A pendulum of mass m with a string of length L swings in a simple harmonic way. has a toughening period of 2 s using a simple harmonic swinging pendulum of mass 2m If you want the swing period to be 1 second, how many times the length of the rope is L?
Answers
Answer:
A simple pendulum consists of a ball (point-mass) m hanging from a (massless) string of length L and fixed at a pivot point P. When displaced to an initial angle and released, the pendulum will swing back and forth with periodic motion. By applying Newton's secont law for rotational systems, the equation of motion for the pendulum may be obtained
τ
=
I
α
⇒
−
m
g
sin
θ
L
=
m
L
2
d
2
θ
d
t
2
and rearranged as
d
2
θ
d
t
2
+
g
L
sin
θ
=
0
If the amplitude of angular displacement is small enough, so the small angle approximation ($\sin\theta\approx\theta$) holds true, then the equation of motion reduces to the equation of simple harmonic motion
d
2
θ
d
t
2
+
g
L
θ
=
0
The simple harmonic solution is
θ
(
t
)
=
θ
o
cos
(
ω
t
)
,
where
θ
o
is the initial angular displacement, and
ω
=
√
g
/
L
the natural frequency of the motion. The period of this sytem (time for one oscillation) is
T
=
2
π
ω
=
2
π
√
L
g
.