144 Higher Engineering Mathematics
α, and a graph ofy=sin(ωt+α)leadsy=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 .239rad.
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 by
i=30 sin( 100 πt+ 0. 27 )amperes. Find the
amplitude, periodic time, frequency and phase
angle (in degrees and minutes).
i=30sin( 100 πt+ 0. 27 )A, henceamplitude=30A
Angular velocityω= 100 π, hence
periodic time,T=
2 π
ω
=
2 π
100 π
=
1
50
=0.02sor20ms
Frequency, f=
1
T
=
1
0. 02
=50Hz
Phase angle,α= 0 .27rad=
(
0. 27 ×
180
π
)◦
=15.47◦or 15 ◦ 28 ′leading
i=30sin(100πt)
Problem 15. An oscillating mechanism has a
maximum displacement of 2.5m and a frequency of
60Hz. At timet=0 the displacement is 90cm.
Express the displacement in the general form
Asin(ωt±α).
Amplitude=maximum displacement= 2 .5m.
Angular velocity,ω= 2 πf= 2 π( 60 )= 120 πrad/s.
Hence displacement= 2 .5sin( 120 πt+α)m.
Whent=0, displacement=90cm= 0 .90m.
Hence 0. 90 = 2 .sin( 0 +α)
i.e. sinα=^0.^90
2. 5
= 0. 36
Hence α=arcsin0. 36 = 21. 10 ◦= 21 ◦ 6 ′
= 0 .368rad
Thusdisplacement=2.5 sin(120πt+0.368)m
Problem 16. The instantaneous value of voltage
in an a.c. circuit at any timetseconds is given by
v=340sin( 50 πt− 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=10ms
(d) the time when the voltage first reaches
200V, and
(e) the time when the voltage is a maximum.
Sketch one cycle of the waveform.
(a) Amplitude=340V
Angular velocity,ω= 50 π
Henceperiodic time,T=
2 π
ω
=
2 π
50 π
=
1
25
=0.04sor40ms
Frequency,f=
1
T
=
1
0. 04
= 25 Hz
Phase angle= 0 .541rad=
(
0. 541 ×
180
π
)
= 31 ◦laggingv=340sin( 50 πt)
(b) Whent= 0 ,
v=340sin( 0 − 0. 541 )=340sin(− 31 ◦)
=−175.1V