202 Basic Engineering Mathematics
Now try the following Practice Exercise
Practice Exercise 88 Trigonometric
waveforms (answers on page 350)
1. A sine wave is given by y = 5 sin3x. State its
peak value.
2. A sine wave is given by y = 4 sin2x. State its
period in degrees.
3. A periodic function is given by y = 30 cos5x.
State its maximum value.
4. A periodic function is given by y = 25 cos3x.
State its period in degrees.
In problems 5 to 11, state the amplitude and period
of the waveform and sketch the curve between 0 ◦
and 360 ◦ .
5. y = cos 3 A
6. y = 2 sin
5x
2
7. y = 3 sin4t
8. y = 5 cos
θ
2
9. y =
7
2
sin
3x
8
10. y = 6 sin(t − 45 ◦ )
11. y = 4 cos(2θ + 30 ◦ )
12. The frequency of a sine wave is 200 Hz.
Calculate the periodic time.
13. Calculate the frequency of a sine wave that
has a periodic time of 25 ms.
14. Calculate the periodic time for a sine wave
having a frequency of 10 kHz.
15. An alternating current completes 15 cycles in
24 ms. Determine its frequency.
16. Graphs of y 1 = 2 sin x and
y 2 = 3 sin(x + 50 ◦ ) are drawn on the same
axes. Is y 2 lagging or leading y 1 ?
17. Graphs of y 1 = 6 sin x and
y 2 = 5 sin(x − 70 ◦ ) are drawn on the same
axes. Is y 1 lagging or leading y 2 ?
22.5 Sinusoidal form: A sin(ωt ± α)
If a sine wave is expressed in the form
y = A sin(ωt ± α) then
(a) A = amplitude.
(b) ω = angular velocity = 2π f rad/s.
(c) frequency, f =
ω
2π
hertz.
(d) periodic time, T =
2π
ω
seconds
i.e. T =
1
f
.
(e) α = angle of lead or lag (compared with
y = A sin ωt ).
Here are some worked problems involving the sinusoidal form A sin(ωt ± α).
Problem 10. An alternating current is given by
i = 30 sin(100πt + 0.35) amperes. Find the
(a) amplitude, (b) frequency, (c) periodic time and
(d) phase angle (in degrees and minutes)
(a) i = 30 sin(100πt + 0.35)A; hence,
amplitude = 30 A.
(b) Angular velocity, ω = 100π, rad/s, hence
frequency, f =
ω
2π
=
100π
2π
= 50 Hz
(c) Periodic time, T =
1
f
=
1
50
= 0.02 s or 20 ms.
(d) 0.35 is the angle in radians. The relationship
between radians and degrees is
360
◦
= 2π radians or 180
◦
= πradians
from which,
1
◦
=
π
180
rad and 1rad =
180 ◦
π
(≈ 57.30
◦
)
Hence, phase angle, α = 0.35 rad
=
0.35 ×
180
π
◦
= 20.05 ◦ or 20 ◦ 3 leading
i = 30 sin(100πt).
Problem 11. 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.