Behaviour in the real world that involves rotation or repetitive motion can be modelled using trigonometric equations like $y=\sin x$y=sinx and $y=\cos x$y=cosx.
We can develop more accurate models by transforming these equations. In this lesson we will focus on phase shifts of the form $y=\sin\left(x-c\right)$y=sin(x−c), and also look at other transformations.
A small disc sitting on a flat surface begins to roll slowly with a constant angular velocity of $1$1 radian per second when a fan is switched on.
As the disc rolls, the height $h$h in centimetres of a point on the disc is described by the equation $h=\cos t+1$h=cost+1, where $t$t is in seconds.
A graph of the equation is shown below. Notice that the period of the equation is $2\pi$2π seconds, or just over $6$6 seconds. This is the time it takes the point on the disc to complete one full revolution.
Graph of the function $h=\cos t+1$h=cost+1.
Two seconds after the first disc starts rolling along the surface, a second disc of the same size is placed in front of the fan. It begins to roll in a similar way, and the height of a point on its surface is described by the equation $h=\cos\left(t-2\right)+1$h=cos(t−2)+1.
The second disc has the same behaviour as the first disc, except it has been translated in time. The graph below shows that the phase shift of $2$2 seconds corresponds to a horizontal translation of the graph of $h=\cos t+1$h=cost+1 to the graph of $h=\cos\left(t-2\right)+1$h=cos(t−2)+1.
Graph of the functions $h=\cos t+1$h=cost+1 and $h=\cos\left(t-2\right)+1$h=cos(t−2)+1.
Suppose now that we want the discs to roll a bit faster. If we crank up the speed of the fan, we can increase the angular velocity of the disc to $4$4 radians per second. The motion of the point on the first disc is then described by the equation $h=\cos\left(4t\right)+1$h=cos(4t)+1, and the point on the second disc has the equation $h=\cos\left(4\left(t-2\right)\right)+1$h=cos(4(t−2))+1.
In the image below we see the consequence of increasing the rolling speed is to decrease the period of each equation. The new period is $\frac{2\pi\text{ radians}}{4\text{ radians per second}}=\frac{\pi}{2}$2π radians4 radians per second=π2 seconds, or about $1.57$1.57 seconds.
Graph of the functions $h=\cos\left(4t\right)+1$h=cos(4t)+1 and $h=\cos\left(4\left(t-2\right)\right)+1$h=cos(4(t−2))+1.
The height of a point on a disc from the scenario above is described by the equation $h=8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$h=8sin(π5(t−4))+8. The graph of the function is shown below.
Graph of the function $h=8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$h=8sin(π5(t−4))+8.
a. Find the initial height of the point.
Think: The initial point in time is the moment when $t=0$t=0 seconds.
Do: Substitute $t=0$t=0 into the equation $h=8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$h=8sin(π5(t−4))+8.
$h$h | $=$= | $8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$8sin(π5(t−4))+8 | |
$=$= | $8\sin\left(\frac{\pi}{5}\left(0-4\right)\right)+8$8sin(π5(0−4))+8 | (Substitute the value of $t$t) | |
$=$= | $8\sin\left(-\frac{4\pi}{5}\right)+8$8sin(−4π5)+8 | (Simplify the product) | |
$=$= | $3.30$3.30 cm (2 d.p.) | (Evaluate the expression) |
The initial height of the point is $3.30$3.30 cm above the surface.
b. Find the maximum height of the point.
Think: The sine function takes values between $-1$−1 and $1$1. The maximum height will correspond to the times at which $\sin\left(\frac{\pi}{5}\left(t-4\right)\right)=1$sin(π5(t−4))=1.
Do: By substituting $\sin\left(\frac{\pi}{5}\left(t-4\right)\right)=1$sin(π5(t−4))=1 into the equation $h=8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$h=8sin(π5(t−4))+8 we see that the maximum height is $h=8\times1+8=16$h=8×1+8=16 cm above the surface.
c. Find the time when the point is first at a height of $12$12 cm.
Think: We have a value of $h$h, and we want to find the corresponding value of $t$t.
Do: Substitute $h=12$h=12 into the equation and rearrange to solve for $t$t.
$h$h | $=$= | $8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$8sin(π5(t−4))+8 | |
$12$12 | $=$= | $8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)+8$8sin(π5(t−4))+8 | (Substitute the value of $h$h) |
$4$4 | $=$= | $8\sin\left(\frac{\pi}{5}\left(t-4\right)\right)$8sin(π5(t−4)) | (Subtract $8$8 from both sides) |
$\frac{1}{2}$12 | $=$= | $\sin\left(\frac{\pi}{5}\left(t-4\right)\right)$sin(π5(t−4)) | (Divide both sides by $8$8) |
$\frac{\pi}{6}$π6 | $=$= | $\frac{\pi}{5}\left(t-4\right)$π5(t−4) | (Take the inverse $\sin$sin of both sides) |
$\frac{5}{6}$56 | $=$= | $t-4$t−4 | (Multiply both sides by $\frac{5}{\pi}$5π) |
$t$t | $=$= | $4.83$4.83 seconds (2 d.p.) | (Add $4$4 to both sides) |
The point on the disk will first reach a height of $12$12 cm when it has rolled for $4.83$4.83 seconds.
Tobias is jumping on a trampoline. Victoria watches him bounce at a regular rate and wants to try to model his height over time. When Victoria starts her stopwatch, Tobias is at a minimum height of $30$30 cm below the trampoline frame. A moment later Victoria records Tobias reaching a maximum height of $50$50 cm above the trampoline frame. She uses the function $H\left(s\right)=a\sin\left(2\pi\left(s-c\right)\right)+d$H(s)=asin(2π(s−c))+d, where $H$H is the height in cm above the trampoline frame and $s$s is the time in seconds.
Find the value of $a$a, the amplitude of the function.
Find the value of $d$d.
Find the value of $H\left(0\right)$H(0).
Find the value of $c$c, if $0
At what time does Tobias first reach a height of $30$30 cm?