In the simple harmonic oscillator model, we talk about the angular frequency and I understand that this tells something about the instantaneous angular velocity of the object. For example, a simple pendulum. But what does this mean when we apply this concept to an object attached to a spring that behaves as a simple harmonic oscillator? There is no instantaneous circular motion for the object, so how can we relate the angular frequency to a system like this?

# [Physics] The meaning of “Angular frequency” of a spring that behaves as a simple harmonic oscilator

complex numbersharmonic-oscillatorspringwaves

#### Related Solutions

In the case of a simple pendulum (also called a mathematical pendulum of simple gravity pendulum), one assumes that *all of the mass* is the bob and the rest of the pendulum is massless. An example of the simple pendulum is given in the image below.

The simple pendulum (see wikipedia or hyperphysics) leads to a simple differential equation by using Newton's second law: $$\ddot{\theta}+\frac{g}{l}\sin(\theta)=0.$$

This pendulum gives the easiest way te look at harmonic motion. The above case is what they call the simple pendulum.

You could add an extrernal source so it would be a driven simple pendulum, of friction so that it becomes a simple pendulum with friction.

If you want to further complexify it, you could drop the assumption that all of the mass is in the bob and add inertia to the picture (so a real-life pendulum). This kind of pendulum is called the physical pendulum (or compound pendulum), the swinging body is no longer considered a point mass, but a mass with finite measurements. An example (compared to the simple pendulum) is given on the figure below.

This also leads to equations using Newton's second law applied on angular momentum (just as the ones which were used to derive the equation for the simple pendulum).

An (undriven) damped harmonic oscillator (satisfying $m\ddot{x} + \gamma \dot{x} + \omega_0^2 x=0$) can be solved by the solution(s) $x_0e^{i \omega t}$. For an underdamped oscillator these solutions represent pure oscillations mixed with exponential decay(/growth). Because both solutions for $\omega$ oscillate with the same period, all combinations of them also oscillate with the same period.

I suppose that your confusion arises from an intuitive idea of why a restoring force leads to periodic motion. In the undamped case the phase space trajectories of the particle are closed (i.e. the particle always returns to the same position(s)). In the damped case the trajectories spiral towards rest. But both of these motions are periodic in the sense that they reach their relative extrema (and zero crossing) after a specific interval of time. How can you see this? Perhaps you can take some inspiration from the undamped case: notice that the period is independent of the size of the orbit. Because the equation is linear any change in amplitude is exactly accounted for by a change in the force. In fact all linear (homogenous) ODE's can be satisfied by the anstaz above (i.e. solutions are periodic, damped, or both).

## Best Answer

The basic equation for the simple harmonic oscillator is $\frac{d^2x}{dt^2} =- \omega^2 x$. One solution that works here is $x=x_0 e^{i\omega t}$. Notice how the complex conjugate, $x^* = x_0^* e^{-i\omega t}$ is also a valid solution. The reason for this is because, while the body may be oscillating in one dimensional motion, it can be thought of as going either clockwise or counterclockwise in circle in the complex plane, where its angular velocity is either $i\omega$ for counterclockwise or $-i\omega$ for clockwise.