Applications of sine and cosine functions including phase shift (radians)

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.

 

Exploration

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.

 

Worked example

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.

 

Practice question