Trigonometric waveforms 203
hence,
0.90 = 2.5 sin(0 + α)
i.e.
sin α =
0.90
2.5
= 0.36
Hence,
α = sin
−1 0.36 = 21.10
◦
= 21
◦ 6
= 0.368 rad.
Thus, displacement = 2.5 sin(120πt + 0.368) m.
Problem 12. 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 the
(a) amplitude, frequency, periodic time and phase
angle (in degrees), (b) value of the voltage when
t = 0, (c) value of the voltage when t = 10 ms,
(d) time when the voltage first reaches 200 V and
(e) time when the voltage is a maximum. Also,
(f ) sketch one cycle of the waveform
(a) Amplitude = 340 V
Angular velocity, ω = 50π
Frequency, f =
ω
2π
=
50π
2π
= 25 Hz
Periodic time, T =
1
f
=
1
25
= 0.04 s or 40 ms
Phase angle = 0.541 rad =
0.541 ×
180
π
◦
= 31
◦ lagging v = 340 sin(50πt)
(b) When t = 0,
v = 340 sin(0 − 0.541)
= 340 sin(−31
◦
) = −175.1V
(c) When t = 10 ms,
v = 340 sin(50π × 10 × 10 −3 − 0.541)
= 340 sin(1.0298)
= 340 sin 59 ◦ = 291.4 volts
(d) When v = 200 volts,
200 = 340 sin(50πt − 0.541)
200
340
= sin(50πt − 0.541)
Hence,
(50πt − 0.541) = sin
−1 200
340
= 36.03 ◦ or 0.628875 rad
50πt = 0.628875 + 0.541
= 1.169875
Hence, when v = 200 V,
time, t =
1.169875
50 π
= 7.448 ms
(e) When the voltage is a maximum, v = 340 V.
Hence
340 = 340 sin(50πt − 0.541)
1 = sin(50πt − 0.541)
50πt − 0.541 = sin
−1 1 = 90
◦ or 1.5708 rad
50πt = 1.5708 + 0.541 = 2.1118
Hence, time, t =
2.1118
50 π
= 13.44 ms
(f ) A sketch of v = 340 sin(50πt − 0.541) volts is
shown in Figure 22.23.
v 5340 sin(50 t 2 0.541)
v 5340 sin 50 t
0
t (ms)
10
30
40
7.448 13.44
2340
2175.1
200
291.4
340
Voltage v
20
Figure 22.23
Now try the following Practice Exercise
Practice Exercise 89 Sinusoidal form
A sin(ωt ± α) (answers on page 350)
In problems 1 to 3 find the (a) amplitude,
(b) frequency, (c) periodic time and (d) phase angle
(stating whether it is leading or lagging sin ωt ) of
the alternating quantities given.
1. i = 40 sin(50πt + 0.29) mA
2. y = 75 sin(40t − 0.54) cm
hence,
0.90 = 2.5 sin(0 + α)
i.e.
sin α =
0.90
2.5
= 0.36
Hence,
α = sin
−1 0.36 = 21.10
◦
= 21
◦ 6
= 0.368 rad.
Thus, displacement = 2.5 sin(120πt + 0.368) m.
Problem 12. 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 the
(a) amplitude, frequency, periodic time and phase
angle (in degrees), (b) value of the voltage when
t = 0, (c) value of the voltage when t = 10 ms,
(d) time when the voltage first reaches 200 V and
(e) time when the voltage is a maximum. Also,
(f ) sketch one cycle of the waveform
(a) Amplitude = 340 V
Angular velocity, ω = 50π
Frequency, f =
ω
2π
=
50π
2π
= 25 Hz
Periodic time, T =
1
f
=
1
25
= 0.04 s or 40 ms
Phase angle = 0.541 rad =
0.541 ×
180
π
◦
= 31
◦ lagging v = 340 sin(50πt)
(b) When t = 0,
v = 340 sin(0 − 0.541)
= 340 sin(−31
◦
) = −175.1V
(c) When t = 10 ms,
v = 340 sin(50π × 10 × 10 −3 − 0.541)
= 340 sin(1.0298)
= 340 sin 59 ◦ = 291.4 volts
(d) When v = 200 volts,
200 = 340 sin(50πt − 0.541)
200
340
= sin(50πt − 0.541)
Hence,
(50πt − 0.541) = sin
−1 200
340
= 36.03 ◦ or 0.628875 rad
50πt = 0.628875 + 0.541
= 1.169875
Hence, when v = 200 V,
time, t =
1.169875
50 π
= 7.448 ms
(e) When the voltage is a maximum, v = 340 V.
Hence
340 = 340 sin(50πt − 0.541)
1 = sin(50πt − 0.541)
50πt − 0.541 = sin
−1 1 = 90
◦ or 1.5708 rad
50πt = 1.5708 + 0.541 = 2.1118
Hence, time, t =
2.1118
50 π
= 13.44 ms
(f ) A sketch of v = 340 sin(50πt − 0.541) volts is
shown in Figure 22.23.
v 5340 sin(50 t 2 0.541)
v 5340 sin 50 t
0
t (ms)
10
30
40
7.448 13.44
2340
2175.1
200
291.4
340
Voltage v
20
Figure 22.23
Now try the following Practice Exercise
Practice Exercise 89 Sinusoidal form
A sin(ωt ± α) (answers on page 350)
In problems 1 to 3 find the (a) amplitude,
(b) frequency, (c) periodic time and (d) phase angle
(stating whether it is leading or lagging sin ωt ) of
the alternating quantities given.
1. i = 40 sin(50πt + 0.29) mA
2. y = 75 sin(40t − 0.54) cm