Phase Angle
In previous articles, we discussed the differential equation that describes oscillatory movement, particularly simple harmonic motion. We know the solution that satisfies the equation is expressed as
$$x=A\sin\left(\omega t+\phi_0\right).$$
Where \(A\) is the amplitude in meters \((\mathrm m)\), \(\omega\) is the angular frequency in radians per second \((\frac{\mathrm{rad}}{\mathrm s})\), and \(\phi_0\) is the initial phase in radians \((\mathrm{rad})\).
The phase angle is the angular component of a periodic wave, such that it is defined as the argument of the sine function, \(\omega t+\phi_0\). By choosing \(\phi_0\), we specify the oscillating object's initial position to make sure we have the correct equation with the oscillator's position, no matter where it may have been located at \(t=0\). We can restate the above equation in terms of the symbol \(\phi\) for the phase angle.
$$\begin{align*}\phi&=\omega t+\phi_0,\\x&=A\sin\left(\phi\right).\end{align*}$$
To determine the initial phase we use the following formula:
$$\phi_0=\sin^{-1}\left(\frac{x_0}A\right),$$
where \(A\) is the amplitude in meters \((\mathrm m)\) and \(x_0\) is the initial position of the object at \(t=0\) in meters \((\mathrm m)\).
A simple harmonic oscillator has an amplitude of \(3.0\;\mathrm{cm}\) and a frequency of \(4.0\;\mathrm{Hz}\). At time \(t=0\), its position is \(y=3.0\;\mathrm{cm}\). Where is it at time \(t=0.3\;\mathrm s\)?
The amplitude is \(A=0.03\;\mathrm m\) and the angular frequency is \(\omega=2\pi f=2\pi(4.0\;\mathrm{Hz})=8\pi\;{\textstyle\frac{\mathrm{rad}}{\mathrm s}}\). Now we can determine the initial phase,
\begin{align*}\phi_0&=\sin^{-1}\left(\frac{y_0}A\right),\\\phi_0&=\sin^{-1}\left(\frac{0.03\;\mathrm m}{0.03\;\mathrm m}\right),\\\phi_0&=\frac\pi2.\end{align*}
Now we know the position of the oscillator at any moment in time,
$$y(t)=0.03\sin\left(8\pi t+\frac\pi2\right).$$
We can find the position of the oscillator at time \(t=0.3\;\mathrm s\),
\begin{align*}y(0.3\;\mathrm s)&=(0.03\;\mathrm m)\sin\left((8\pi\;{\textstyle\frac{\mathrm{rad}}{\mathrm s}})(0.3\;\mathrm s)\;+\;\frac\pi2\;\mathrm{rad}\right),\\y(0.3\;\mathrm s)&=0.0093\;m.\end{align*}
The position of an oscillator is given by the equation:
$$y=(0.04\;\mathrm m)\sin\left((6\pi\;{\textstyle\frac{\mathrm{rad}}{\mathrm s}})t-\frac\pi2\;\mathrm{rad}\;\right).$$
Where is the oscillator at time \(t=0\)?
\begin{align*}y(0\;\mathrm s)&=(0.04\;\mathrm m)\sin\left((6\pi\;{\textstyle\frac{\mathrm{rad}}{\mathrm s}})(0\;\mathrm s)-\frac\pi2\;\mathrm{rad}\;\right),\\y(0\;\mathrm s)&=-0.04\;\mathrm m.\end{align*}
The initial phase will determine if a sine or cosine function is used to describe the position of the oscillating object. For example, if \(\phi_0=\frac\pi2\) we can use a cosine function instead of a sine function with the initial phase. This is due to the trigonometric identity, \(\sin\left(\frac\pi2+\theta\right)=\cos\left(\theta\right)\). The table below clarifies how the two expressions yield the same results at any time.
Equation | \(t=0\) | \(t=\frac\pi{2\omega}\) |
\(\sin\left(\omega t+\;\frac\pi2\right)\) | 1 | 0 |
\(\cos\left(\omega t\right)\) | 1 | 0 |
As a side note, the phase angle plays a very important role in experimental physics, especially in electronics where there is a direct relationship between the voltage and sinusoidal functions. In electronics, the phase angle refers to the angular displacement between the voltage and current waveforms in an alternating current circuit.
Understanding the Initial Phase in a Graph
We have covered the theoretical definition of the phase angle and the initial phase. How do we understand the impact of changing the initial phase of a sinusoidal function? It is easier to understand if we actually represent the sinusoidal functions in a graph.
Fig. 1 - Different examples of initial phases to visualize the impact of adjusting the initial phase of a sinusoidal function.
From the image above we see that at the initial value \(x=0\), \(f(0)=\sin\left(0\right)=0\). For the same sine function with an initial phase \(\phi_0=\frac{-\pi}4\), \(f(0)=\sin\left(0-\frac\pi4\right)=-\frac{\sqrt2}2\) and \(f(\frac\pi4)=\sin\left(\frac\pi4-\frac\pi4\right)=0\). We notice that the sine function has shifted horizontally to the right by an amount of \(\frac\pi4\). If we change the initial phase to \(\phi_0=-\pi\), we notice that the sine function shifts to the right by an amount of \(\pi\). We notice a pattern here, a negative initial phase will shift the function horizontally to the right, while a positive initial phase will shift the function horizontally to the left. This is visually represented in the figure below.
Fig. 2 - Sine function: case when the initial phase is zero.
Fig. 3 - Effect of a positive initial phase on a sinusoidal function.
Fig. 4 - Effect of a negative initial phase on a sinusoidal function.
Phase Angle - Key takeaways
- The solution for the differential equation that describes oscillatory movement, particularly simple harmonic motion is, \(x=A\sin\left(\omega t+\phi_0\right)\).
- The phase angle is the angular component of a periodic wave, such that it is defined as the argument of the sine function, \(\omega t+\phi_0\).
- We use the initial phase to specify the oscillating object's initial position to make sure we have the correct equation with the oscillator's position, no matter where it may have been located at \(t=0\), \(\phi_0=\sin^{-1}\left(\frac{x_0}A\right)\).
- A negative initial phase will shift the function horizontally to the right, while a positive initial phase will shift the function horizontally to the left.
References
- Fig. 1 - Different examples of initial phases to visualize the impact of adjusting the initial phase of a sinusoidal function, StudySmarter Originals
- Fig. 2 - Sine function: case when the initial phase is zero, StudySmarter Originals
- Fig. 3 - Effect of a positive initial phase on a sinusoidal function, StudySmarter Originals
- Fig. 4 - Effect of a negative initial phase on a sinusoidal function, StudySmarter Originals