Applications of sine and cosine functions including phase shift (degrees)
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^\circ$1° 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 $360$360 seconds, or $6$6 minutes. 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.
Thirty 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 $h=\cos\left(t-30^\circ\right)+1$h=cos(t−30°)+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 $30$30 seconds corresponds to a horizontal translation of the graph of $h=\cos t+1$h=cost+1 to the graph of $h=\cos\left(t-30^\circ\right)+1$h=cos(t−30°)+1.
Graph of the functions $h=\cos t+1$h=cost+1 and $h=\cos\left(t-30^\circ\right)+1$h=cos(t−30°)+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 $15^\circ$15° per second. The motion of the point on the first disc is then described by the equation $h=\cos\left(15t\right)+1$h=cos(15t)+1, and the point on the second disc has the equation $h=\cos\left(15\left(t-30\right)\right)+1$h=cos(15(t−30))+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{360^{\circ}}{15^{\circ}\text{ per second}}=24$360∘15∘ per second=24 seconds.
Graph of the functions $h=\cos\left(15t\right)+1$h=cos(15t)+1 and $h=\cos\left(15\left(t-30\right)\right)+1$h=cos(15(t−30))+1.
Worked example
The height of a point on a disc from the scenario above is described by the equation $h=10\sin\left(20\left(t-7\right)\right)+10$h=10sin(20(t−7))+10. The graph of the function is shown below.
Graph of the function $h=10\sin\left(20\left(t-7\right)\right)+10$h=10sin(20(t−7))+10.
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 $h=10\sin\left(20\left(t-7\right)\right)+10$h=10sin(20(t−7))+10.
| $h$h | $=$= | $10\sin\left(20\left(t-7\right)\right)+10$10sin(20(t−7))+10 | |
| $=$= | $10\sin\left(20\left(0-7\right)\right)+10$10sin(20(0−7))+10 | (Substitute the value of $t$t) | |
| $=$= | $10\sin\left(20\times\left(-7\right)\right)+10$10sin(20×(−7))+10 | (Simplify the product) | |
| $=$= | $10\sin\left(-140^\circ\right)+10$10sin(−140°)+10 | (Evaluate the multiplication) | |
| $=$= | $3.57$3.57 cm (2 d.p.) | (Evaluate the expression) |
The initial height of the point is $3.57$3.57 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 $\sin\left(20\left(t-7\right)\right)=1$sin(20(t−7))=1.
Do: By substituting $\sin\left(20\left(t-7\right)\right)=1$sin(20(t−7))=1 into the equation $h=10\sin\left(20\left(t-7\right)\right)+10$h=10sin(20(t−7))+10 we see that the maximum height is $h=10\times1+10=20$h=10×1+10=20 cm above the surface.
c. Find the time when the point is first at a height of $15$15 cm.
Think: We have a value of $h$h, and we want to find the corresponding value of $t$t.
Do: Substitute $h=15$h=15 into the equation and rearrange to solve for $t$t.
| $h$h | $=$= | $10\sin\left(20\left(t-7\right)\right)+10$10sin(20(t−7))+10 | |
| $15$15 | $=$= | $10\sin\left(20\left(t-7\right)\right)+10$10sin(20(t−7))+10 | (Substitute the value of $h$h) |
| $5$5 | $=$= | $10\sin\left(20\left(t-7\right)\right)$10sin(20(t−7)) | (Subtract $10$10 from both sides) |
| $\frac{1}{2}$12 | $=$= | $\sin\left(20\left(t-7\right)\right)$sin(20(t−7)) | (Divide both sides by $10$10) |
| $30$30 | $=$= | $20\left(t-7\right)$20(t−7) | (Take the inverse $\sin$sin of both sides) |
| $\frac{3}{2}$32 | $=$= | $t-7$t−7 | (Divide both sides by $20$20) |
| $t$t | $=$= | $8.5$8.5 seconds | (Add $7$7 to both sides) |
The point on the disk will first reach a height of $15$15 cm when it has rolled for $8.5$8.5 seconds.
Practice question
The height above the ground of a rider on a Ferris wheel can be modelled by the function $h\left(t\right)=25\cos\left(3\left(t-60\right)\right)+30$h(t)=25cos(3(t−60))+30, where $h\left(t\right)$h(t) is the height in metres, $t$t is the time in seconds, and the angular velocity of $3^\circ$3° per second is the speed at which the wheel rotates.
Draw the graph of $y=h\left(t\right)$y=h(t).
Loading Graph...What is the maximum height of the rider?
What is the minimum height of the rider?
At what height is the rider after $85$85 seconds? Give your answer to two decimal places.
How long does it take the rider to complete one full revolution?