Harmonic oscillations

A function of the form

f(x)=A\cdot \sin{\large(}u(x-v){\large)}+b

is called harmonic oscillation. $A$ , $u$ , $v$ and $b$ are parameters that are typically specific to a given problem. The harmonic oscillation often occurs in physics and describes the position of oscillating objects (in the simplest case a pendulum) as a function of time. $x$ then takes the role of time, and $y=A\cdot \sin(u(x-v))+b$ describes the deflection from the starting position.

In the following, we will examine the meaning of the parameters in more detail. We will discover many parallels to the parameters we have used to shift and stretch a quadratic function or, more generally, power functions.

Let's start our investigation by drawing some harmonic oscillations.

Exercise 1

Using the unit circle, sketch the following graphs and compare with the simple sine function $h(x)=\sin(x)$ :

$f(x)=2\sin(x)$ und $g(x)=0.5\sin(x)$
$f(x)=\sin(2x)$ und $g(x)=\sin(0.5x)$
$f(x)=\sin(x+\frac{\pi}{4})$ und $g(x)=\sin(x-\frac{\pi}{4})$
$f(x)=\sin(x)+0.5$ und $g(x)=\sin(x)-0.5$

Solution

The following can be seen from the sketches. Let $h(x)=\sin(x)$ and $f(x)=A\sin(u(x-v))+b$ . The parameters $A, u, v$ and $b$ describe geometric transformations to get from the graph $h$ to the graph $f$ , where applies:

$A$ stretches the graph of $h$ by the factor $A$ in the $y$ -direction.
1. $\sin(x) \rightarrow 2\sin(x)$ : Graph is stretched by a factor of $2$ .
2. $\sin(x) \rightarrow \frac{1}{2}\sin(x)$ : Graph is stretched by the factor $\frac{1}{2}$ (compression).
$u$ stretches the graph of $h$ by the factor $\frac{1}{u}$ in the $x$ -direction.
1. $\sin(x) \rightarrow \sin(2x)$ : Graph is stretched by the factor $\frac{1}{2}$ (compression).
2. $\sin(x) \rightarrow \sin(\frac{1}{2}x)$ : Graph is stretched by a factor of $2$ .
$v$ shifts the graph of $h$ by $v$ units in the $x$ direction.
1. $\sin(x) \rightarrow \sin(x-\frac{\pi}{4})$ : Shift by $\frac{\pi}{4}$ units to the right.
2. $\sin(x) \rightarrow \sin(x+\frac{\pi}{4})$ : Shift $\frac{\pi}{4}$ units to the left.
$b$ shifts the graph of $h$ by $b$ units in the $y$ direction.
1. $\sin(x) \rightarrow \sin(x)+\frac{1}{2}$ : Shift by $\frac{1}{2}$ units upwards.
2. $\sin(x) \rightarrow \sin(x)-\frac{1}{2}$ : Shift by $\frac{1}{2}$ units to down. Das gleiche gilt auch für die Graphen der Funktionen $\cos(x)$ und $\tan(x)$ . Und in der Tat gilt es für den Graphen jeder Funktion.

Indeed, we have the following general theorem:

Theorem 1

The interpretation of the parameters $A, u, v$ and $b$ in the function

f(x)=A\sin(u(x-v))+b

are as follows:

$A$ : stretch $\sin(x)$ by factor $A$ in the $y$ direction
$u$ : stretch the new graph by the factor $\frac{1}{u}$ in the $x$ -direction
$v$ : Shift the resulting graph by $v$ to the right ( $v>0$ ) or to the left ( $v<0$ ).
$b$ : Move the resulting graph up ( $b>0$ ) or down ( $b<0$ ) by $b$ .

The above sequence of transformations must be observed in order to get from the graph of $h(x)=\sin(x)$ to the graph of $f(x)=A\sin(u(x-v))+b$ .

Example 1

We illustrate this using the graph of the function

$f(x)=\sin(0.5(x-\frac{\pi}{2}))$

This is the blue solid line in the sketch below. It is $A=1, u=0.5, v=\frac{\pi}{2}$ and $b=0$ . We can see that the graph $\sin(x)$ is not stretched in $y$ -direction. It is stretched by the factor $\frac{1}{u}=2$ in the $x$ direction (blue dashed line), then shifted to the right by $\frac{\pi}{2}$ . There is no shift in $y$ -direction.

We can also write the function like this:

$f(x)=\sin(0.5x-\frac{\pi}{4})$

In this form, however, we cannot read off the shift in the $x$ direction directly.

Exercise 2

Find the transformations that lead from the sine curve $\sin(x)$ to the dashed line and write the functional equation of the dashed line. The dashed line intersects the $x$ -axis at $\frac{pi}{4}, \frac{3pi}{4}, ...$ .

Solution

Stretch the sine curve $\sin(x)$ by the factor $\frac{1}{2}$ in the $x$ direction ( $\rightarrow u=\frac{1}{1/2}=2$ ) and shift to the right by $\frac{\pi}{4}$ .

The following therefore applies $f(x)=\sin(2(x-\frac{\pi}{4}))$ .

Exercise 3

Which transformations are used to transform the graph of the function $h(x)=\sin(x)$ into the graph of the function $f(x)=3\sin(0.5x-\frac{\pi}{4})+1$ ?

Then draw the graph of $f$ by applying the transformations to the graph of $h$ .

Solution

Bring to the correct format: $f(x)=3\sin(0.5x-\frac{\pi}{4})+1=3\sin(0.5(x-\frac{\pi}{2}))+1$ , i.e. $A=3, u=0.5, v=\frac{\pi}{2}, b=1$ , i.e. starting from $\sin(x)$ :

stretch by a factor of $A=3$ in the $y$ direction, then
stretch by a factor of $\frac{1}{u}=2$ in the $x$ -direction, then
shift by $\frac{\pi}{2}$ units to the right, then
shift upwards by one unit.

Exercise 4

Find the $x$ -intercepts of the function $f(x)=\sin(0.2x-2)$ in two different ways:

by calculation.
by considering how the $x$ -intercepts of the function $\sin(x)$ are transformed.

Solution

$\sin(x)$ has the $x$ -intercepts at $..., -\pi, 0, \pi, 2\pi, ...$ . So find $x$ with $0.2x-2=-\pi$ , $0.2x-2=0, 0.2x-2=\pi, 0.2x-2=2\pi, ...$ , and we see that $x=10-5\pi, 10, 10+5\pi, 10+10\pi, ...$ .
because $f(x)=\sin(0.2x-2)=\sin(0.2(x-10))$ we see that $\sin(x)$ is stretched by the factor $\frac{1}{u}=\frac{1}{0.2}=5$ in the $x$ -direction, and then shifted to the right by $v=10$ . The points $..., -\pi, 0, \pi, 2\pi, ...$ are therefore first stretched by $5$ , i.e. $-5\pi, 0, 5\pi, 10\pi, ...$ , and then shifted to the right by $10$ , i.e. $-5\pi+10, 10, 10+5\pi, 10+10\pi, ...$ .

Exercise 5

If not explicitly mentioned, all problems are to be solved without calculator.

Describe the geometric transformations to get from the graph of the sine or cosine to the graph of the function $f$ . Then sketch the graph of $f$ in which the transformations are carried out.
1. $h(x)=\sin(x) \rightarrow f(x)=4\sin(x)+3$
2. $h(x)=\cos(x) \rightarrow f(x)=1.5\cos(0.25 x)$
3. $h(x)=\sin(x) \rightarrow f(x)=0.25\sin(4x + 1)-1$
The graph of the function $h(x)=\cos(x)$ is stretched by a factor of $0.75$ in both the $x$ and $y$ directions and then shifted to the left by 2 units. Determine the functional equation $f$ of the resulting graph.
The graph of the function $h(x)=\sin(x)$ is first stretched by a factor of $2$ in the $x$ -direction and then shifted to the right by $\frac{\pi}{4}$ units. Sketch the graph by applying the transformations and determine the functional equation of the resulting graph.
Repeat the transformation from above, but in reverse order. Sketch the graph by applying the transformations and compare with above. Do the two graphs match? Determine the functional equation.
How can the transformation for $A=-1$ also be called? And for $u=-1$ ?
Use sketches to determine which equations are correct. If an equation is not correct, correct the right-hand side so that it is correct.
1. $\sin(-x)=-\sin(x)$
2. $\cos(-x)=-\cos(x)$
3. $-\sin(x-\frac{\pi}{2})=-\cos(x)$
4. $\cos(x+\pi)=-\cos(x)$
Determine at least two $x$ -intercepts of the function $f(x)=10\sin(3x-2)$ .
Determine at least two $x$ -intercepts of the function $f(x)=10\sin(3x-2)+5$ .
The graph of the function $h(x)=\tan(x)$ is to be mirrored about the horizontal line at height $y=2. Determine the functional equation of the mirrored graph.
Determine the functional equations of the following graphs:
What transformations are needed to convert the graph of the function $\sin(x)$ to the graph of the function $f(x)=3\sin(1.5x-\frac{\pi}{6})+1$ ? Sketch the graph of $f$ .
The following transformations are successively applied to the graph of the function $\cos(x)$ :
- stretches in the $x$ -direction by a factor of $3$
- stretching in the $y$ -direction by a factor of $1.5$
- shift to the left by $6$
- shift downwards by $2$
Determine the function equation of the resulting graph $f$ .

Solution

we have
1. stretching in $y$ -direction with factor $4$ , then shifting upwards by 3.
2. stretching in $y$ -direction with factor $1.5$ , then stretching in $x$ -direction with factor $1/u=1/0.25=4$ .
3. stretching in $y$ -direction with factor $0.25$ , then stretching in $x$ -direction with factor $1/u=1/4$ , then shifting by $v=1/4$ to the left and shifting by $1$ downwards.
$A=0.75$ , $1/u=0.75 \rightarrow u=4/3$ , $v=-2$ . Also $f(x)=0.75\cos(\frac{4}{3}(x+2))$ .
$1/u=2\rightarrow u=1/2$ , $v=\frac{\pi}{4} \rightarrow f(x)=\sin(\frac{1}{2}(x-\frac{\pi}{4}))$ .
If the graph of $\sin(x)$ is first shifted to the right by $\pi/4$ and then stretched by a factor of $2$ in the $x$ -direction (so not in the usual order), we obtain the following graph (shown below). It is not the same graph as the one shown in exercise 3. Starting from $\sin(x)$ and executing the transformations in the right order, we obtain the graph below by stretching it by a factor of $2$ in the $x$ -direction and then shifting it to the right by $\pi/2$ . So $u=1/2$ and $v=\frac{\pi}{2}$ and therefore $f(x)=\sin(\frac{1}{2}(x-\frac{\pi}{2}))$ .
It is
1. $A=-1$ , thus $f(x)=-\sin(x)$ : the graph $\sin(x)$ is reflected about the $x$ -axis.
2. $u=-1$ , thus $f(x)=\sin(-x)$ : the graph $\sin(x)$ is reflected about the $y$ -axis.
Sketch the left and right functions, and compare the two graphs to see if the are the same:
1. $\sin(-x)=-\sin(x)$ is correct
2. $\cos(-x)=-\cos(x)$ is false, it is $\cos(-x)=\cos(x)$
3. $-\sin(x-\frac{\pi}{2})=-\cos(x)$ is false, it is $-\sin(x-\frac{\pi}{2})=\cos(x)$
4. $\cos(x+\pi)=-\cos(x)$ is false, it is $\cos(x+\pi)=-\sin(x)$
Find $x$ with $f(x)=10\sin(3x-2)=0$ . The sine function has the $x$ -intercepts $...,-\pi, 0, \pi, 2\pi, ...$ , thus find $x$ with
$\begin{array}{lllll} 3x-2 & = & -\pi & \rightarrow & x =\frac{-\pi+2}{3}\\[0.5em] 3x-2 & = & 0 & \rightarrow & x =\frac{2}{3}\\[0.5em] 3x-2 & = & \pi & \rightarrow & x =\frac{\pi+2}{3}\\[0.5em] 3x-2 & = & 2\pi & \rightarrow & x =\frac{2\pi+2}{3}\\[0.5em] ... & & & & \end{array}$
Find $x$ with $f(x)=10\sin(3x-2)+5=0$ , thus
$\sin(\underbrace{3x-2}_{s})=-0.5$
We therefore have to find out for which radians $s$ the sine equals $-0.5$ (see unit circle below). We can calculate a value using the arcsine
$s_1=\arcsin(-0.5)=-0.524$
Another value is
$s_2=\pi+0.524=3.665$
(see unit circle below). Thus we have $3x-2=s_1=-0.524$ and therefore $x_1=\underline{0.492}$ , and $3x-2=s_2=3.665$ , and it follows $x_2=\underline{1.888}$ . Let's check: $f(-0.524)=10\sin(3\cdot (-0.524)-2)+5=0.003\approx 0$ (rounding error), and $f(1.888)=10\sin(3\cdot 1.888-2)+5=0.010\approx 0$ (rounding error).
Idea: First draw the mirrored graph and then think about which transformations are needed to get from $\tan(x)$ to the mirrored graph. It then follows that we can first mirror $\tan(x)$ on the $y$ -axis, $A=-1$ , and then shift it upwards by $4$ ( $b=4$ ). We therefore have $f(x)=\underline{-\tan(x)+4}$ .
We start with $\sin(x)$ in each case, and think about how we get to the graph shown by stretching and then shifting.
1. stretch in the $y$ -direction by a factor of $1.5$ ( $A=1.5$ ), and in the $x$ -direction by a factor of $0.25$ ( $u=1/0.25=4$ ). So $f(x)=\underline{1.5\sin(4x)}$ .
2. stretch in the $x$ -direction by a factor of $0.5$ ( $\frac{1}{u}=0.5\rightarrow u=2$ ), mirror on the $x$ -axis ( $A=-1$ ), shift upwards by $1$ ( $b=1$ ). We therefore have $f(x)=\underline{-\sin(2x)+1}$ (see image below).
We write $f$ in a better form:
$f(x)=3\sin(1.5x-\frac{\pi}{6})+1==3\sin(1.5(x-\frac{\pi}{9})+1$
So $A=3, u=1.5, v= \frac{\pi}{9}, b=1$ . It follows:
1. stretch $\sin(x)$ in $y$ -direction by $3$ , then
2. stretch in the $x$ -direction by $\frac{1}{u}=\frac{1}{1.5}=\frac{2}{3}$ , then
3. shift to the right by $v\frac{\pi}{9}$ , then
4. shift upwards by $1$
It is $A=1.5, \frac{1}{u}=3 \rightarrow u=\frac{1}{3}, v=-6 \rightarrow v=-6$ , and $b=-2$ . So

f(x)=1.5\cos(\frac{1}{3}(x+6))-2