A special kind of parametric equations goes by the name of Lissajous figures. These curves describe harmonic motions in physics.
A Lissajous figure is a curve #\orange{C}# described by a parametric equation of the form
\[\begin{array}{rcl}\blue{x(t)}&=& \blue{A \sin ( at + b)}\\\green{y(t)}&=& \green{B \sin (ct +d) }\end{array}\]
where #A, B, a, b, c# and #d# are numbers.
The period of the curve is the length of the smallest interval of #t# such that the full curve is displayed.
More examples
The variables in the definition can be changed to obtain various figures. You can use the sliders to change the values and see what happens.
In the definition of a Lissajous curve, we demand that it is a parametric curve with parametric equations of a specific form i.e. involving sine and cosine functions of a specific type. Using trigonometric identities we can rewrite parametric equations to see if they are also Lissajous figures but do not necessarily look the same as in the definition.
Example
The parametric equations
\[\begin{array}{rcl}\green{x(t)}&=& \green{2\sin (t) \cos(t)} \\\blue{y(t)}&=& \blue{\cos (t - \frac{\pi}{2})} \end{array}\]
can be rewritten as
\[\begin{array}{rcl}\green{x(t)}&=& \green{\sin (2t)} \\\blue{y(t)}&=& \blue{\sin (t )} \end{array}\]
The most basic example of a Lissajous figure is the unit circle, which is described by \[\begin{array}{rcl}\green{x(t)}&=&\green{\cos(t)}\\\blue{y(t)}&=&\blue{\sin(t)},\end{array}\]where #t# is in the interval #\ivcc{0}{2\pi}#.
The period of a Lissajous figure is the least common multiple of the periods of #\blue{x(t)}# and #\green{y(t)}#. These are #\frac{2\pi}{|a|}# and #\frac{2\pi}{|c|}# respectively. This was treated in the theory page on the greatest common divisor and least common multiple.
The period of the Lissajous figure can, in most cases, be calculated as follows. Note that the least common multiple and greatest common divisor are technically only defined on integers, but the calculation below gives an idea of how to calculate the period:
\[\begin{array}{rcl}\text{period}&=&\mathrm{lcm}\left(\displaystyle\frac{2\pi}{|a|},\displaystyle\frac{2\pi}{|c|}\right)\\&=&2\pi\cdot \mathrm{lcm}\left(\displaystyle\frac{1}{|a|},\displaystyle\frac{1}{|c|}\right)\\&=&2\pi\cdot \mathrm{lcm}\left(\displaystyle\frac{|c|}{|a\cdot c|},\displaystyle\frac{|a|}{|a\cdot c|}\right)\\&=&2\pi\cdot \displaystyle\frac{\mathrm{lcm}(|c|,|a|)}{|a\cdot c|}\\&=&\displaystyle\frac{2\pi}{\gcd(|c|,|a|)}\end{array}\]
Two examples are given below.
Example 1
The period of the following Lissajous figure
\[\begin{array}{rcl}\blue{x(t)}&=& \blue{5 \sin ( 2t + 3)}\\\green{y(t)}&=& \green{4 \sin (6t ) }\end{array}\]
can be calculated with the formula given above:
\[\begin{array}{rcl}\mathrm{lcm}\left(\displaystyle\frac{2\pi}{|a|},\displaystyle\frac{2\pi}{|c|}\right)&=&\displaystyle\frac{2\pi}{\gcd(|c|,|a|)}\\&=& \displaystyle\frac{2\pi} {\gcd(6,2)}\\&=&\displaystyle\frac{2\pi}{2}\\&=&\pi\end{array}\]
Example 2
The period of the following Lissajous figure
\[\begin{array}{rcl}\blue{x(t)}&=& \blue{ \sin ( \frac{t}{2})}\\\green{y(t)}&=& \green{5 \sin (\frac{t}{3} +1) }\end{array}\]
can again be calculated with the formula given above:
\[\begin{array}{rcl}\mathrm{lcm}\left(\displaystyle\frac{2\pi}{|a|},\displaystyle\frac{2\pi}{|c|}\right)&=&\mathrm{lcm}(4\pi,6\pi)\\&=&12\pi \end{array}\]
Determine whether the parametric curve given by equations
\[\begin{array}{rcl}\blue{x(t)}&=& \blue{\sin(t) \cos(2) + \cos(t) \sin(2) }\\ \green{ y(t)}&=& \green{4 \sin( \frac{2t}{\pi} + \pi) } \end{array}\]
is a Lissajous figure.
Yes it is.
We use trigonometric identities to rewrite the equations. For #\blue{x(t)}# we use the identity #\sin( \alpha + \beta ) = \sin(\alpha) \cos (\beta) + \cos(\alpha) \sin(\beta)#. For #\blue{x(t)}# we get
\[\begin{array}{rcl}\blue{ x(t)} & = & \blue{\sin(t) \cos(2) + \cos(t) \sin(2) }\\ & = & \sin(t + 2). \end{array}\]
We see that #\green{y(t)}# is already of the desired form.
Question
Let #\orange P# be a point whose orbit is described by parametric equations
\[\begin{array}{rcl}\blue{x(t)}&=&\blue{ 2\sin ( \pi t )}\\ \green{y(t)}&=& \green{2 \cos (t)} \end{array}\]
where #t# lies between #- \pi # and # \pi#. Determine the values of #t# where #\orange P# passes through the #y#-axis.
Solution
We solve #\blue{x(t)} = 0# for #t \in \ivcc{- \pi}{\pi}#. We get
\[\begin{array}{rcl}\blue{2 \sin (\pi t )} &=&0 \\\pi t &=& k\cdot \pi \\t &= & k\end{array}\]
where #k# is an integer.
Therefore, for #t = -3, -2, -1, 0, 1, 2, 3# in the interval the point #\orange P# passes through the #y#-axis.
Here is a picture of the curve
\[\begin{array}{rcl}\blue{x(t)}&=&\blue{ 2\sin ( \pi t )}\\ \green{y(t)}&=& \green{2 \cos (t)}. \end{array}\]
The value of #t# can be adjusted to move the point # \orange P#. As is apparent from the picture the values #t = -3, -2, -1, 0, 1, 2, 3# are the only values where #\orange P# passes the #y#-axis.
Consider the point #P# given by equations
\[\begin{cases}
x(t) & = 2\cdot \sin(t+{{\pi}\over{4}}), \\
y(t) & = 2\cdot \sin({{t}\over{4}}+{{\pi}\over{16}}).
\end{cases}\] where #t# ranges from #- \pi# to #\pi#.
Determine how often the point #P# passes through the origin.
The point #P# passes through the origin once.
To determine how often the point #P# passes through the origin we simultaneously have to solve #x(t) = 0# and #y(t) = 0#. We have that #x(t) = 2\cdot \sin(t+{{\pi}\over{4}}) = 0# if and only if #t+{{\pi}\over{4}} # is a multiple of #\pi#. We get
\begin{array}{rcl}
t+{{\pi}\over{4}} & = & k \cdot \pi \\
t & = & \pi\cdot k-{{\pi}\over{4}}
\end{array}
for all integers #k#. The values of #t# which lie in the interval are those in the list \[\left[ t={{3\cdot \pi}\over{4}} , t=-{{\pi}\over{4}} \right] \]
The ones that also have #y(t) = 0# are
\[ \left[ t=-{{\pi}\over{4}} \right] \]
So #P# passes through the origin once.
Lissajous figures
