusoidal oscillation can be viewed as the projection of a 2-D circular motion onto one axis.

ProfRob
  • 130,455
  • Thank you for your answer.Reciprocrating motion can be obtained from the projection of object moving in a circular path. I wanna understand that whether they take every harmonic oscillations are analogous to circular motion and device a mathematical equation? – Karthikeyan s Nov 11 '14 at 11:58
0

More context might help, but I can think of two possible interpretations:

  1. The angular frequency as defined by $\omega = 2\pi f = 2\pi/T$ can be applied to any periodic motion ($T$ is the period and $f=1/T$ the “regular” frequency).

  2. The definition as the time derivative of the angle (as a coordinate) $\omega = \dot\phi = d\phi/dt$ could be applied to any motion. (Though it might more properly be called angular velocity.)

xebtl
  • 357
  • If you consider a reciprocating motion there is no circular motion involved so that no angle present. Why should we have an angular frequency which is rad/s to interpret the motion? my question is Is it an easy way to interpret the period motion with circular motion? – Karthikeyan s Nov 11 '14 at 11:55
  • As @Rob writes in his answer, an expression like $\cos\omega t$ can arise in contexts where no circles are involved (e.g. an oscillating mass on a spring might have position $x \sim \cos\omega t$. If the core of your question is, why is it “$\omega$” instead of “$f$”, the answer is simply the $2\pi$-periodicity of $\sin$ and $\cos$. – xebtl Nov 11 '14 at 14:34
  • I think in case of an oscillation of a mass on a spring, the angular velocity terms comes when we are going to represent the motion of the mass graphically. Since the motion of a oscillating mass on spring is similar to the motion of the foot on any coordinate axis of a particle moving in a circular path, hence like circular motion we can also represent the motion of the oscillating mass on spring. So compering the spring motion with a circular motion we introduce the angular velocity even in non-circular motion also . – Rajesh Sardar Nov 11 '14 at 15:23