0
$\begingroup$

I've seen in many books, internet and video lectures, that equation of SHM: $$x(t)=A\cos(\omega t+\phi)$$ where they just say that (without telling anything about $\omega$): $$\omega=\frac{2\pi}{T}$$

where $\omega$ is Angular Frequency, $A$ is the amplitude, $\phi $ is the phase angle and $T$ is time-period.

But I want to know from where this formula ($\omega =\frac{2\pi}{T}$) came (or how to derive it)? And what's the intuition behind angular frequency?

Note: I don't want an explanation/answer based on/related to circular motion.

$\endgroup$
1
  • $\begingroup$ I now found the derivation of this formula in HRK Vol 1 $\endgroup$
    – Ishika
    Apr 12, 2022 at 14:54

4 Answers 4

3
$\begingroup$

The cosine is a periodic function. So, for the function:$A\cos(\omega t + \phi)$, when $t = 0$ its value is $A\cos(\phi)$. After a time $T$ the function repeats itself when $T$ is such that: $A\cos(2\pi + \phi)$. It happens when, $$2\pi = \omega T \implies \omega = \frac{2\pi}{T} $$

$\endgroup$
2
  • $\begingroup$ thanks, it cleared my first question, but can you please also include the answer to my second question(intution behind angular frequency)? so, I now realise that this answered my second question too, that angular frequency tells us the number of times one complete oscillation is repeated, right? $\endgroup$
    – Ishika
    Jan 31, 2022 at 12:17
  • $\begingroup$ @Ishika yes, exactly. $\endgroup$ Jan 31, 2022 at 14:59
0
$\begingroup$

Based on the general theory of the linear ordinary differential equations with constant coefficients, the most general solution to equation $$\ddot{x}(t)+\omega_0^2x(t)=0$$ (i.e., the equation describing the simple harmonic oscilator) is given by $$x(t)=c_1\cos(\omega_0 t) + c_2\sin(\omega_0 t).$$ This can also be written as $$x(t)=A\cos(\omega_0 t + \phi),$$ where $$ c_1=A\cos\phi, c_2=-A\sin\phi.$$

We see that the solution is epriodic, repeating itself every time the argument of the trigonometric functions, $\omega_0 t$ changes by $2\pi$, i.e., every time when time $t$ changes by $2\pi/\omega_0$. As this is the definition of period, we have $$T=\frac{2\pi}{\omega_0}.$$

$\endgroup$
0
$\begingroup$

Unit analysis might also help your intuition:

  • $\phi$ is in radian so $\omega t$ must also be in radian since they're being added together.
  • Since $t$ is in seconds, $\omega$ must be in radian per seconds so that their product, $\omega t$, is in radian.
  • So $\omega$ needs to be something in radian divided by something in seconds.
  • $\frac{2 \pi}{T}$ makes sense since $2 \pi$ is the period in radian and $T$ is the period in seconds.
$\endgroup$
-1
$\begingroup$

You can also derive that equation from the fact that sine wave is periodic with time period $T$ and angle $2 \pi$

$$x(t) = A \sin(\omega t + \varphi) = A \sin(\omega(t+T) + \varphi) = A \sin(\omega t + \varphi + \omega T)$$

The above expression is true only if $\omega T = 2 k \pi$, where $k \in \mathbb{Z}$, hence

$$\omega = \frac{2 \pi}{T}$$


Angular velocity

Imagine a particle is rotating at tangential speed $v$ at distance $r$ from axis of rotation and it takes time $T$ to complete one full revolution.

The distance the particle travels in one revolution is

$$s = 2 \pi r$$

and we can relate this to tangential speed as

$$v = \frac{s}{T} = \frac{2 \pi}{T} r = \omega_\text{av} r$$

where $\omega_\text{av}$ is newly introduced variable known as average angular velocity. It denotes time it takes particle to complete one full revolution or $2\pi \text{ rad}$. Common units of measure for angular velocity are revolutions-per-minute ($\text{rpm}$) and radians-per-second ($\text{rad/s}$).

To get instantaneous angular velocity, you need to include limits and derivatives. The distance particle covers in time $\Delta t$ is

$$\Delta s = r \cdot (\theta(t + \Delta t) - \theta(t))$$

and the velocity is

$$v(t) = \lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t} = r \frac{d \theta(t)}{dt} = r \omega(t)$$

where both $v$ and $\omega$ are now instantaneous velocities rather than average velocities.

$\endgroup$
1
  • $\begingroup$ I've asked for Angular Frequency and not Angular Velocity(I saw on Youtube that they're different though represented by same symbol, let me know if I'm wrong). Also, you answered it based on rotation, I had said in the question that I don't want an answer based on circular motion. $\endgroup$
    – Ishika
    Jan 31, 2022 at 11:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.