Higher Engineering Mathematics

158 GEOMETRY AND TRIGONOMETRY

by angleα, and a graph ofy=sin(ωt+α)leads

y=sinωtby angleα.

(The angle ωt is measured in radians (i.e.

ω

rad

s

)

(ts)=ωtradians) hence angleαshould

also be in radians.

The relationship between degrees and radians is:

360 ◦= 2 πradians or 180 ◦=πradians

Hence 1 rad=

180

π

= 57. 30 ◦ and, for example,

71 ◦= 71 ×

π

180

= 1 .239 rad.

Given a general sinusoidal function

y=Asin(ωt±α), then

(i)A=amplitude

(ii)ω=angular velocity= 2 πfrad/s

(iii)

2 π

ω

=periodic timeTseconds

(iv)

ω

2 π

=frequency,fhertz

(v)α=angle of lead or lag (compared with

y=Asinωt)

Problem 14. An alternating current is given

byi=30 sin(100πt+ 0 .27) amperes. Find the

amplitude, periodic time, frequency and phase

angle (in degrees and minutes).

i=30 sin(100πt+ 0 .27) A, henceamplitude=30 A

Angular velocityω= 100 π, hence

periodic time,T=

2 π

ω

=

2 π

100 π

=

1

50

=0.02 sor20 ms

Frequency,f=

1

T

=

1

0. 02

=50 Hz

Phase angle,α= 0 .27 rad=

(

0. 27 ×

180

π

)◦

=15.47◦or 15 ◦ 28 ′leading

i=30 sin(100πt)

Problem 15. An oscillating mechanism has

a maximum displacement of 2.5 m and a

frequency of 60 Hz. At timet=0 the displace-

ment is 90 cm. Express the displacement in the

general formAsin(ωt±α).

Amplitude=maximum displacement= 2 .5m.

Angular velocity,ω= 2 πf= 2 π(60)= 120 πrad/s.

Hence displacement= 2 .5 sin(120πt+α)m.

Whent=0, displacement=90 cm= 0 .90 m.

Hence 0. 90 = 2 .5 sin (0+α)

i.e. sinα=

0. 90

2. 5

= 0. 36

Hence α=arcsin 0. 36 = 21. 10 ◦= 21 ◦ 6 ′

= 0 .368 rad

Thusdisplacement=2.5 sin(120πt+0.368) m

Problem 16. The instantaneous value of volt-

age in an a.c. circuit at any timetseconds is given

byv=340 sin(50πt− 0 .541) volts. Determine:

(a) the amplitude, periodic time, frequency and

phase angle (in degrees)

(b) the value of the voltage whent= 0

(c) the value of the voltage whent=10 ms

(d) the time when the voltage first reaches

200 V, and

(e) the time when the voltage is a maximum.

Sketch one cycle of the waveform.

(a)Amplitude=340 V

Angular velocity,ω= 50 π

Henceperiodic time,T=

2 π

ω

=

2 π

50 π

=

1

25

=0.04 sor40 ms

Frequency,f=

1

T

=

1

0. 04

= 25 Hz

Phase angle= 0 .541rad=

(

0. 541 ×

180

π

)

= 31 ◦laggingv=340 sin (50πt)

(b)Whent= 0 ,

v=340 sin(0− 0 .541)=340 sin (− 31 ◦)

=−175.1 V