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Standard XII
Physics
Time Period Independent of Amplitude
Question
A simple harmonic motion is represented by
y
=
5
(
sin
3
π
t
+
√
3
cos
3
π
t
)
c
m
The amplitude and time period of the motion are :
10
c
m
,
2
3
s
10
c
m
,
3
2
s
5
c
m
,
2
3
s
5
c
m
,
3
2
s
A
5
c
m
,
2
3
s
B
10
c
m
,
2
3
s
C
10
c
m
,
3
2
s
D
5
c
m
,
3
2
s
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Solution
Verified by Toppr
y
=
5
(
sin
3
π
t
+
√
3
cos
3
π
t
)
y
=
5
×
2
(
1
2
sin
3
π
t
+
√
3
2
cos
3
π
t
)
y
=
10
(
cos
π
3
sin
3
π
t
−
sin
π
3
cos
3
π
t
)
y
=
10
sin
(
3
π
t
+
π
3
)
Comparing it with
y
=
A
sin
(
ω
t
+
ϕ
)
⇒
A
=
10
i.e., amplitude is
10
cm
⇒
T
=
2
π
ω
=
2
π
3
π
=
2
3
sec. i.e, time period is
2
3
sec.
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Similar Questions
Q1
A simple harmonic motion is represented by
y
=
5
(
sin
3
π
t
+
√
3
cos
3
π
t
)
c
m
The amplitude and time period of the motion are :
View Solution
Q2
A simple motion is represented by:
y
=
5
(
sin
3
π
t
+
√
3
cos
3
π
t
)
c
m
The amplitude and time period of the motion are:
View Solution
Q3
A simple harmonic motion is represented by:
y
=
5
(
sin
3
π
t
+
√
3
cos
3
π
t
)
cm
The amplitude and time period of the motion are:
View Solution
Q4
A simple harmonic motion is given by the equation
y
=
5
(
sin
3
π
t
+
√
3
cos
3
π
t
)
. What is the amplitude of motion if y is in m ?
View Solution
Q5
A particle executing simple harmonic motion has an angular frequency of
6.28
s
−
1
and amplitude of
10
c
m
. The time period of the SHM is :
View Solution