1 Applications of trigonometry to waves

1.1 Two-dimensional motion

Suppose that a wheel of radius $R$ is rotating anticlockwise as shown in Figure 38.

Figure 38

Assume that the wheel is rotating with an angular velocity $\omega$ radians per second about $O$ so that, in a time $t$ seconds, a point $\left(x,y\right)$ initially at position $A$ on the rim of the wheel moves to a position $B$ such that angle $AOB=\omega t$ radians.

Then the coordinates $\left(x,y\right)$ of $B$ are given by

$x=OP=Rcos\omega t$

$y=OQ=PB=Rsin\omega t$

We know that both the standard sine and cosine functions have period $2\pi$ . Since the angular velocity is $\omega$ radians per second the wheel will make one complete revolution in $\frac{2\pi }{\omega }$ seconds.

The time $\frac{2\pi }{\omega }$ (measured in seconds in this case) for one complete revolution is called the period of rotation of the wheel. The number of complete revolutions per second is thus $\frac{1}{T}=f$ say which is called the frequency of revolution. Clearly $f=\frac{1}{T}=\frac{\omega }{2\pi }$ relates the three quantities

introduced here. The angular velocity $\omega =2\pi f$ is sometimes called the angular frequency .

1.2 One-dimensional motion

The situation we have just outlined is two-dimensional motion. More simply we might consider one-dimensional motion.

An example is the motion of the projection onto the $x$ -axis of a point $B$ which moves with uniform angular velocity $\omega$ round a circle of radius $R$ (see Figure 39). As $B$ moves round, its projection $P$ moves to and fro across the diameter of the circle.

Figure 39

The position of $P$ is given by

$x=Rcos\omega t$ (1)

Clearly, from the known properties of the cosine function, we can deduce the following:

1. $x$ varies periodically with $t$ with period $T=\frac{2\pi }{\omega }$ .
2. $x$ will have maximum value $+R$ and minimum value $-R$ .

   (This quantity $R$ is called the amplitude of the motion.)

Task!

Using (1) write down the values of $x$ at the following times:

$t=0$ , $t=\frac{\pi }{2\omega }$ , $t=\frac{\pi }{\omega }$ , $t=\frac{3\pi }{2\omega },$ $t=\frac{2\pi }{\omega }$ .

Answer

Using (1) this ’to and fro’ or ‘vibrational’ or ‘oscillatory’ motion between $R$ and $-R$ continues indefinitely. The technical name for this motion is simple harmonic . To a good approximation it is the motion exhibited (i) by the end of a pendulum pulled through a small angle and then released (ii) by the end of a hanging spring pulled down and then released. See Figure 40 (in these cases damping of the pendulum or spring is ignored).

Figure 40

Task!

Using your knowledge of the cosine function and the results of the previous Task sketch the graph of $x$ against $t$ where

$x=Rcos\omega t$ for $t=0$ to $t=\frac{4\pi }{\omega }$

Answer

We know that the shape of the cosine graph and the sine graph are identical but offset by $\frac{\pi }{1.7}$ radians horizontally. Bearing this in mind, attempt the following Task.

Task!

Write the equation of the wave $x\left(t\right)$ , part of which is shown in the following graph. You will need to find the period $T$ and angular frequency $\omega$ .

Answer

The quantity $x$ , a function of $t$ , is referred to as the displacement of the wave.

1.3 Phase of a wave

We recall that $cos\left(\theta -\frac{\pi }{2}\right)=sin\theta$ which means that the graph of $x=sin\theta$ is the same shape as that of $x=cos\theta$ but is shifted to the right by $\frac{\pi }{2}$ .

Suppose now that we consider the waves

$x_1=Rcos2t{x}_2=Rsin2t$

Both have amplitude $R$ , angular frequency $\omega =2\text{rad}{\text{s}}^{-1}$ . Also

$x_2=Rcos\left(2t-\frac{\pi }{2}\right)=Rcos\left[2\left(t-\frac{\pi }{4}\right)\right]$

The graphs of $x_1$ against $t$ and of $x_2$ against $t$ are said to have a phase difference of $\frac{\pi }{4}$ . Specifically $x_1$ is ahead of, or ‘leads’ $x_2$ by $\frac{\pi }{4}$ radians.

More generally, consider the following two sine waves of the same amplitude and frequency:

$x_1\left(t\right)=Rsin\omega t$

$x_2\left(t\right)=Rsin\left(\omega t-\alpha \right)$

Now $x_1\left(t-\frac{\alpha }{\omega }\right)=Rsin\left[\omega \left(t-\frac{\alpha }{\omega }\right)\right]=Rsin\left(\omega t-\alpha \right)=x_2\left(t\right)$

so it is clear that the waves $x_1$ and $x_2$ are out of phase by $\frac{\alpha }{\omega }$ . Specifically $x_1$ leads $x_2$ by $\frac{\alpha }{\omega }$ .

Task!

Calculate the phase difference between the waves

where the time $t$ is in seconds.

Answer

1.4 Combining two wave equations

A situation that arises in some applications is the need to combine two trigonometric terms such as

$Acos\theta +Bsin\theta$ where $A$ and $B$ are constants.

For example this sort of situation might arise if we wish to combine two waves of the same frequency but not necessarily the same amplitude and with a phase difference. In particular we wish to be able to deal with an expression of the form

$R_{1}cos\omega t+R_{2}sin\omega t$

where the individual waves have, as we have seen, a phase difference of $\frac{\pi }{2\omega }$ .

Consider an expression   $Acos\theta +Bsin\theta .$  We seek to transform this into the single form
  $Ccos\left(\theta -\alpha \right)$  (or $Csin\left(\theta -\alpha \right)$ ), where $C$ and $\alpha$ have to be determined. The problem is easily solved with the aid of trigonometric identities.

We know that

$Ccos\left(\theta -\alpha \right)\equiv C\left(cos\theta cos\alpha +sin\theta sin\alpha \right)$

Hence if $Acos\theta +Bsin\theta =Ccos\left(\theta -\alpha \right)$ then

$Acos\theta +Bsin\theta =\left(Ccos\alpha \right)cos\theta +\left(Csin\alpha \right)sin\theta$

For this to be an identity (true for all values of $\theta$ ) we must be able to equate the coefficients of $cos\theta$ and $sin\theta$ on each side.

Hence

$A=Ccos\alpha \text{and}B=Csin\alpha$ (2)

Task!

By squaring and adding the Equations (2), obtain $C$ in terms of $A$ and $B$ .

Answer

Task!

By eliminating $C$ from Equations (2) and using the result of the previous Task, obtain $\alpha$ in terms of $A$ and $B$ .

Answer

In terms of waves, using Key Point 21 we have

$R_{1}cos\omega t+R_{2}sin\omega t=Rcos\left(\omega t-\alpha \right)$

where $R=\sqrt{R_1^2+R_2^2}\text{and}tan\alpha =\frac{R_2}{R_1}$ .

The form $Rcos\left(\omega t-\alpha \right)$ is said to be the amplitude/phase form of the wave.

Example 5

Express in the form $Ccos\left(\theta -\alpha \right)$ each of the following:

1. $3cos\theta +3sin\theta$
2. $-3cos\theta +3sin\theta$
3. $-3cos\theta -3sin\theta$
4. $3cos\theta -3sin\theta$

Solution

In each case $C=\sqrt{A^2+B^2}=\sqrt{9+9}=\sqrt{18}$

1. $tan\alpha =\frac{B}{A}=\frac{3}{3}=1$ gives $\alpha =45^{\circ }$ ( $A$ and $B$ are both positive so the first quadrant is the correct one.) Hence $3cos\theta +2sin\theta =\sqrt{18}cos\left(\theta -45^{\circ }\right)=\sqrt{18}cos\left(\theta -\frac{\pi }{4}\right)$
2. The angle $\alpha$ must be in the second quadrant as $A=-3<0$ , $B=+3>0$ . By calculator : $tan\alpha =-1$ gives $\alpha =-45^{\circ }$ but this is in the 4th quadrant. Remembering that $tan\alpha$ has period $\pi$ or $180^{\circ }$ we must therefore add $180^{\circ }$ to the calculator value to obtain the correct $\alpha$ value of $135^{\circ }$ . Hence

   $-3cos\theta +3sin\theta =\sqrt{18}cos\left(\theta -135^{\circ }\right)$

3. Here $A=-3,B=-3$ so $\alpha$ must be in the 3rd quadrant. $tan\alpha =\frac{-3}{-3}=1$ giving $\alpha =45^{\circ }$ by calculator. Hence adding $180^{\circ }$ to this tells us that

   $-3cos\theta -3sin\theta =\sqrt{18}cos\left(\theta -225^{\circ }\right)$

4. Here $A=3B=-3$ so $\alpha$ is in the 4th quadrant. $tan\alpha =-1$ gives us (correctly) $\alpha =-45^{\circ }$ so

   $3cos\theta -3sin\theta =\sqrt{18}cos\left(\theta +45^{\circ }\right).$

   Note that in the amplitude/phase form the angle may be expressed in degrees or radians.

Task!

Write the wave form $x=3cos\omega t+4sin\omega t$ in amplitude/phase form. Express the phase in radians to 3 d.p..

Answer

Exercises

1. Write down the amplitude and the period of $y=\frac{5}{2}sin2\pi t$ .
2. Write down the amplitude, frequency and phase of
   1. $y=3sin\left(2t-\frac{\pi }{3}\right)$
   2. $y=15cos\left(5t-\frac{3\pi }{2}\right)$
3. The current in an a.c. circuit is   $i\left(t\right)=30sin120\pi t\text{amp}$  where $t$ is measured in seconds.

   What is the maximum current and at what times does it occur?

4. The depth $y$ of water at the entrance to a small harbour at time $t$ is $y=asinb\left(t-\frac{\pi }{2}\right)+k$

   where $k$ is the average depth. If the tidal period is 12 hours, the depths at high tide and low tide are 18 metres and 6 metres respectively, obtain $a,b,k$ and sketch two cycles of the graph of $y$ .

5. The Fahrenheit temperature at a certain location over 1 complete day is modelled by

   $F\left(t\right)=60+10sin\frac{\pi }{12}\left(t-8\right)0\le t\le 24$

   where $t$ is in the time in hours after midnight.

   1. What are the temperatures at 8.00 am and 12.00 noon?
   2. At what time is the temperature $60^{\circ }$ F?
   3. Obtain the maximum and minimum temperatures and the times at which they occur.
6. In each of the following write down expressions for shifted sine and shifted cosine functions that satisfy the given conditions:
   1. Amplitude 3, Period $\frac{2\pi }{3}$ , Phase shift $\frac{\pi }{3}$
   2. Amplitude 0.7, Period 0.5, Phase shift 4.
7. Write the a.c.current   $i=3cos5t+4sin5t$  in the form $i=Ccos\left(5\pi -\alpha \right).$
8. Show that if   $Acos\omega t+Bsin\omega t=Csin\left(\omega t+\alpha \right)$  then

   $C=\sqrt{A^2+B^2},cos\alpha =\frac{B}{C},sin\alpha =\frac{A}{C}.$

9. Using Exercise 8 express the following in the amplitude/phase form $Csin\left(\omega t+\alpha \right)$
   1. $y=-\sqrt{3}sin2t+cos2t$
   2. $y=cos2t+\sqrt{3}sin2t$
10. The motion of a weight on a spring is given by $y=\frac{2}{3}cos8t-\frac{1}{6}sin8t.$

    Obtain $C$ and $\alpha$ such that $y=Csin\left(8t+\alpha \right)$

11. Show that for the two a.c.currents

    $i_1=sin\left(\omega t+\frac{\pi }{3}\right)\text{and}{i}_2=3cos\left(\omega t-\frac{\pi }{6}\right)$  then   $i_1+i_2=4cos\left(\omega t-\frac{\pi }{6}\right).$

12.   Show that the power $P=\frac{v^2}{R}$ in an electrical circuit where $v=V_{0}cos\left(\omega t+\frac{\pi }{4}\right)$ is

    $P=\frac{V_0^2}{2R}\left(1-sin2\omega t\right)$

13.   Show that the product of the two signals

    $f_1\left(t\right)=A_{1}sin\omega t{f}_2\left(t\right)=A_{2}sin\left\{\omega \left(t+\tau \right)+\varphi \right\}$ is given by

    $f_1\left(t\right)f_2\left(t\right)=\frac{A_1{A}_2}{2}\left\{cos\left(\omega \tau +\varphi \right)-cos\left(2\omega t+\omega \tau +\varphi \right)\right\}$ .

Answer