qcf level: 4 credit value: 15 outcome 2 - · pdf file©d.j.dunn 2 1. radian in engineering...

© D.J.Dunn www.freestudy.co.uk 1

UNIT 1: ANALYTICAL METHODS FOR ENGINEERS

Unit code: A/601/1401

QCF Level: 4

Credit value: 15

OUTCOME 2 - TRIGONOMETRIC METHODS

TUTORIAL 1 –SINUSOIDAL FUNCTION

2 Be able to analyse and model engineering situations and solve problems using trigonometric

methods

Sinusoidal functions: review of the trigonometric ratios; Cartesian and polar co-ordinate

systems; properties of the circle; radian measure; sinusoidal functions

Applications: 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.

©D.J.Dunn www.freestudy.co.uk 2

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 2πor 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

The two systems are clearly linked as we can convert from one to the

other using trigonometry and Pythagoras’ theorem.

y = R sin x = R cos

y/x = tan θ

R = (x2 + y

2)

½

WORKED EXAMPLE No. 1

The x, y coordinates of a point is 4, and 6. Calculate the polar coordinates.

SOLUTION

R = (42 + 6

2)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 74o)

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?

(31.623 108.4o)

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 AB

CB

Hypotenuse

Opposite 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

The ratio AB

AC

Hypotenuse

Adjacent is called the cosine of the angle A.

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 AC

BC

Adjacent

Opposite 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

b

sinA

a

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

c

sinB

b

sinA

a


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

SOLUTION

a = 50 mm A = 30o B = 45

o C = 180

o - 30

o - 45

o = 105

o

sinC

c

sinB

b

sinA

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


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 70

o F1 = 205.2 N

F2/sin 70o = 300/sin 70

o F1 = 300 N


COSINE RULE

Consider the diagram. Using Pythagoras we have:

h2 = a

2 – (b – x)

2 and h

2 = c

2 – x

2

a2 – (b

2 + x

2 – 2bx) = c

2 – x

2

a2 – b

2 - x

2 + 2bx = c

2 – x

2

a2 = b

2 + x

2 – 2bx + c

2 – x

2

a2 = b

2 + c

2– 2bx substitute x = c cos(A) a

2 = b

2 + c

2– 2bc cos(A)

2bc

acbcos(A)

222

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

2ca

baccos(B)

222

2ab

cbacos(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.


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(60

222o

222o c07060))(2)(60)(7cos(60

2c85002004

c2 = 4300 c = 65.574 mm

61.0)574.65)(70)(2(

60574.6570

2bc

acbcos(A)

222222

A = 52.4o

B = 180 – 52.4 – 60 = 126.5 o


Find the resultant of the two forces shown.

SOLUTION

The addition of the two force is done as shown.

0)(2)(100(10

R010010)cos(135

222o

222o R010010100))(2)(100)(cos(135

-14142.1 = 20000 – R2

R2 = 34142.13 R = 184.78 N



1. Find the resultant of the two forces shown. (Answer 328.8 N)

2. Vector A has polar coordinates 12 60

o 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 49

o and 59

o)

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, 3.59 T and 2.93 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(t)

Velocity v = dx/dt = R cos(t)

Acceleration a = dv/dt = -2R sin(t) = -2 x

Anything that obeys these equations is said to have SIMPLE HARMONIC MOTION

The starting point of the oscillation

could be at any angle so in that

case the equations become:

Displacement

x = R sin(t + )

Velocity

v = dx/dt = R cos(t + )

Acceleration

a = dv/dt = -2R sin(t + ) = -2 x

The plots show the displacement,

velocity and acceleration for = 0 on

the left and a negative 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 m

k

2π

1f

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) = mω

2R sin(ωt)



The displacement of a body performing simple harmonic motion is described by the following

equation

x = A sin (t + ) where A is the amplitude, is the natural frequency and is the phase angle.

Given A = 20 mm, = 50 rad/s and = /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 = /2 = 50/2 = 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 18.4820sin1.9638

π1.5720sinx

8

π0.314)x (50sin 20x

Next write down the equations for v and a

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

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

φωt20sinx

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



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 0.0899f

1T

Hz 11.252π

ωf

rad/s 70.714

20000

M

kω

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 = xocost

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 = -xosin t

v = -xosin t = -70.71 (-8)sin(70.71 x 0.05)

v = -217 mm/s


Differentiate again to get the acceleration.

a = -2xocost and since x = xocost a = -

2x

a = -70.712 x 7.387 = -36 934 mm/s

2




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 the frequency is 2 Hz and

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(t + ) at time t=0

Ans. x= 0.0311 cos(4t - 50o)

v = -0.3914 sin(4t - 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 + ) where is

the starting angle.

RESISTANCE

When a sinusoidal voltage is applied across a resistor

the current is sinusoidal and in phase with the voltage.

R

ωtsin Vi

CAPACITANCE

When a sinusoidal voltage is applied across a capacitor C the current

is given by:

ωtcos ωCVdt

dvCi

INDUCTANCE

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

given by: ωL

ωtcos Vi

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.



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 = 0.0025 s? (106 V)

4. Determine the following from the graph shown.

The amplitude.

The offset displacement.

The periodic time.

The frequency.

The angular frequency.

(Answers 5, 2, 1.57 s, 0.637 Hz, 4 rad/s)

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.

2

2000tcos32000tsin 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 t or i = I sin t

The sinusoidal voltage formula is then v = V sin(2ft) 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(ω7

V t)sin5(ω

5

V t)sin3(ω

3

V 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(7ω49

V t)sin(5ω

25

Vπ) tsin(3ω

9

V t)Vsin(ωv

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

qcf level: 4 credit value: 15 outcome 2 - · pdf file©d.j.dunn 2 1. radian in engineering...

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