# How to visualize the angular frequency in SHM?

Can anyone define how can I visualize the angular frequency ($$\omega$$) in a simple harmonic motion $$y(t)=R \sin(\omega t+\phi)$$ (where $$\omega=2\pi/T$$). Because we can visualize the frequency ($$f=1/T$$) as the number of times the process is repeated in 1 sec so how to define this angular frequency $$\omega$$?

Also in SHM ($$a=–kx/m$$) then here why do we define $$k/m = \omega^2$$?

So please tell me how to visualize the angular frequency in SHM.

You can relate it to $$\omega$$ in uniform circular motion(the angular velocity) as SHM can be visualised as projection of uniform circular motion.

• The constructions sometimes goes by the name "reference circle" which might help in searching for it. Commented May 14, 2019 at 17:48

Consider looking at a rotating disk from above its axis of rotation. Lets say we have a point marked on the perimeter of the disk. The time it takes to complete a full rotation is the angular period and the inverse is the angular frequency of the disk. In other words the angular frequency is the number of rotations completed per unit of time. When the disk is looked at 'edge on' from the side, the marked point just appears to going vertically up and down. This motion of the marked point seen from the side is SHM. The number of times it goes up and down per unit time, is equal to the number of rotations of the disk per unit time. The angular aspect comes from the rotation of the disk and this is the best way to visualise the angular frequency of SHM.

For your first question regarding how to "visualize" $$\omega$$, I understand what your asking is, "can we draw an intuitive understanding of the angular frequency $$\omega$$ in much the same way that we can for the frequency $$f$$?". The answer is yes, though I think using a visualization is a little overkill and will leave you looking back at videos of points rotating on circular paths trying to remember what part of the motion $$\omega$$ dictated. The following argument is long winded, but don't be overwhelmed by it, it's length is not because of any complexity but simply because I tried my best to make it as intuitive as possible without forcing you to accept any one equation at face value. As to your first question, I'll have to leave that to someone else for now...hope this helps

I think your intuition for the frequency is a better jumping off point; like you said the frequency $$f = \frac{1}{T}$$ describes the number of full oscillations that are completed in one second. But more fundamentally, the frequency is a ratio, designed to summarize how far along one continuous process has come (a full oscillation) after a certain other continuous process has completed (The passage of time i.e. one second). This is the basic intuition that should underlie all these quantities, and there are more then a few. Velocity $$v = \frac{\Delta d}{\Delta t}$$, describing how much distance is covered in one second, is the most prominent one that comes to mind. Again, velocity is simply quantifying how far along an object has come in it's continuous motion after one second has elapsed.

Let's get back to considering frequency. The important thing to notice, is that these ratios are constructed as $$s = \frac{\Delta a(t)}{\Delta b(t)}$$ where $$a(t_{0})$$ and $$b(t_{0})$$ describe two continuous processes in time (the passage of time is of course a continuous process itself $$t(t) = t$$). In the case of frequency, we are measuring number of oscillations after a time $$T$$ (known as the period) has elapsed, against the total amount of elapsed time. In other words $$f = \frac{NumOscillations(T)}{\Delta t(T)}$$. We can now evaluate both of these functions in this expression because the period $$T$$ is defined as the time it takes for the oscillator to complete one oscillation. By definition then, $$NumOscillations(T) = 1$$, and $$\Delta t(T) = T$$, so we get $$f = \frac{1}{T}$$.

Now finally let's consider angular frequency. You wrote $$\omega = \frac{2\pi}{T}$$. Following the above logic we know this ratio is trying to summarize how much of one process (measured by the top number) has completed after another process (the bottom number) has passed. The intuition for the angular frequency is gained by simply asking, "what two continuous processes are being measured?". The bottom process is easy, it's the same as with the frequency, the passage of time. But what is the top process? Let's return to our SHM function you wrote, to get a sense of what the top number is measuring $$y(t) = R \sin (\omega t + \phi)$$. More generally we can write $$y(t) = R \sin (\theta(t))$$ where $$\theta$$ is the "phase" that is passed into the sin function to retrieve $$y(t)$$. I will say now that the top process is the change in the phase during one period. Using our "intuition" we can then write $$\omega = \frac{\Delta\theta(T)}{t(T)}$$. We can evaluate the bottom function as before $$t(T) = T$$. To evaluate the top function we use our knowledge of the sin function and the definition of the period $$T$$: we know that the sin function has a period of $$2 \pi$$ so that $$y(t_1 + T) = R \sin (\theta(t_1 + T)) = R \sin (\theta(t_1) + 2 \pi)$$. Here we see that $$\Delta \theta(T) = \theta(t_1 + T) - \theta(t_1) = 2\pi$$ and that is why the angular frequency is given by $$\omega = \frac{\Delta \theta(T)}{t(T)} = \frac{2\pi}{T}$$.