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Standard XII
Physics
Relative Acceleration
81. Why is th...
Question
81. Why is the time period of simple pendulum equal to 2π\sqrt{}(l/g)?
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Similar questions
Q.
The bob of mass
m
of a simple pendulum, is attached to a horizontal spring of spring constant
k
. The pendulum will undergo simple harmonic motion with period
T
equal to
Q.
The periodic time
(
T
)
of a simple pendulum of length
(
L
)
is given by
T
=
2
π
√
L
g
. What is the dimensional formula of
T
=
√
g
L
?
Q.
A simple pendulum of length
l
is made to oscillate with amplitude of
45
degrees. The acceleration due to gravity is
g
. Let
T
0
=
2
π
√
l
/
g
. The time period of oscillation of this pendulum will be.
Q.
Show that the expression of the time period
T
of a simple pendulum of length
l
given by
T
=
2
π
√
l
g
is dimensionally correct.
Q.
The time period of a simple pendulum is given by the equation
T
=
2
π
√
l
g
. The equation of motion of the pendulum will be
d
2
x
d
t
2
:
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