The standard equation for any object moving in a linear simple harmonic motion (SHM) is $$x=A\sin(\omega t +\delta)$$ where $A$ is the amplitude, $\omega$ is the angular velocity.

Likewise, converting the linear variables in the above equation to angular variables, we get the equation of angular SHM as $$\theta =\theta_o\sin(\omega t + \delta)$$

On differentiating this equation with respect to time, we get, $$\Omega = \theta_o\omega\cos(\omega t +\delta)$$

This is where I got confused with the difference between the term $\Omega$ and $\omega$.

enter image description here

I know that the rate of change of angular displacement $\theta$ is the angular velocity $\omega$, which means that the equation of angular velocity of SHM is $\omega = \theta_o\omega\cos(\omega t +\delta)$, which also does not seem to make sense as the equation of $\omega$ cannot be in terms of $\omega$ itself.

So, my question is, What exactly is the difference between the terms $\omega$ and $\Omega$ in angular SHM?

  • $\begingroup$ Your definition and introduction of the angular variables are unclear. Does the angular variable come from the Lagrangian mechanics? $\endgroup$ – Vadim Sep 29 '20 at 7:42

You seem to have goofed up in the drama of variables. Now to simplify this, lets begin with one dimensional simple harmonic motion along the x-axis. Have a look first at the following animation: enter image description here

As you can see that when you project a circular motion on either of the axes, you get a simple harmonic motion. Now the $\omega$ that you have put inside the the sine function, i.e., the coefficient of time(t), is nothing but the angular velocity of the particle in this circle whose projection on the x/y axis is the simple harmonic motion.

Now you do not convert anything to angular variables. That's a different case of the simple harmonic motion known as ANGULAR SIMPLE HARMONIC MOTION . Suppose you consider the following pendulum, enter image description here

Here when you solve for the torque of this pendulum it comes out to be: $$\tau\propto\, -\alpha$$ with $\alpha$ being the angle from the vertical at any given point of time. Here, an approximation has been used to arrive at the above propotionality which is that the amplitude must be small so that $$sin\,\alpha\,\approx\,\alpha$$ Solving this further like we solve most cases we arrive at the differential equation:

$$\frac{d^2{\beta}}{dt^2} + q^2\alpha\,=0\,\,\,(1)$$

where $\beta$ is the angular acceleration of the pendulum. Now this differential equation is the same as the differential equation of a normal SHM as shown in the first case above, $$\frac{d^2a}{dt^2} + {\omega}^2x\,=0\,\,\,(2)$$ where $\omega$ what I have explained in the first case. The same differential equation would quite obviously give the same solution. Therefore the solving is often skipped and we directly write the solution with the new "Angular Variables". Now you can see that $q$ in equation (1) above corresponds to $\omega$ in equation (2) above. This is is nothing but a simple constant that appears while solving the differential equation which can be expressed in terms of mass, length and acceleration due to gravity.

Now the capital $\Omega$ that you use is nothing but the angular velocity of the pendulum. $$\Omega\,=\frac{d\alpha}{dt}$$ Now everything should be just fine from here on, with the time period being $$T= \frac{2\pi}{q}$$ which can be defined just like the period of any mathematical periodic function is defined. And by now you should be clear about the two variables. But physics is not only about the mathematics and this is mechanics!! Study of our everday motion! So if you still want to visualise the constant $q$ like we did in the first case, do the following:

  1. Consider the angular SHM to be like a linear simple harmonic motion by projecting the center of mass of the pendulum bob on the horizontal axis.

  2. Now just look at this projection.

  3. Think this projection to be another projection of a circular motion with angular velocity $q$.

So I hope that all sort of confusion is clear after reading this. Should you struggle, feel free to comment and sort it out!

  • $\begingroup$ @rash, if you liked the answer, please take a moment off to accept it. $\endgroup$ – Tesla's Coil Sep 30 '20 at 3:22

The rate of change of angular displacement is called angular velocity which is usually denoted by $\omega$. $$\omega=\frac{d\theta}{dt}$$

In SHM, the term $\omega$ does not mean the same as above. Here it is used to denote angular frequency with respective to the phase at any instant in SHM. Now I am assuming you know the meaning of phase. So angular frequency means how many phases can be completed in a second or even simply how fast the object is oscillating. Hence it appears in the velocity term for SHM. Let Time Period of SHM be T. Then$$\omega = \frac{2\pi}{T}$$

In angular SHM , $$\theta=\theta_0\sin(\omega t)$$ We have a problem with denoting angular velocity by $\omega$ as it has already been used for angular frequency . Hence we take up another for angular velocity $\Omega$. Now differentiating above with time. $$\Omega= \theta_0 \omega \sin(\omega t)$$

Note here $\Omega$ measures angular velocity - how fast is angular displacement changing (which is not constant). Whereas $\omega$ measures angular frequency - how any oscillations can the object undergo in one unit time (which is a constant)


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