Chapter 26 : Helical Springs and Pendulums „„„„„ 547

Now the force (P) acting at G will produce an acceleration (a) such that,

P = ma or

P

a

m

= ...(i)

where m is the mass of the body.

Similarly, the turning moment (P × l) will produce an angular acceleration (α) such that,

G

Pl

I

×

α=

where IG is the moment of inertia of the body about the axis parallel to the axis of rotation and
passing through G. Now the corresponding linear acceleration of O,

1

G

Plh

ah

I

=α = ...

G

Pl

I

⎛⎞

⎜⎟α=

⎝⎠

Q

(^12)
G
Plh
a
mk
= ...(ii)
Now, if the axis through O is not to have any impulse, then a should be equal to a 1. Therefore
equating (i) and (ii),
2
G
PPlh
m mk

∴ kG^2 = l h ...(iii)
We know that I 0 = IG + m h^2
∴ m k 02 = m kG^2 + m h^2
or k 02 = kG^2 + h^2
∴ kG^2 = k 02 – h^2 ...(iv)
Equating equations (iii) and (iv),
k 02 – h^2 = l h
∴ k 02 = l h + h^2 = h (l + h) = OG × OC
or
2
OG k^0
OC

It is thus obvious, that centres of suspension (O) and oscillation (C) are interchangeable. The
distance between these two centres (i.e., OC) known as the length of equivalent simple pendulum.
∴
222
LhlhkhkGG
hh

• =+=+ =
  Now time period in terms of the length of equivalent simple pendulum (L),
  2 2
  22 G
  L hk
  t
  ggh


• 
  

  =π =π