outcome2t1

© D.J.Dunn www.freestudy.co.uk 1

EDEXCEL ANALYTICAL METHODS FOR ENGINEERS H1

UNIT 2 - NQF LEVEL 4

OUTCOME 2 - TRIGONOMETRIC METHODS

TUTORIAL 1 –SINUSOIDAL FUNCTION

Sinusoidal functions: review of the trigonometric ratios; Cartesian and polar co-ordinate systems; properties of the circle; radian measure; sinusoidal functions Applications such as: angular velocity; angular acceleration; centripetal force; frequency; amplitude; phase; the production of complex waveforms using sinusoidal graphical synthesis; AC waveforms and phase shift Trigonometric identities: relationship between trigonometric and hyperbolic identities; double angle and compound angle formulae and the conversion of products to sums and differences; use of trigonometric identities to solve trigonometric equations and simplify trigonometric expressions You should judge your progress by completing the self assessment exercises.

Trigonometry has been covered in the NC Maths module and should have been studied prior to this module. This tutorial provides further studies and applications of that work.

1. RADIAN In Engineering and Science, we use the radian to measure angle as well as degrees. This is defined as the angle created by placing a line of length 1 radius around the edge of the circle as shown. In mathematical words it is the angle subtended by an arc of length one radius. This angle is called the RADIAN. The circumference of a circle is 2πR. It follows that the number of radians that make a complete circle is

R

R 2π or 2π.

There are 2π radians in one revolution so 360o =2π radian 1 radian = 360/2π = 57.296o

2. TWO DIMENSIONAL COORDINATE SYSTEMS CARTESIAN In a two dimensional system the vertical direction is usually y (positive up) and the horizontal is direction is x (positive to the right). Other letters may be used to designate an axis and they don’t have to be vertical and horizontal. The origin ‘o’ is where the axis cross at x = 0 and y = 0 A point p on this plane has coordinates x, y and this is usually written as p (x,y) POLAR If a line is drawn from the origin to point p it is a radius R and forms an angle θ with the x axis. The angle is positive measured from the x axis in a counter clockwise direction. A vector with polar coordinates is denoted R∠ θ CONVERSION

©D.J.Dunn www.freestudy.co.uk 2

The two systems are clearly linked as we can convert from one to the other using trigonometry and Pythagoras’ theorem. y = R sin θ x = R cos θ y/x = tan θ R = (x2 + y2) ½

WORKED EXAMPLE No. 1 The x, y coordinates of a point is 4, and 6. Calculate the polar coordinates. SOLUTION R = (42 + 62)1/2 = 7.211 θ = tan-1 (6/4) = 56.31o

SELF ASSESSMENT EXERCISE No. 1 1. Convert 60o to radian. (1.0472 rad) 2. Convert π/6 radian into degrees. (30o) 3. The x, y Cartesian coordinates of a vector are 2 and 7.Express the vector in Polar coordinates. (7.28 74∠ o) 4. A vector with Cartesian coordinates 10, 20 is added to a vector with coordinates -20,10. What

are the polar coordinates of the resulting vector? (31.623 108.4∠ o) 3. REVIEW OF TRIGONOMETRIC RATIOS The ratios of the lengths of the sides of a right angle triangle are always the same for any given angle θ. These ratios are very important because they allow us to calculate lots of things to do with triangles. In the following the notation above is used with the corners denoted AB and C SINE

The ratio ABCB

HypotenuseOpposite

= is called the sine of the angle A. (note we usually drop the e on sine)

Before the use of calculators, the values of the sine of angles were placed in tables but all you have to do is enter the angle into your calculator and press the button shown as sin. For example if you enter 60 into your calculator in degree mode and press sin you get 0.8660 If you enter 0.2 into your calculator in radian mode and press sin you get 0.1987 Note that sin(θ) = -sin (-θ) and sin (180 – θ) = sin(θ) COSINE

ABAC

HypotenuseAdjacent

= is called the cosine of the angle A. The ratio

On your calculator the button is labelled cos. For example enter 60 into your calculator in degree mode and press the cos button. You should obtain 0.5 If you enter 0.2 into your calculator in radian mode and press cos you get 0.9800 Note that cos(θ) = cos (-θ) and cos(180 – θ) = -cos(θ) TANGENT

The ratio ACBC

AdjacentOpposite

= is called the tangent of the angle A.

On your calculator the button is labelled tan. For example enter 60 into your calculator in degree mode and press the tan button and you should obtain 1.732. If you enter 0.2 into your calculator in radian mode and press tan you get 0.2027 Note that sin(θ)/cos(θ) = tan(θ) and tan(90 – θ) = 1/tan(θ)


INVERSE FUNCTIONS Some people find it useful to use the inverse functions which are as follows. cosec (θ) = sin-1(θ) sec (θ) = cos-1(θ) cot (θ) = tan-1(θ) 4. SINE AND COSINE RULE The following work enables us to solve triangles other than right angles triangles. SINE RULE Consider the diagram. h = b sin A = a sin B

It follows that sinB

bsinA

a=

If we did the same for another perpendicular to side b or a we could show that sinC

csinB

bsinA

a==

WORKED EXAMPLE No. 2


Find the length of the two unknown side in the triangle shown.

SOLUTION a = 50 mm A = 30o B = 45o C = 180 o - 30o - 45o = 105o

sinC

csinB

bsinA

a==

sin(105)c

sin(45)b

sin(30)50

==

mm 70.711(30)sin

sin(45) 50b == mm 96.593(30)sin

sin(105) 50c ==

WORKED EXAMPLE No. 3 A weight of 300 N is suspended on two ropes as shown. Calculate the length of the ropes Draw the vector diagram for the three forces in equilibrium. Calculate the forces in the ropes.

SOLUTION The third internal angle is 110o. 4/sin 110 = L1/sin 50 = L2/sin 20 L1 = 3.261 m and L2 = 1.456 m Next draw the triangle of forces as shown. F1/sin 40o = 300/sin 70o F1 = 205.2 N F2/sin 70o = 300/sin 70o F1 = 300 N

COSINE RULE Consider the diagram. Using Pythagoras we have: h2 = a2 – (b – x)2 and h2 = c2 – x2

a2 – (b2 + x2 – 2bx) = c2 – x2

a2 – b2 - x2 + 2bx) = c2 – x2

a2 = b2 + x2 – 2bx + c2 – x2

a2 = b2 + c2– 2bx substitute x = c cos(A) a2 = b2 + c2– 2bc cos(A) = c2

2bcacbcos(A)

222 −+=

If we repeated the process with h drawn normal to the other sides we could show that :

2cabaccos(B)

222 −+=

2abcbacos(C)

222 −+=

You can see a pattern for remembering the formulae. This is a useful formula for solving a triangle with three known sides or two known sides and the angle opposite the unknown side. WORKED EXAMPLE No. 4


Find the length of the unknown side in the triangle shown. Find the other internal angles. SOLUTION

2ab

cbacos(C)222 −+

= (2)(60(70)

c0706)cos(60222

o −+=

222o c07060))(2)(60)(7cos(60 −+= 2c85002004 −= c2 = 4300 c = 65.574 mm

61.0)574.65)(70)(2(

60574.65702bc

acbcos(A)222222=

−+=

−+=

A = 52.4o

B = 180 – 52.4 – 60 = 126.5 o

WORKED EXAMPLE No. 5 Find the resultant of the two forces shown. SOLUTION The addition of the two force is done as shown.

0)(2)(100(10R010010)cos(135

222o −+=

222o R010010100))(2)(100)(cos(135 −+= -14142.1 = 20000 – R2

R2 = 34142.13 R = 184.78 N

SELF ASSESSMENT EXERCISE No. 2 1. Find the resultant of the two forces shown. (Answer 328.8 N)

2. Vector A has polar coordinates 12∠ 60o and vector B has polar coordinates 5 20∠ o

Find the resultant in polar form. (16.15∠ 48.5o ) 3. The diagram shows a weight suspended from two ropes. Calculate the angles of the ropes to the

horizontal support.

(Answers 49o and 59o) 4. A weight of 4 Tonne is suspended on two ropes as shown. Calculate the length of the ropes and

the forces in them.

(Answers 1.034 m, 1.464 m, 3.172 T and 2.928 T)


5. SINUSOIDAL FUNCTIONS In Nature and in Engineering there are many things that oscillate in some form or other and produce a repetitive change of some quantity with respect to time. Examples are mechanical oscillations and alternating electricity. In many cases a plot of the quantity against time produces a sinusoidal graph and the change is said to be sinusoidal. MECHANICAL EXAMPLES SCOTCH YOKE and ECCENTRIC CAM

The Scotch Yoke is a device that produces up and down motion when the wheel is rotated. The displacement of the yoke from the horizontal position is x = R sin θ = x R sin(ωt). Plotting x against time or angle will produce a sinusoidal graph. The eccentric cam is really another version of this. In all cases we should remember that velocity is the first derivative of displacement and acceleration is the second derivative. It follows that: Displacement x = R sin(ωt) Velocity v = dx/dt = ωR cos(ωt) Acceleration a = dv/dt = -ω2R sin(ωt) = -ω2 x Anything that obeys these equations is said to have SIMPLE HARMONIC MOTION The starting point of the oscillation could be at any angle φ so in that case the equations become: Displacement x = R sin(ωt + φ) Velocity v = dx/dt = ωR cos(ωt + φ) Acceleration a = dv/dt = -ω2R sin(ωt + φ) = -ω2 x The plots show the displacement, velocity and acceleration for φ = 0 on the left and a negative φ on the right.


MASS ON A SPRING


A mass on the end of a spring will oscillate up and down and produce identical motion to the Scotch Yoke without the rotation of a wheel. The displacement is x = xo sin (ωt) The velocity of the mass is v = ω xo sin(ωt) The acceleration of the mass is a = -ω2 xo cos(ωt) It can be shown for the frictionless case that ω = (k/m)1/2 where k is the spring stiffness in N/m and m the mass in kg. This is called the natural frequency.

The natural frequency of oscillation is hence mk

2π1f =

CENTRIFUGAL FORCE When a mass rotates at radius R the centrifugal force is given by:

CF = mω2R If this is a machine mounted on a platform as shown that can only move in one direction, the force acting in that direction is the component of the force in that direction. In this example the force exerted on the spring is:

mω2R sin(θ) = mω2R sin(ωt)

WORKED EXAMPLE No. 6 The displacement of a body performing simple harmonic motion is described by the following

equation x = A sin (ωt + φ) where A is the amplitude, ω is the natural frequency and φ is the phase angle.

Given A = 20 mm, ω = 50 rad/s and φ = π/8 radian, calculate the following. i. The frequency. ii. The periodic time. iii. The displacement, velocity and acceleration when t = T/4. Sketch the graphs of x, v and a and confirm your answers. SOLUTION First deduce the frequency. f = ω/2π = 50/2π = 7.96 Hz. Next deduce the periodic time. T = 1/f = 0.126 s Next deduce the time t. t = T/4 = 0.0314 s Next write out the equation for displacement and solve x at t = 0.0314 s

mm 18.4820sin1.963

8π1.5720sinx

8π0.314)x (50sin 20x

==⎟⎠⎞

⎜⎝⎛ +=

⎟⎠⎞

⎜⎝⎛ +=

Next write down the equations for v and a

( )( ) ( )( ) 222 mm/s 46203- sin(1.963) 50x 20- φωtsin20ω- a

mm/s 382.2- 1.963cosx 50x 20 φωtcos 20ω vφωt20sinx

==+=

==+=+=

The plots of x, v and a confirm these answers.


WORKED EXAMPLE No. 7 A spring of stiffness 20 kN/m supports a mass of 4 kg. The mass is pulled down 8 mm and

released to produce linear oscillations. Calculate the frequency and periodic time. Sketch the graphs of displacement, velocity and acceleration. Calculate the displacement, velocity and acceleration 0.05 s after being released.

SOLUTION

s 0.0899f1T

Hz 11.252πω

rad/s 70.714

20000Mkω

==

==

===

f

The oscillation starts at the bottom of the cycle so xo = -8 mm. The resulting graph of x against time will be a negative cosine curve with an amplitude of 8 mm.

The equations describing the motion are as follows. x = xocosωt When t = 0.05 seconds x = -8 cos(70.71 x 0.05) x = 7.387 mm. (Note angles are in radian) This is confirmed by the graph. If we differentiate once we get the equation for velocity. v = -ωxosin ωt v = -ωxosin ωt = -70.71 (-8)sin(70.71 x 0.05) v = -217 mm/s This is confirmed by the graph. Differentiate again to get the acceleration. a = -ω2xocosωt and since x = xocosωt a = -ω2x a = -70.712 x 7.387 = -36 934 mm/s2

This is confirmed by the graph.


SELF ASSESSMENT EXERCISE No. 3 1. Calculate the frequency and periodic time for the oscillation produced by a mass – spring

system given that the mass is 0.5 kg and the spring stiffness is 3 N/mm. (12.3 Hz, 0.081 s). 2. A mass of 4 kg is suspended from a spring and oscillates up and down at 2 Hz. Determine the

stiffness of the spring. (631.6 N/m). The amplitude of the oscillation is 5 mm. Determine the displacement, velocity and

acceleration 0.02 s after the mass passes through the mean or rest position in an upwards direction. (1.243 mm, 60.86 mm/s and -196.4 mm/s2)

3. From recordings made of a simple harmonic motion, it is found that at a certain point in the

motion the velocity is 0.3 m/s and the displacement is 20 mm, both being positive downwards in direction. Determine the amplitude of the motion and the maximum velocity and acceleration. Write down the equations of motion.

Note that the data given is at time t = 0. You will have to assume that x = xocos(ωt + φ) at time t=0 Ans. x= 0.0311 cos(ωt - 50o) v = -0.3914 sin(ωt - 50o) a = -157.9 x


ELECTRICAL EXAMPLES ALTERNATING ELECTRICITY Electricity is generated by rotating a conductor relative to a magnetic field at angular velocity ω rad/s (very simplified case shown). The voltage generated is directly proportional to the angle of rotation.

This explains why the voltage in our mains electrical system is sinusoidal. The voltage at any moment in time is given by the equation v = V sin(ωt) where V is the maximum voltage (amplitude) in the cycle and ω the angular velocity or frequency. If we choose to measure the angle from a different starting point then v = V sin(ωt + φ) where φ is the starting angle. RESISTANCE When a sinusoidal voltage is applied across a resistor the current is sinusoidal and in phase with the voltage.

( )Rωtsin Vi φ+

=

CAPACITANCE When a sinusoidal voltage is applied across a capacitor C the current is given by:

( )φ+ωtcos == ωCVdtdvCi

INDUCTANCE When a sinusoidal voltage is applied across an inductor L the current is

given by: ( )ωLωtcos Vi φ+

−=


If we plot these we see that the current in the capacitor is displaced -90o from that in the resistor and the current in the inductor is displaced +90o. This is a similar relationship to that of the displacement, velocity and acceleration in a mechanical system.

SELF ASSESSMENT EXERCISE No. 4 1. Mains electricity has a frequency of 50 Hz. What is the periodic time and angular frequency?

(0.02 s and 314 rad/s) 2. An alternating current has a periodic time of 0.0025 s. What is the frequency? (400 Hz) 3. A alternating voltage has a peak to peak amplitude of 300 V and frequency of 50 Hz. What is

the amplitude? (150 V) What is the voltage at t = 0.02 s? (16.4 V) 4. Determine the following from the graph shown.


The amplitude. The offset displacement. The periodic time. The frequency. The angular frequency. The phase angle.

(Answers 5, 2, 1.57 s, 4 rad/s, 0.637 Hz, 0.2 radian or 11.5o) 5. A resistor of value 10 Ω, a capacitor of value 40 µF and an inductor of value 10 mH are all connected in parallel to a voltage source as shown. The voltage is 50 sin(2000t). Determine an expression for the current drawn from the source. Determine the peak current. Determine the phase of the source current.

( ) ( )22000tcos32000tsin 5i +=

(5.22 A, -0.29 radian)

6. COMPLEX WAVEFORMS

FUNDAMENTAL FREQUENCY

A sinusoidal voltage or current is described by the mathematical formula v = V sin ωt or i = I sin ωt

The sinusoidal voltage formula is then v = V sin(2πft) In this formula f is the fundamental frequency.

HARMONICS A harmonic is a multiple of the fundamental frequency.

2f is the second harmonic. 3f is the third harmonic nf is the nth harmonic.

SYNTHESISING COMPLEX WAVES Waveforms with shapes that are not sinusoidal may be synthesised from one common sinusoidal waveform. The proof of this is not given here but the following is mathematically correct. This graph shows the result of adding the first and third harmonic with equal amplitudes.

In reality the amplitude of the harmonic is likely to be less than the amplitude of the fundamental. This graph shows the affect of adding the third harmonic with 1/3 of the amplitude.

GENERATION OF HARMONICS Harmonics are generated when a sinusoidal signal passes through a non-linear amplifier. An ideal amplifier increases a sinusoidal signal perfectly.


SQUARE WAVES Square waveforms are really d.c. levels that suddenly change from plus to minus. It can be shown that the following formula relates voltage and time. The formula is an infinite series.

........ t)sin7(ω7V t)sin5(ω

5V t)sin3(ω

3V t)Vsin(ωv ++++=

TRIANGULAR WAVES It can be shown that the following formula relates voltage and time. The formula is an infinite series.

........π) tsin(7ω49V t)sin(5ω

25Vπ) tsin(3ω

9V t)Vsin(ωv +−++−+=

Note that in this series, a phase shift of π radians is added to each


outcome2t1

Documents