144 Higher Engineering Mathematics
α, and a graph of y = sin(ωt + α) leads y = sin ωt by
angle α.
The angle ωt is measured in radians (i.e.
ω
rad
s
(t s) = ωt radians) hence angle α should also
be in radians.
The relationship between degrees and radians is:
360
◦
= 2π radians or 180
◦
= π radians
Hence 1 rad =
180
π
= 57.30 ◦ and, for example,
71 ◦ = 71 ×
π
180
= 1.239 rad.
Given a general sinusoidal function
y = A sin(ω t ± α), then
(i) A = amplitude
(ii) ω = angular velocity = 2π f rad/s
(iii)
2π
ω
= periodic time Tseconds
(iv)
ω
2π
= frequency, f hertz
(v) α = angle of lead or lag (compared with
y = A sin ωt )
Problem 14. An alternating current is given by
i = 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, hence amplitude =30 A
Angular velocity ω = 100π, hence
periodic time, T =
2π
ω
=
2π
100π
=
1
50
= 0.02 s or 20 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 time t = 0 the displacement is 90 cm.
Express the displacement in the general form
A sin(ωt ± α).
Amplitude= maximum displacement = 2.5 m.
Angular velocity, ω = 2π f = 2π(60) = 120π rad/s.
Hence displacement = 2.5 sin(120πt + α) m.
When t = 0, displacement = 90 cm = 0.90 m.
Hence
0.90 = 2. sin(0 + α)
i.e.
sin α =
0.90
2.5
= 0.36
Hence
α = arcsin 0.36 = 21.10
◦
= 21
◦ 6
= 0.368 rad
Thus displacement = 2.5 sin(120πt + 0.368) m
Problem 16. The instantaneous value of voltage
in an a.c. circuit at any time t seconds is given by
v = 340 sin(50πt − 0.541) volts. Determine:
(a) the amplitude, periodic time, frequency and
phase angle (in degrees)
(b) the value of the voltage when t = 0
(c) the value of the voltage when t = 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 = 50π
Hence periodic time, T =
2π
ω
=
2π
50π
=
1
25
= 0.04 s or 40 ms
Frequency, f =
1
T
=
1
0.04
= 25 Hz
Phase angle = 0.541rad =
0.541 ×
180
π
= 31
◦ lagging v = 340 sin(50πt )
(b) When t = 0,
v = 340 sin(0 − 0.541) = 340 sin(−31
◦
)
= −175.1 V
α, and a graph of y = sin(ωt + α) leads y = sin ωt by
angle α.
The angle ωt is measured in radians (i.e.
ω
rad
s
(t s) = ωt radians) hence angle α should also
be in radians.
The relationship between degrees and radians is:
360
◦
= 2π radians or 180
◦
= π radians
Hence 1 rad =
180
π
= 57.30 ◦ and, for example,
71 ◦ = 71 ×
π
180
= 1.239 rad.
Given a general sinusoidal function
y = A sin(ω t ± α), then
(i) A = amplitude
(ii) ω = angular velocity = 2π f rad/s
(iii)
2π
ω
= periodic time Tseconds
(iv)
ω
2π
= frequency, f hertz
(v) α = angle of lead or lag (compared with
y = A sin ωt )
Problem 14. An alternating current is given by
i = 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, hence amplitude =30 A
Angular velocity ω = 100π, hence
periodic time, T =
2π
ω
=
2π
100π
=
1
50
= 0.02 s or 20 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 time t = 0 the displacement is 90 cm.
Express the displacement in the general form
A sin(ωt ± α).
Amplitude= maximum displacement = 2.5 m.
Angular velocity, ω = 2π f = 2π(60) = 120π rad/s.
Hence displacement = 2.5 sin(120πt + α) m.
When t = 0, displacement = 90 cm = 0.90 m.
Hence
0.90 = 2. sin(0 + α)
i.e.
sin α =
0.90
2.5
= 0.36
Hence
α = arcsin 0.36 = 21.10
◦
= 21
◦ 6
= 0.368 rad
Thus displacement = 2.5 sin(120πt + 0.368) m
Problem 16. The instantaneous value of voltage
in an a.c. circuit at any time t seconds is given by
v = 340 sin(50πt − 0.541) volts. Determine:
(a) the amplitude, periodic time, frequency and
phase angle (in degrees)
(b) the value of the voltage when t = 0
(c) the value of the voltage when t = 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 = 50π
Hence periodic time, T =
2π
ω
=
2π
50π
=
1
25
= 0.04 s or 40 ms
Frequency, f =
1
T
=
1
0.04
= 25 Hz
Phase angle = 0.541rad =
0.541 ×
180
π
= 31
◦ lagging v = 340 sin(50πt )
(b) When t = 0,
v = 340 sin(0 − 0.541) = 340 sin(−31
◦
)
= −175.1 V