Trigonometric waveforms 203
hence,^0.^90 =^2 .5sin(^0 +α)
i.e. sinα=^0.^90
2. 5
= 0. 36Hence, α=sin−^10. 36 = 21. 10 ◦
= 21 ◦ 6 ′= 0 .368rad.Thus,displacement= 2 .5sin( 120 πt+ 0. 368 )m.
Problem 12. The instantaneous value of voltage
in an a.c. circuit at any timetseconds is given by
v=340sin( 50 πt− 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 whent=10 ms,
(d) time when the voltage first reaches 200V and
(e) time when the voltage is a maximum. Also,
(f ) sketch one cycle of the waveform(a) Amplitude=340V
Angular velocity,ω= 50 πFrequency,f=ω
2 π=50 π
2 π=25HzPeriodic time,T=1
f=1
25=0.04s or 40msPhase angle= 0 .541rad=(
0. 541 ×180
π)◦= 31 ◦laggingv=340sin( 50 πt)(b) Whent= 0 ,
v=340sin( 0 − 0. 541 )
=340sin(− 31 ◦)=− 175 .1V(c) Whent=10ms,v=340sin( 50 π× 10 × 10 −^3 − 0. 541 )=340sin( 1. 0298 )=340sin59◦=291.4 volts(d) Whenv=200 volts,200 =340sin( 50 πt− 0. 541 )200
340
=sin( 50 πt− 0. 541 )Hence, ( 50 πt− 0. 541 )=sin−^1200
340
= 36. 03 ◦or 0.628875rad
50 πt= 0. 628875 + 0. 541
= 1. 169875
Hence, whenv=200V,
time,t=1. 169875
50 π= 7 .448ms(e) When the voltage is a maximum,v=340V.Hence 340 =340sin( 50 πt− 0. 541 )1 =sin( 50 πt− 0. 541 )50 πt− 0. 541 =sin−^11 = 90 ◦or 1.5708 rad50 πt= 1. 5708 + 0. 541 = 2. 1118Hence,time,t=2. 1118
50 π= 13 .44ms(f ) A sketch ofv=340sin( 50 πt− 0. 541 )volts is
shown in Figure 22.23.v 5 340 sin(50t 2 0.541)
v 5 340 sin 50t(^0103040) t (ms)
7.448 13.44
2340
2 175.1
200
291.4
340
Voltage v
20
Figure 22.23
Now try the following Practice Exercise
PracticeExercise 89 Sinusoidal form
Asin(ωt±α)(answers on page 350)
In problems 1 to 3 find the (a) amplitude,
(b)frequency, (c) periodictime and (d)phase angle
(stating whether it is leading or lagging sinωt)of
the alternating quantities given.
- i=40sin( 50 πt+ 0. 29 )mA
- y=75sin( 40 t− 0. 54 )cm