268
6 Oscillations
d 2 θ
dt 2 +
Mg D
I
θ = 0
( 3 )
d 2 θ
dt 2 +
3
2
g
L
θ = 0
ω
2
=
3
2
g
L
T =
2π
ω
= 2π
2L
3g
(4)
For a simple pendulum
T = 2π
l
g
(5)
Comparing (4) and (5), the equivalent length of a simple pendulum is l =
2
3
L.
6.33 From the results of prob. (6.31) the time period of a physical pendulum is
given by
T = 2π
I
Mg D
(1)
where I is the moment of inertia about the pivot A, Fig. 6.9.
Now I = I C + M D
2 and I C = Mk
2
(2)
where k is the radius of gyration. Formula (1) then becomes
T = 2π
k 2 + D 2
g D
(3)
and the length of the simple equivalent pendulum is D +
k 2
D
.
If a point B be taken on AG such that AB = D +
k 2
D
, A and B are known as
the centres of suspension and oscillation, respectively. Here G is the centre of
mass (CM) of the physical pendulum.
Suppose now the body is suspended at B, then the time of oscillation is
obtained by substituting
k 2
D
for D in the expression