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JEE Main Physics Class 11 Oscillations Part 1 Questions
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© examsnet.com
Question : 9
Total: 100
The function of time representing a simple harmonic motion with a period of
π
ω
is
[18 Mar 2021 Shift 2]
sin
(
ω
t
)
=
cos
(
ω
t
)
cos
(
ω
t
)
+
cos
(
2
ω
t
)
+
cos
(
3
ω
t
)
sin
2
(
ω
t
)
3
cos
(
π
4
−
2
ω
t
)
Validate
Solution:
Time period,
T
=
2
π
ω
′
Given, time period,
T
=
π
ω
2
π
ω
′
=
π
ω
⇒
ω
′
=
2
ω
Options (a) and (b) are incorrect.
Option (c),
sin
2
ω
t
=
1
2
(
2
sin
2
ω
t
)
=
1
2
(
1
−
cos
2
ω
t
)
Hence, the angular frequency is
2
ω
.
Option (d),
3
cos
(
π
4
−
2
ω
t
)
Angular frequency of
S
H
M
is
2
ω
.
Option (d) is the correct answer.
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