All oscillating motions – the movement of a guitar string, a rod vibrating after being struck, or the bouncing of a weight on a spring – have a natural frequency. The basic situation for calculation involves a mass on a spring, which is a simple harmonic oscillator. For more complicated cases, you can add the effects of damping (the slowing of the oscillations) or build up detailed models with driving forces or other factors taken into account. However, calculating the natural frequency for a simple system is easy.

The Natural Frequency of a Simple Harmonic Oscillator Defined

Imagine a spring with a ball attached to the end with mass ​m​. When the setup is stationary, the spring is partially stretched out, and the whole setup is at the equilibrium position where the tension from the extended spring matches the force of gravity pulling the ball downward. Moving the ball away from this equilibrium position either adds tension to the spring (if you stretch it downwards) or gives gravity the opportunity to pull the ball down without the tension from the spring counteracting it (if you push the ball upward). In both cases, the ball starts oscillating around the equilibrium position.

The natural frequency is the frequency of this oscillation, measured in hertz (Hz). This tells you how many oscillations happen per second, which depends on the properties of the spring and the mass of the ball attached to it. Plucked guitar strings, rods struck by an object and many other systems oscillate at a natural frequency.

Calculating the Natural Frequency

The following expression defines the natural frequency of a simple harmonic oscillator:

(f=frac{omega}{2pi})

Where ​ω​ is the angular frequency of the oscillation, measured in radians/second. The following expression defines the angular frequency:

(omega=sqrt{frac{k}{m}})

So this means:

(f=frac{sqrt{k/m}}{2pi})

Here, ​k​ is the spring constant for the spring in question and ​m​ is the mass of the ball. The spring constant is measured in Newtons/meter. Springs with higher constants are stiffer and take more force to extend.

To calculate the natural frequency using the equation above, first find out the spring constant for your specific system. You can find the spring constant for real systems through experimentation, but for most problems, you are given a value for it. Insert this value into the spot for ​k​ (in this example, ​k​ = 100 N/m), and divide it by the mass of the object (for the example, ​m​ = 1 kg). Then, take the square root of the result, before dividing this by 2π. Going through the steps:

(begin{aligned} f&=frac{sqrt{k/m}}{2pi}&=frac{sqrt{100/1}}{2pi}&=frac{10}{2pi}&=1.6text{ Hz}end{aligned})

In this case, the natural frequency is 1.6 Hz, which means the system would oscillate just over one and a half times per second.