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Wize University Physics Textbook (Master) > Periodic Motion: Oscillations

Finding the Phase Constant

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Finding the Phase Constant


It is sometimes tricky to find the phase constant of an oscillation. There are many different ways to find the phase constant. Here we will see how we can use the information given to us in a problem to find the phase constant.

As a reference, let's consider the following formulas for displacement, velocity and acceleration of an oscillator:

x(t)=Acos⁡(ωt+φ)v(t)=−Aωsin⁡(ωt+φ)a(t)=−Aω2cos⁡(ωt+φ)\begin{array}{c}x\left(t\right)=A\cos\left(\omega t+\varphi\right)\\\\v\left(t\right)=-A\omega\sin\left(\omega t+\varphi\right) \\\\a\left(t\right)=-A\omega^2\cos\left(\omega t+\varphi\right) \end{array}

Wize Concept
Here is the general recipe to find the phase constant from position, velocity or acceleration graphs:
  1. Use the information at time equal to zero to find two possible solutions for the phase constant by using the above equations.
  2. Use the slope of tangent line to the curve at the same moment to pick the right angle.

To understand above recipe let's look at three different situations. But before that we need a quick review of trigonometric circle (unit circle):

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Quick Review of Trigonometry

In the unit circle, sin⁡θ=y\sin\theta=y and cos⁡θ=x\cos\theta=x


• Quadrant 1: All trig ratios are positive• Quadrant 2: Only sin ratio is positive• Quadrant 3: Only tan ratio is positive• Quadrant 4: Only cos ratio is positive\begin{array}{l} •\ \text{Quadrant 1: All trig ratios are positive}\\\\ •\ \text{Quadrant 2: Only sin ratio is positive}\\\\ •\ \text{Quadrant 3: Only tan ratio is positive}\\\\ •\ \text{Quadrant 4: Only cos ratio is positive} \end{array}




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Finding the Phase Constant from x(t)\colorOne{x(t)}graph

Example:








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Finding the Phase Constant from v(t)\colorOne{v(t)}graph

Example:





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Finding the Phase Constant from a(t)\colorOne{a(t)}graph

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Example: Drawing Position Graph From Velocity Graph


The velocity (cm/scm/s) vs. time (ss) of a mass-spring on a SHM is shown in the graph.

Find the equation representing position against time, and draw that graph.


Solution:

v(t)=−Aωsin⁡(ωt+φ)v\left(t\right)=-A\omega\sin\left(\omega t+\varphi\right)
vmax⁡=Aω =0.4 cmv_{\max}=A\omega\ =0.4\ cm from the plot!


T=πT=\pi (for example from t=−16πt=-\frac{1}{6}\pi to t=56πt=\frac{5}{6}\pi)
⇒ω=2πT=2 rad/s⇒A=0.2 cm\Rightarrow\omega=\frac{2\pi}{T}=2\ rad/s\Rightarrow A=0.2\ cm


The last thing we need to find is the phase constant:
v(0)=−Aωsin⁡(ω(0)+φ)=−0.35v\left(0\right)=-A\omega\sin\left(\omega\left(0\right)+\varphi\right)=-0.35
⇒sin⁡(φ)=0.350.4=78\Rightarrow\sin\left(\varphi\right)=\frac{0.35}{0.4}=\frac{7}{8}
This gives us two possible answers (61 degrees or 119 degrees). Which is it?

To figure that out, we will look at the acceleration function:

a(t)=−Aω2cos⁡(ωt+φ)a\left(t\right)=-A\omega^2\cos\left(\omega t+\varphi\right)
⇒a(0)=−Aω2cos⁡(ω(0)+φ)\Rightarrow a\left(0\right)=-A\omega^2\cos\left(\omega\left(0\right)+\varphi\right)
→        −cos⁡(φ)<0      →     cos⁡(φ)>0\rightarrow\ \ \ \ \ \ \ \ -\cos\left(\varphi\right)<0\ \ \ \ \ \ \rightarrow\ \ \ \ \ \cos\left(\varphi\right)>0
We know the acceleration at t=0 is negative from the graph (slope of the tangent tine to v(t)v\left(t\right) at t=0t=0 is negative!)
⇒cos⁡(φ)>0\Rightarrow\cos\left(\varphi\right)>0
Only the 61 degree answer satisfies this condition.



So, we can now plot x(t)=Acos⁡(ωt+φ)=0.2cos⁡(2t+1.06)cmx\left(t\right)=A\cos\left(\omega t+\varphi\right)=\boxed{0.2\cos\left(2t+1.06\right)cm}

Note: We need to convert the 61 degrees to 1.06 radians because the "2t" is also in radians. (The 2 is in radians per second.)

The plot below is in seconds on the horizontal axis.




The displacement graph of a simple harmonic motion over time is shown below. Write the equation for displacement as a function of time.