>Graphical Representation>Transforming into a Harmonic Oscillator

Transforming into a Harmonic Oscillator

You might come across situations where you need to transform functions that contain both sines and cosines into harmonic oscillators. Then you will need to use this formula:

Formula

Transforming into a Harmonic Oscillator

\begin{array}{l} a \sin c x + b \cos c x & = A \sin (c x + ϕ) \\ = A \cos (c x + ϕ^{*}) \end{array}

where

\begin{array}{l} A & = \sqrt{a^{2} + b^{2}} \\ \tan ϕ & = \frac{b}{a} \\ ϕ^{*} & = ϕ - \frac{π}{2} \end{array}

and $ϕ$ is in the same quadrant as the point $(a, b)$ . You can decide which quadrant an angle $ϕ$ is in by comparing it to $\frac{π}{2}$ , $π$ , $\frac{3 π}{2}$ and $2 π$ .

Note! It’s often useful to draw the unit circle to be sure that the value you have found for

ϕ

is in the correct quadrant.

When you are solving equations in the form

a \sin c x + b \cos c x = d

it’s a good idea to rewrite the left-hand side to fit the form $A \sin (c x + ϕ)$ .

Example 1

Rewrite the expression $\sin 2 x - 4 \cos 2 x$ into a harmonic oscillator with sine as the base function: $f (x) = A \sin (c x + ϕ)$

To rewrite the equation, you have to find the amplitude $A$ and the phase $ϕ$ first. The amplitude $A$ is

A = \sqrt{a^{2} + b^{2}} = \sqrt{1^{2} + {(- 4)}^{2}} = \sqrt{17} .

When you’re finding the phase, you need to take the signs of $a$ ( $x$ -axis) and $b$ ( $y$ -axis) into account. As $a > 0$ and $b < 0$ , you can see that $(a, b)$ is in the fourth quadrant. That means you find the phase like this:

\begin{array}{l} \tan ϕ & = \frac{b}{a} = \frac{- 4}{1} \\ ϕ & = \tan^{- 1} (- 4) \approx - 1.33 + n π \end{array}

As $- \frac{π}{2} < - 1.33 < 0$ , $- 1.33$ is the value for $ϕ$ you’re looking for. That makes the equation look like this on the form of a harmonic oscillator:

\sqrt{17} \sin (2 x - 1.33)

Example 2

Solve the equation

3 \sin 2 x + 4 \cos 2 x = 1 x \in ℝ

First, you find $A$ :

A = \sqrt{a^{2} + b^{2}} = \sqrt{3^{2} + 4^{2}} = 5

Then you find $ϕ$ : $\begin{array}{l} \tan ϕ & = \frac{b}{a} = \frac{4}{3}, \\ ϕ & = \tan^{- 1} (\frac{4}{3}) \approx 0.93 + n π \end{array}$

You get that $ϕ \approx 0.93 + n \cdot π$ . As $a > 0$ and $b > 0$ , $ϕ$ has to be in the first quadrant, and because $0 < 0.93 < \frac{π}{2} \approx 1.57$ , you get $ϕ = 0.93$ . That gives you

\begin{array}{l} 5 \sin (2 x + 0.93) & = 1 \\ \sin (2 x + 0.93) & = 0.2 \end{array}

This equation has the solutions

\begin{array}{l} 2 x_{1} + 0.93 & = \sin^{- 1} 0.2 + n \cdot 2 π \\ \approx 0.2 + n \cdot 2 π \\ 2 x_{2} + 0.93 & = (π - \sin^{- 1} (0.2)) + n \cdot 2 π \\ \approx (π - 0.2) + n \cdot 2 π \end{array}

That gives you these two solutions:

\begin{array}{l} x_{1} & \approx - 0.365 + n \cdot π \\ x_{2} & \approx 1.01 + n \cdot π \end{array}

Example 3

Rewrite the expression $2 \sin x + 3 \cos x$ as a harmonic oscillator with the cosine as the base function: $f (x) = A \cos (c x + ϕ^{*})$

Note! When you want to make a cosine function, you need to subtract $\frac{π}{2}$ from the value you find for $ϕ$ with the $\tan$ function. You get $ϕ^{*} = ϕ - \frac{π}{2}$ . First, you find the amplitude $A$ :

A = \sqrt{a^{2} + b^{2}} = \sqrt{2^{2} + 3^{2}} = \sqrt{13}

When you’re finding the phase, you need to take the sign of $a$ ( $x$ -axis) and $b$ ( $y$ -axis) into account. As $a > 0$ and $b > 0$ , you are in the first quadrant, and you can find the phase like this:

\begin{array}{l} \tan ϕ & = \frac{b}{a} = \frac{3}{2}, \\ ϕ & = \tan^{- 1} (\frac{3}{2}) \approx 0.98 + n π \end{array}

Because $0 < 0.98 < \frac{π}{2}$ , $0.98$ is the value for $ϕ$ you need to subtract $\frac{π}{2}$ from. You get

ϕ^{*} = 0.98 - \frac{π}{2} \approx 0.59

That means the harmonic oscillator is

\sqrt{13} \cos (x - 0.59)

Example 4

Solve the equation

4 \sin 2 x - 2 \cos 2 x = 1 x \in ℝ

First, you find $A$ and $ϕ^{*}$ to let you use the formula for the harmonic oscillator:

\begin{array}{l} A & = \sqrt{a^{2} + b^{2}} = \sqrt{4^{2} + {(- 2)}^{2}} \\ = \sqrt{20} \\ \tan ϕ & = \frac{b}{a} = \frac{- 2}{4} = - \frac{1}{2} \\ ϕ & = \tan^{- 1} (- \frac{1}{2}) \approx - 0.46 + n π \end{array}

\begin{array}{l} A & = \sqrt{a^{2} + b^{2}} = \sqrt{4^{2} + {(- 2)}^{2}} = \sqrt{20} \\ \tan ϕ & = \frac{b}{a} = \frac{- 2}{4} = - \frac{1}{2} \\ ϕ & = \tan^{- 1} (- \frac{1}{2}) \approx - 0.46 + n π \end{array}

a > 0

and

b < 0

ϕ

is in the fourth quadrant, and because

- \frac{π}{2} \approx - 1.57 < - 0.46 < 0

you can use

ϕ^{*} = - 0.46 - \frac{π}{2} \approx - 2.03

You get

\begin{array}{l} \sqrt{20} \cos (2 x - 2.03) & = 1 | \div \sqrt{20} \\ \cos (2 x - 2.03) & = 0.22 \\ 2 x - 2.03 & = \cos^{- 1} (0.22) \end{array}

This base equation has the solutions

\begin{array}{l} 2 x_{1} - 2.03 & = \cos^{- 1} (0.22) + n \cdot 2 π \\ \approx 1.35 + n \cdot 2 π \\ 2 x_{2} - 2.03 & = - \cos^{- 1} (0.22) + n \cdot 2 π \\ \approx - 1.35 + n \cdot 2 π \end{array}

Solving these equations gives you

\begin{array}{l} x_{1} & \approx 1.69 + n π \\ x_{2} & \approx 0.34 + n π \end{array}

These are the solutions to the equation, as there are no restrictions on $x$ .

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