trigonometry

158 GEOMETRY AND TRIGONOMETRY
by angle 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 )
(ts) = ωt radians) hence angle α should
s
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 = 1.239 rad.
180
Given a general sinusoidal function
y = A sin(ω t ± α), then
(i) A = amplitude
(ii) ω = angular velocity = 2πf rad/s
(iii) 2π = periodic time T seconds
ω
ω
(iv) = frequency, f hertz
2π
(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 = 50 Hz
0.02 (
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 = 2.5m.
Angular velocity, ω = 2πf = 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 = 2.5 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
200V, 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 = 25 Hz
0.04
(
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