# Questions about $\omega, f, \omega_{0}$ in harmonic oscillator

The motion equation for a mass-spring system is

$$x(t)=A\sin({\omega t+\phi})$$

where $$A$$ is amplitude and $$\omega$$ is vibration frequency. We have frequency value with $$f$$ and its inverse $$T$$. What is the difference between $$\omega$$ and $$f$$?

If there is an example for these terms, It would be nice.

Another questions is about the damped harmonic oscillator. In the damped harmonic oscillator, $$\omega^2=\omega_0^2-\beta^2$$. My teacher told that $$\omega$$ and $$\beta$$ are constant. I don't understand if $$\omega$$ is varying with time, how it can be constant?

• I don't understand if $\omega$ is varying with time, how it can be constant? Why do you think that $\omega$ is varying with time? It’s not. Feb 19, 2020 at 17:18
• Why did you say “distance ($\omega_0$)”? The frequency $\omega_0$ is not a distance. Feb 19, 2020 at 17:33
• Isn't that correct? No, not for what we call damped harmonic motion. The amplitude decreases but the period does not change. Either your senses are untrustworthy or you observed a system that is not accurately modeled by the equations for a damped harmonic oscillator. Feb 19, 2020 at 17:35
• In the future, please stick to asking one question per post. You shifted from talking about an undamped oscillator to a damped one, never defined what $\beta$ and $\omega_0$ are, never wrote the equations for damped motion, etc. Feb 19, 2020 at 17:40
• Are you using your senses or your imagination? These are very different things. Feb 19, 2020 at 17:42

The angular frequency, $$\omega$$ is related to the time period $$T$$ via $$\omega = 2\pi / T$$. Evidently then we could also write $$\omega = 2\pi f$$.

To try and justify this, consider the trajectory for simple harmonic motion

$$x = A\sin{(\omega t + \phi)} \, .$$

If we increase $$t$$ by $$2\pi / \omega$$, then

$$x_2 = A\sin \left(\omega \left(t + \frac{2\pi}{\omega} \right) + \phi \right) = A\sin{(\omega t + \phi + 2\pi)}$$

Increasing $$t$$ by $$2\pi / \omega$$ has increase the phase by $$2\pi$$, so the object has returned to its previous position. This time increment is then interpreted as the time period of the oscillation.

As for your second question, to understand why the oscillator behaves the way it does under driving forces and/or damping, you really need to try and solve the differential equation for the motion. If you "underdamp" the oscillator, then you do indeed obtain the relationship $$\omega^{2} = \omega_{0}^{2} - \beta^{2}$$ where $$\beta$$ is the damping ratio.

• We have solved the differential equation for the harmonic oscillator under a friction force that varies by the velocity. In the example the friction force was $F_{s}=-b\dot{x}$. So, $F_{\text{net}}=M\ddot{x}=-b\dot{x}-Cx \Rightarrow M\ddot{x}+b\dot{x}+Cx=0 \Rightarrow \ddot{x}+\frac{b}{M}\dot{x}+\frac{C}{M}x=0 \Rightarrow \ddot{x}+\frac{1}{\tau}\dot{x}+w_0^2 x=0$. We didn't solve this differential equation in a mathematician way, we solved it by solution proposal $x(t)=e^{-\beta t}x_{0}\sin(\omega t+\phi)$. Feb 19, 2020 at 16:30
• By doing this I have found the expression $\omega^2=\omega_0^2-\beta^2$. But sometimes I can't be able to understand the meaning of the equations. But, thanks for your help! Feb 19, 2020 at 16:32
• Physically, it just means that $\omega_{0}$ is the natural frequency of the oscillator in the absence of damping (using your notation, $\omega_{0} = \sqrt{\frac{C}{M}}$). Also, $\beta$ is proportional to the coefficient of velocity (in your example, $b$) and inversely proportional to the square roots of mass and the spring constant. Intuitively, the more damping in the system, the lower the natural frequency of your oscillator! Feb 19, 2020 at 16:37