Consider the series of graphs in Figure 2 and the way each change to the equation changes the image.

Figure 2. (a) The basic graph of $\text{ }y=\sin x\text{ }$ (b) Changing the amplitude from 1 to 2 generates the graph of $\text{ }y=2\sin x.\text{ }$ (c) The period of the sine function changes with the value of $\text{ }B$, such that $\text{ period}=\frac{2\pi }{B}$. Here we have $\text{ }B=4$, which translates to a period of $\text{ }\frac{\pi }{2}$. The graph completes one full cycle in $\text{ }\frac{\pi }{2}\text{ }$ units. (d) The graph displays a horizontal shift equal to $\text{ }\frac{C}{B}$, or $\text{ }\frac{\frac{\pi }{2}}{4}=\frac{\pi }{8}$. (e) Finally, the graph is shifted vertically by the value of $\text{ }D$. In this case, the graph is shifted up by 2 units.

We will solve these problems according to the models.

1. $y=2\sin \left(\frac{1}{4}x\right)\text{ }$ involves sine, so we use the form
   $y=A\sin \left(Bt+C\right)+D$

   We know that $\text{ }|A|\text{ }$ is the amplitude, so the amplitude is 2. Period is $\text{ }\frac{2\pi }{B}$, so the period is

   $\begin{align}\frac{2\pi }{B}&=\frac{2\pi }{\frac{1}{4}} \\ &=8\pi \end{align}$

   See the graph in Figure 3.

   

   Figure 3

2. $y=-3\sin \left(2x+\frac{\pi }{2}\right)\text{ }$ involves sine, so we use the form
   $y=A\sin \left(Bt-C\right)+D$

   Amplitude is $\text{ }|A|$, so the amplitude is $\text{ }|-3|=3$. Since $\text{ }A\text{ }$ is negative, the graph is reflected over the x-axis. Period is $\text{ }\frac{2\pi }{B}$, so the period is

   $\frac{2\pi }{B}=\frac{2\pi }{2}=\pi$

   The graph is shifted to the left by $\text{ }\frac{C}{B}=\frac{\frac{\pi }{2}}{2}=\frac{\pi }{4}\text{ }$ units.

   

   Figure 4

3. $y=\cos x+3\text{ }$ involves cosine, so we use the form
   $y=A\cos \left(Bt\pm C\right)+D$

   Amplitude is $\text{ }|A|,\text{ }$ so the amplitude is 1. The period is $\text{ }2\pi .\text{ }$ This is the standard cosine function shifted up three units.

   

   Figure 5

The amplitude is $|-4|=4.$ The period is $\text{ }\frac{2\pi }{\omega }=\frac{2\pi }{\pi }=2.\text{ }$ (Recall that we sometimes refer to $\text{ }B\text{ }$ as $\omega$.) One cycle of the graph can be drawn over the interval $\text{ }\left[0,2\right].\text{ }$ To find the key points, we divide the period by 4. Make a table similar to the one below, starting with $\text{ }x=0\text{ }$ and then adding $\text{ }\frac{1}{2}\text{ }$ successively to $\text{ }x\text{ }$ and calculate $\text{ }y.\text{ }$ See the graph in Figure 7.

Figure 7

Recall that amplitude is found using the formula

$A=\frac{\text{largest value }-\text{smallest value}}{2}$

Thus, the amplitude is

$\begin{align} |A|&=\frac{69 - 42.5}{2} \\ &=13.25 \end{align}$

The data covers a period of 12 months, so $\frac{2\pi }{B}=12$ which gives $B=\frac{2\pi }{12}=\frac{\pi }{6}$.

The vertical shift is found using the following equation.

$D=\frac{\text{highest value}+\text{lowest value}}{2}$

Thus, the vertical shift is

$\begin{align}D&=\frac{69+42.5}{2} \\ &=55.8\end{align}$

So far, we have the equation $y=13.3\sin \left(\frac{\pi }{6}x-C\right)+55.8$.

To find the horizontal shift, we input the $x$ and $y$ values for the first month and solve for $C$.

$\begin{gathered}42.5=13.3\sin \left(\frac{\pi }{6}\left(1\right)-C\right)+55.8 \\ -13.3=13.3\sin \left(\frac{\pi }{6}-C\right) \\ -1=\sin \left(\frac{\pi }{6}-C\right) \\ \sin \theta =-1\to \theta =-\frac{\pi }{2}\end{gathered}$

$\begin{align} \frac{\pi }{6}-C&=-\frac{\pi }{2}\\ \frac{\pi }{6}+\frac{\pi }{2}&=C \\ &=\frac{2\pi }{3} \end{align}$

We have the equation $y=13.3\sin \left(\frac{\pi }{6}x-\frac{2\pi }{3}\right)+55.8$. See the graph in Figure 8.

Figure 8

Begin by making a table of values as shown in the table below.

To model an equation, we first need to find the amplitude.

$\begin{align}|A|&=\left\rvert\frac{78 - 30}{2}\right\rvert \\ &=24 \end{align}$

The clock’s cycle repeats every 12 hours. Thus,

$\begin{align}B&=\frac{2\pi }{12} \\ &=\frac{\pi }{6} \end{align}$

The vertical shift is

$\begin{align}D&=\frac{78+30}{2} \\ &=54 \end{align}$

There is no horizontal shift, so $C=0$. Since the function begins with the minimum value of $y$ when $x=0$ (as opposed to the maximum value), we will use the cosine function with the negative value for $A$. In the form $y=A\cos \left(Bx\pm C\right)+D$, the equation is

$y=-24\cos \left(\frac{\pi }{6}x\right)+54$

Figure 9

As the water level varies from 7 ft to 15 ft, we can calculate the amplitude as

$\begin{align}|A|&=\left\rvert\frac{\left(15 - 7\right)}{2}\right\rvert \\ &=4 \end{align}$

The cycle repeats every 12 hours; therefore, $B$ is

$\frac{2\pi }{12}=\frac{\pi }{6}$

There is a vertical translation of $\frac{\left(15+8\right)}{2}=11.5$. Since the value of the function is at a maximum at $t=0$, we will use the cosine function, with the positive value for $A$.

$y=4\cos \left(\frac{\pi }{6}\right)t+11$

Figure 10

The period is given by

$\begin{align}\frac{2\pi }{\omega }&=\frac{2\pi }{160\pi } \\ &=\frac{1}{80} \end{align}$

In a blood pressure function, frequency represents the number of heart beats per minute. Frequency is the reciprocal of period and is given by

$\begin{align}\frac{\omega }{2\pi }&=\frac{160\pi }{2\pi }\\ &=80\end{align}$

See the graph in Figure 11.

The blood pressure reading on the graph is $\frac{120}{80}\text{ }\left(\frac{\text{maximum}}{\text{minimum}}\right)$.

Figure 11

Analysis of the Solution

Blood pressure of $\frac{120}{80}$ is considered to be normal. The top number is the maximum or systolic reading, which measures the pressure in the arteries when the heart contracts. The bottom number is the minimum or diastolic reading, which measures the pressure in the arteries as the heart relaxes between beats, refilling with blood. Thus, normal blood pressure can be modeled by a periodic function with a maximum of 120 and a minimum of 80.

1. $y=5\sin \left(3t\right)$
   1. The maximum displacement is equal to the amplitude, $|a|$, which is 5.
   2. The period is $\frac{2\pi }{\omega }=\frac{2\pi }{3}$.
   3. The frequency is given as $\frac{\omega }{2\pi }=\frac{3}{2\pi }$.
   4. See Figure 12. The graph indicates the five key points.
      

      Figure 12

2. $y=6\cos \left(\pi t\right)$
   1. The maximum displacement is $6$.
   2. The period is $\frac{2\pi }{\omega }=\frac{2\pi }{\pi }=2$.
   3. The frequency is $\frac{\omega }{2\pi }=\frac{\pi }{2\pi }=\frac{1}{2}$.
   4. See Figure 13.
      

      Figure 13.

3. $y=5\cos \left(\frac{\pi }{2}\right)t$
   1. The maximum displacement is $5$.
   2. The period is $\frac{2\pi }{\omega }=\frac{2\pi }{\frac{\pi }{2}}=4$.
   3. The frequency is $\frac{1}{4}$.
   4. See Figure 14.
      

      Figure 14

At time $t=0$, the displacement is the maximum of 10 cm, which calls for the cosine function. The cosine function will apply to both models.

We are given the frequency $f=\frac{\omega }{2\pi }$ of 0.5 cycles per second. Thus,

$\begin{align}\frac{\omega }{2\pi }&=0.5 \\ \omega &=\left(0.5\right)2\pi \\ &=\pi \end{align}$

The first spring system has a damping factor of $c=0.5$. Following the general model for damped harmonic motion, we have

$f\left(t\right)=10{e}^{-0.5t}\cos \left(\pi t\right)$

Figure 15 models the motion of the first spring system.

Figure 15

The second spring system has a damping factor of $c=0.1$ and can be modeled as

$f\left(t\right)=10{e}^{-0.1t}\cos \left(\pi t\right)$

Figure 16 models the motion of the second spring system.

Figure 16

Analysis of the Solution

Notice the differing effects of the damping constant. The local maximum and minimum values of the function with the damping factor $c=0.5$ decreases much more rapidly than that of the function with $c=0.1$.

Substitute the given values into the model. Recall that period is $\frac{2\pi }{\omega }$ and frequency is $\frac{\omega }{2\pi }$.

1. $y=20{e}^{-0.05t}\cos \left(\frac{\pi }{2}t\right)$. See Figure 17.
   

   Figure 17

2. $y=2{e}^{-1.5t}\cos \left(6\pi t\right)$. See Figure 18.
   

   Figure 18

Calculate the value of $\omega$ and substitute the known values into the model.

1. As period is $\frac{2\pi }{\omega }$, we have
   $\begin{gathered}\frac{\pi }{6}=\frac{2\pi }{\omega } \\ \omega \pi =6\left(2\pi \right) \\ \omega =12 \end{gathered}$

   The damping factor is given as 10 and the amplitude is 7. Thus, the model is $y=7{e}^{-10t}\sin \left(12t\right)$. See Figure 19.

   

   Figure 19

2. As frequency is $\frac{\omega }{2\pi }$, we have
   $\begin{gathered}20=\frac{\omega }{2\pi } \\ 40\pi =\omega \end{gathered}$

   The damping factor is given as $0.2$ and the amplitude is $0.3$. The model is $y=0.3{e}^{-0.2t}\sin \left(40\pi t\right)$. See Figure 20.

   

   Figure 20

Analysis of the Solution

A comparison of the last two examples illustrates how we choose between the sine or cosine functions to model sinusoidal criteria. We see that the cosine function is at the maximum displacement when $t=0$, and the sine function is at the equilibrium point when $t=0$. For example, consider the equation $y=20{e}^{-0.05t}\cos \left(\frac{\pi }{2}t\right)$ from Example 9. We can see from the graph that when $t=0,y=20$, which is the initial amplitude. Check this by setting $t=0$ in the cosine equation:

$\begin{align}y&=20{e}^{-0.05\left(0\right)}\cos \left(\frac{\pi }{2}\right)\left(0\right) \\ &=20\left(1\right)\left(1\right) \\ &=20\end{align}$

Using the sine function yields

$\begin{align}y&=20{e}^{-0.05\left(0\right)}\sin \left(\frac{\pi }{2}\right)\left(0\right) \\ &=20\left(1\right)\left(0\right) \\ &=0 \end{align}$

Thus, cosine is the correct function.

The amplitude begins at 5 in. and deceases 30% each second. Because the spring is initially compressed, we will write A as a negative value. We can write the amplitude portion of the function as

$A\left(t\right)=5{\left(1 - 0.30\right)}^{t}=$

$A\left(t\right)=5{\left(0.70\right)}^{t}=$

We put ${\left(0.70\right)}^{t}$ in the form ${e}^{ct}$ as follows:

$\begin{gathered}0.7={e}^{c} \\ c=\mathrm{ln}(0.7) \\ c=-0.357 \end{gathered}$

Now let’s address the period. The spring cycles through its positions every 3 seconds, this is the period, and we can use the formula to find omega.

$\begin{gathered} 3=\frac{2\pi }{\omega }\\ \omega =\frac{2\pi }{3}\end{gathered}$

The natural length of 10 inches is the midline. We will use the cosine function, since the spring starts out at its maximum displacement. This portion of the equation is represented as

$y=\cos \left(\frac{2\pi }{3}t\right)+10$

Finally, we put both functions together. Our the model for the position of the spring at $t$ seconds is given as

$y=-5{e}^{-0.357t}\cos \left(\frac{2\pi }{3}t\right)+10$

See the graph in Figure 21.

Figure 21

The displacement factor represents the amplitude and is determined by the coefficient $a{e}^{-ct}$ in the model for damped harmonic motion. The damping constant is included in the term ${e}^{-ct}$. It is known that after 3 seconds, the local maximum measures one-half of its original value. Therefore, we have the equation

$a{e}^{-c\left(t+3\right)}=\frac{1}{2}a{e}^{-ct}$

Use algebra and the laws of exponents to solve for $c$.

$\begin{align} a{e}^{-c\left(t+3\right)}&=\frac{1}{2}a{e}^{-ct} \\ {e}^{-ct}\cdot {e}^{-3c}&=\frac{1}{2}{e}^{-ct} && \text{Divide out }a.\\ {e}^{-3c}&=\frac{1}{2}&& \text{Divide out }{e}^{-ct}. \\ {e}^{3c}&=2 && \text{Take reciprocals}. \end{align}$

Then use the laws of logarithms.

$\begin{align}{e}^{3c}&=2 \\ 3c&=\mathrm{ln}2 \\ c&=\frac{\mathrm{ln}2}{3} \end{align}$

The damping constant is $\frac{\mathrm{ln}2}{3}$.

The graph produced by this function will be shown in two parts. The first graph will be the exact function $f\left(x\right)$, and the second graph is the exact function $f\left(x\right)$ plus a bounding function. The graphs look quite different.

Figure 22

Figure 23

Analysis of the Solution

The curves $y=\cos \left(2\pi x\right)$ and $y=-\cos \left(2\pi x\right)$ are bounding curves: they bound the function from above and below, tracing out the high and low points. The harmonic motion graph sits inside the bounding curves. This is an example of a function whose amplitude not only decreases with time, but actually increases and decreases multiple times within a period.