4.5. Applications of Trigonometry to Waves. Introduction. Prerequisites. Learning Outcomes

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1 Applications of Trigonometry to Waves 4.5 Introduction Waves and vibrations occur in many contexts. The water waves on the sea and the vibrations of a stringed musical instrument are just two everyday examples. If the vibrations are simple back and fro oscillations they are referred to as sinusoidal which implies that a knowledge of trigonometry, particularly of the sine and cosine functions, is a necessary pre-requisite for dealing with their analysis. In this Section we give a brief introduction to this topic. Prerequisites Before starting this Section you should... Learning Outcomes After completing this Section you should be able to... 1 have a knowledge of the basics of trigonometry be aware of the standard trigonometric identities use simple trigonometric functions to describe waves be able to combine two waves of the same frequency as a single wave in amplitudephase form

2 1. Applications of trigonometry to waves Suppose that a wheel of radius R is rotating anticlockwise as shown in Figure. y Q B O ωt P A x Figure Assume that the wheel is rotating with an angular velocity ω radians per second so that, in a time t seconds, a point A on the rim of the wheel moves to a point B such that angle AOB = ωt radians. Then the coordinates of the point B are x = OP = R cos ωt 4a y = OQ = PB = R sin ωt from which x + y = R cos ωt + sin ωt =R We know that both the standard sine function and cosine function have period π. Hence the wheel will complete one complete revolution in a time t 1 such that 4b ωt 1 = ωt +π or t 1 = t + π ω The time t 1 t = π ω for one complete revolution is called the period of rotation of the wheel. The number of complete revolutions per second is thus 1 T = f say which is called the frequency of revolution. Clearly f = 1 T = ω relates the three quantities introduced here. The angular π velocity ω = πf is sometimes called the angular frequency. The situation we have just outlined is a two-dimensional one. More simply we might consider one-dimensional motion where x = R cos ωt 5 x being the co-ordinate of a moving particle. Clearly, from the known properties of the cosine function, we can deduce the following: 1. x varies periodically with t with period T = π ω.. x will have maximum value R, minimum value R, This quantity R is called the amplitude of the motion. In other words 5 is a description of a common back and fro motion between R and R. HELM VERSION 1: March 18, 004: Workbook Level 1

3 Using 5 write down the values of x at the following times: t =0,t = π ω, t = π ω, t = 3π ω,t= π. Describe in words the form of the motion: ω Your solution t x = R cos ωt 0 R π ω π ω 3π ω 0 R 0 π ω R Using 5 this back and fro or vibrational or oscillatory motion between R and R continues indefinitely. The technical name for this motion is simple harmonic. Toagood 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 7 in these cases damping of the pendulum or spring is ignored. Figure 7 3 HELM VERSION 1: March 18, 004: Workbook Level 1

4 Using your knowledge of the cosine function and the results of the previous exercise plot the graph of x against t if x = R cos ωt for t =0tot = 4π ω Your solution R x = R cos ωt period π ω 4π ω t R Figure 8 This graph shows part of a cosine wave specifically two periods of oscillation. The shape of the graph suggests that the term wave is indeed an appropriate description. We know that the shape of the cosine graph and the sine graph are identical but just offset by π radians horizontally. Bearing this in mind, attempt the following guided exercise. Write the equation of the wave, part of which is shown in the following graph. You will need to find the period T and angular frequency ω. 5 x 4 8 t secs 5 Figure 9 HELM VERSION 1: March 18, 004: Workbook Level 1 4

5 Your solution From the shape of the graph we have a sine wave rather than a cosine wave. The amplitude is 5. The period T =4s so the angular frequency ω = π 4 = π is πt x =5sin.. Hence the equation of the wave The quantity x, a function of t, is referred to as the displacement of the wave. Phase of a wave We recall that cos θ π = sin θ which simply means that the graph of x = sin θ is the same shape as that of x = cos θ but is shifted to the right by π. Suppose now that we consider the waves x 1 = R cos t x = R sin t Both have amplitude R, angular frequency ω =rads s 1. Also x = R cos t π [ = R cos t π ] 4 The graphs of x 1 against t and of x against t are said to have a phase difference of π 4. Specifically x 1 is ahead of, or leads x by π 4 radians. More generally consider the following sine waves of the same amplitude and frequency: x 1 t = R sin ωt x t = R sinωt α Now x 1 t α [ = R sin ω t α ] = R sinωt α =x t ω ω so it is clear that the waves x 1 and x are out of phase by α ω.specifically x 1 leads x by α ω. Calculate the phase difference between the waves The time t is in seconds. x 1 = 3cos10πt x = 3cos 10πt + π 4 5 HELM VERSION 1: March 18, 004: Workbook Level 1

6 Your solution Note firstly that the waves have the same amplitude 3 and angular frequency 10π corresponding to a common period π 10π = 1 5 s Now cos = cos 10π t πt + π 4 so x 1 t = x t. In other words the phase difference is 1 40 s, the wave x leads the wave x 1 by this amount. Alternatively we could say that x 1 lags x by 1 40 s. The equations x = R cos ωt Key Point x = R sin ωt both represent waves of amplitude R and period π ω. The phase difference between these waves is π [ ω because cos ω t π ] = sin ωt. ω Combining two wave equations A situation that arises in some applications is the need to combine two trigonometric terms such as A cos θ + B sin θ. 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 ωt + R sin ωt where the individual waves have, as we have seen, a phase difference of π ω. HELM VERSION 1: March 18, 004: Workbook Level 1

7 General Theory Consider an expression A cos θ + B sin θ. We seek to transform this into the single form C cosθ α or C sinθ α, where C and α have to be determined. The problem is easily solved with the aid of trigonometric identities. We know that C cosθ α =Ccos θ cos α + sin θ sin α Hence if A cos θ + B sin θ = C cosθ α then A cos θ + B sin θ =Ccos α cos θ +C sin α sin θ For this to be an identity true for all values of θ wemust be able to equate the coefficients of cos θ and sin θ on each side. Hence A = C cos α B = C sin α By squaring and adding these equations, obtain C in terms of A and B. Also, by eliminating C, obtain α in terms of A and B. Your solution A = C cos α B = C sin α gives A + B = C cos α + C sin α = C cos α + sin α = C... C = A + B We take the positive square root. Also, by division, B A = C sin α C cos α = tan α so α is obtained by solving tan α = B A quadrant for α.. However, care must be taken to obtain the correct 7 HELM VERSION 1: March 18, 004: Workbook Level 1

8 Key Point If A cos θ + B sin θ = C cosθ α then C = A + B and tan α = B A In terms of waves we have where R = R 1 + R tan α = R R 1 R 1 cos ωt + R sin ωt = R cosωt α The form R cosωt α issaid to be the amplitude/phase form of the wave. Note that we can write Since A cos θ + B sin θ = { } A A + B A + B cos θ + B A + B sin θ A A + B and B cannot be greater than 1 in magnitude we can write A + B A A + B = cos α B = sin α a, b A + B So where C = A + B as before and A cos θ + B sin θ = A + B cos α cos θ + sin α sin θ = C cosθ α tan α = sin α cos α = B A +B = B A A A +B again as before. 1. A>0, B>0: 1st quadrant. A<0, B>0: nd quadrant 3. A<0, B<0: 3rd quadrant 4. A>0, B<0: 4th quadrant HELM VERSION 1: March 18, 004: Workbook Level 1 8

9 Example Express in the form C cosθ α each of the following: a 3 cos θ +3sin θ b 3 cos θ +3sin θ c 3 cos θ 3 sin θ d 3 cos θ 3 sin θ Solution In each case : C = A + B = 9+9= 18 a tan α = B A = 3 =1gives α =45 A 3 is the correct one. Hence and B are both positive so the first quadrant 3 cos θ +sin θ = 18 cosθ 45 = 18 cos θ π 4 b The angle α must be in the second quadrant as A = 3 < 0, B =+3> 0. By calculator : tan α = 1 gives α = 45 but this is in the 4th quadrant. Remembering that tan α has period π or 180 we must therefore add 180 to the calculator value to obtain the correct α value of 135. Hence 3 cos θ +3sin θ = 18 cosθ 135 c Here A = 3, B = 3 soα must be in the 3rd quadrant. tan α = 3 3 =1giving α =45 by calculator. Hence adding 180 to this tells us that 3 cos θ 3 sin θ = 18 cosθ 5 d Here A =3B = 3 soα is in the 4th quadrant. tan α = 1 gives us correctly α = 45 so 3 cos θ 3 sin θ = 18 cosθ +45. Note that in the amplitude/phase form the angle may be expressed in degrees or radians. Write the wave form x =3cos ωt +4sin ωt in amplitude/phase form. Express the phase in radians. 9 HELM VERSION 1: March 18, 004: Workbook Level 1

10 Your solution We have x = R cosωt α where 3 +4 =5 tan the calculator in radian mode, α =0.97 radians. This is in the first quadrant 0 <α< π is correct since A =3and B =4are both positive. Hence x =5cosωt R = α = 4 3 from which, using which HELM VERSION 1: March 18, 004: Workbook Level 1 10

11 Exercises 1. Write down the amplitude and the period of y = 5 sin πt.. Write down the amplitude, frequency and phase of a y =3sin t π 3 b y =15cos 5t 3π 3. The current in an a/c circuit is it = 30sin 10πt amps Here 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 = a sin b t π + k where k is the average depth. If the tidal period is 1 hours, the depths at high tide and low tide are 18 metres and 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 day is modelled by F t =0+10sin π t 8 0 t 4 1 where t is in the time in hours after midnight. a What are the temperatures at 8.00 am and 1.00 noon? b At what time is the temperature 0 F? c Obtain the maximum and minimum temperatures and the times at which they occur.. In each of the following write down expressions for shifted sine and shifted cosine functions that satisfy the given conditions: a Amplitude 3, Period π 3, Phase shift π 3 b Amplitude 0.7, Period 0.5, Phase shift Write the a/c current in the form i = C cos5π α. i =3cos 5t +4sin 5t 11 HELM VERSION 1: March 18, 004: Workbook Level 1

12 Exercises 8. Show that if then A cos ωt + B sin ωt = C sinωt + α C = A + B, cos α = B C, sin α = A C. 9. Using Question 8 express the following in the amplitude/phase form C sinωt+ α a y = 3 sin t + cos t b y = cos t + 3 sin t 10. The motion of a weight on a spring is given by Obtain C and α such that 11. Show that for the two a/c currents i 1 = sin ωt + π 3 y = 3 cos 8t 1 sin 8t. y = C sin8t + α i =3cos ωt π then i 1 + i =4cos ωt π. 1. Show that the power P = v R in an electrical circuit where v = V 0 cos ωt + π 4 P = V 0 1 sin ωt R 13. Show that the product of the two signals f 1 t =A 1 sin ωt is given by f t =A sin [ωt + τ+φ] is f 1 tf t = A 1A [cosωτ + φ cosωt + ωτ + φ]. HELM VERSION 1: March 18, 004: Workbook Level 1 1

13 1. y = 5 sin πt has amplitude 5 π. The period is π =1. Check: yt +1= 5 sinπt + 1 = 5 sinπt +π =5 sin πt = yt. a Amplitude 3, Period π = π. Writing y =3sin t π 0 we see that there is a phase shift of π radians in this wave compared with y =3sin t. b Amplitude 15, Period π 5. Clearly y =15cos 5 t 3π so there is a phase shift 10 of 3π 10 compared with y =15cos 5t. 3. Maximum current = 30 amps at a time t such that 10πt = π. i.e. t = 1 40 s. This maximum will occur again at 4. y = a sin [ b t π ] n 0 + h. The period is π b s, n =1,, 3,... =1hr... b = π hr 1. Also since y max = a + k y[ min = a + k we have a + k =18 a + k = so k =1 π m, a =m. i.e. y =sin t π ] F t =0+10 sin π 1 t 8 0 t<4 a At t =8: temp = 0 F. At t = 1: temp = sin π 3 =8.7 F b F t =0when π t 8 =0,π, π,... giving t 8=0, 1, 4,...hours so 1 t =8, 0, 3,... hours i.e. in 1 day at t =88.00 am and t =08.00 pm c Maximum temperature is 70 F when π 1 t =8=π i.e. at t =14.00 pm. Minimum temperature is 50 F when π 1 t 8 = 3π i.e. at t =.00 am.. a y =3sin3t π y =3cos3t π b y =0.7 sin4πt 1 y =0.7 cos4πt 1 7. C = 3 +4 =5tan α = 4 3 and α must be in the first quadrant since A =3, B =4 are both positive.... α = tan =0.973 rads... i =5cos5t Since sinωt + α =sin ωt cos α + cos ωt sin α then A = C sin α coefficients of cos ωt B = C cos α coefficients of sin ωt from which C = A + B, sin α = A C, cos α = B C 13 HELM VERSION 1: March 18, 004: Workbook Level 1

14 9. a C = 3+1 = ; cos α = 3 sin α 1 so α is in the second quadrant, α = 5π... y =sin t + 5π b y =sin t + π 10. C = = 17 3 so C = 17 cos α = 1 17 so α is in the second quadrant. α = radians. 11. Since sin x = cos x π sin ωt + π = cos ωt + π 3 3 π... i 1 + i = cos ωt π +3cos ωt π =4cos ωt π 1. v = V 0 cos ωt + π = V 0 cos ωt cos π 4 4 sin ωt sin π 4 and hence... v = V 0 cos ωt + sin ωt sin ωt cos ωt = V 0 = 1 sin α = 17 = cos 3 17 ωt π = 4 17 = V0 cos ωt sin ωt 1 sin ωt P = v R = V 0 1 sin ωt. R 13. Since the required answer involves the difference of two cosine functions we use the identity A + B B A cos A cos B =sin sin Hence with A + B B A = ωt, ωt + ωτ + φ. We find, by adding these equations B =ωt + ωτ + φ and by subtracting A = ωτ φ. Hence sinωt sinωt + ωτ + φ = 1 [cosωτ + φ cosωt + ωτ + φ]. Recall that cos x = cos x. The required result then follows immediately. HELM VERSION 1: March 18, 004: Workbook Level 1 14

EDEXCEL NATIONAL CERTIFICATE UNIT 28 FURTHER MATHEMATICS FOR TECHNICIANS OUTCOME 3 TUTORIAL 1 - TRIGONOMETRICAL GRAPHS

EDEXCEL NATIONAL CERTIFICATE UNIT 28 FURTHER MATHEMATICS FOR TECHNICIANS OUTCOME 3 TUTORIAL 1 - TRIGONOMETRICAL GRAPHS CONTENTS 3 Be able to understand how to manipulate trigonometric expressions and apply