What is angular frequency in simple Harmonic Motion?

If simple harmonic motion is a linear to and fro motion then whose angular frequency are we talking about? A linearly moving body cannot have an angular frequency right? I know that we compare it with uniform circular motion, but then how can we associate angular quantities for linear translation? Is the frequency associated to the air particle/surrounding particles that disturbed due to that SHM?

And why was the motion compared to UCM only?


2 Answers 2


Simple harmonic motion is not just to and fro motion - it is to and fro motion where the displacement from the central position over time follows a sine wave. There are other types of to and fro motion which are not simple harmonic motion - for example, where the displacement from the central position over time follows a triangle wave rather than a sine wave.

The connection with uniform circular motion is that simple harmonic motion is the projection on the x-axis of the position of an object moving with uniform circular motion in the x-y plane. If the object is moving in a circle of radius $R$ with frequency $f$ then at time $t$ its position is

$\left(R\sin (2\pi f t), R\cos (2\pi f t)\right)$

so the projection of its position on the x-axis is $R \sin(2\pi f t)$. A particle moving with simple harmonic motion $x(t)=R \sin(2\pi f t)$ is said to be moving with frequency $f$ because it takes $\frac 1 f$ seconds to move from one extreme point $x=R$ to the opposite extreme point $x=-R$ and back again.

  • $\begingroup$ Thank you. But what exactly is angular frequency ? What does it describe about the motion if we just look at it, more in general terms ? ( what exactly does angular frequency tell us about the SHM) ? $\endgroup$ Jan 23, 2023 at 12:54
  • $\begingroup$ @TrishaShah The reciprocal of the frequency, $\frac 1 f$, is the period of the SHM - the time taken to complete one full cycle of the motion. $\endgroup$
    – gandalf61
    Jan 23, 2023 at 15:10

Simple harmonic motion is a clockwise rotation about the unit circle in the scaled phase space coordinates $(x/x_{xmax},\quad v/x_{max}\omega)$. If the oscillator is a mass $m$ on a spring constant $k$, then $\omega=\sqrt{\frac{k}{m}}$.

$$ \begin{align} x &=x_{max}\quad\cos(\omega t) \\ v=\frac {dx}{dt} &=-x_{max}\omega\quad sin(\omega t) \\ \end {align} $$

Notice that

$$ (\frac{x}{x_{max}})^2 + (\frac{v}{x_{max}\omega})^2=1 \\ $$

$$ \begin{bmatrix} x/x_{max}\\ v/x_{max}\omega\\ \end{bmatrix}_t= \begin{bmatrix} cos(\omega t) & sin(\omega t) \\ -sin(\omega t) & cos(\omega t) \\ \end{bmatrix} \begin{bmatrix} 1\\ 0\\ \end{bmatrix}_{t=0} $$

  • $\begingroup$ But isn't that just a convenient way of expressing SHM , rather than its actual motion , which is kind of kind of linear to and fro? $\endgroup$ Jan 23, 2023 at 12:56
  • $\begingroup$ It is even more than convenient!! Instead of writing down a second order differential equation in x that balances forces (Newton thinking), you can write down first order differential equations in x and v (Hamilton thinking using "Hamilton's equations"). Solving for x and v directly gives how the point (x, v) moves around in phase space as a function of time. As shown above, that path for the SHM is a closed circle, with an angle $\omega t$ specifying where the mass is. For a discussion of Hamiltonian mechanics, please see a mechanics book (like Goldstein). $\endgroup$ Jan 23, 2023 at 21:36
  • $\begingroup$ Evolution of objects often result in rotations in phase space or regular space. $\endgroup$ Jan 23, 2023 at 21:39

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