From the expression for the time period of oscilations of a pendulum the following laws are enunciated.
( i) The law of length
(ii)The law of acceleration
(iii) The law of mass
(iv)The law of amplitude

**Laws of pendulum **

From the
expression for the time period of oscilations of a pendulum the following laws
are enunciated.

**(
i) The law of length **

The period of a
simple pendulum varies directly as the square root of the length of the
pendulum. (i.e) T α l

**(ii)The law of acceleration **

The period of a
simple pendulum varies inversely as the square root of the acceleration due to
gravity.

(i.e) T α 1 g

**(iii) The law of mass**

The time period of a simple pendulum is
independent of the mass and material of the bob.

**(iv)The law of
amplitude**

The period of a
simple pendulum is independent of the amplitude provided the amplitude is
small.

Note : The
length of a seconds pendulum is 0.99 m whose period is two seconds.

2 = 2π.rt(l/g)

l= 9.81x 4 / 4 π^{2}=0.99 m

Oscillations of simple
pendulum can also be regarded as a case of angular SHM.

Let θ be the
angular displacement of the bob B at an instant of time. The bob makes rotation
about the horizontal line which is perpendicular to the plane of motion as
shown in Fig..

Restoring torque
about O is τ = − mg l sin θ

τ = −m g l θ [ ∵
θ is sm all] ...(1)

Moment of
inertia about the axis = m l ^{2} ...(2)

If the amplitude
is small, motion of the bob is angular simple harmonic. Therefore angular
acceleration of the system about the axis of rotation is

α = r/l = -mgl θ
/ ml^{2}

α = - (g/l)
θ ?.(3)

We know that α =
−ω ^{2} θ ??.(4)

Comparing (3)
and (4)

−ω ^{2}θ
= -(g/l) θ

angular
frequency ω = rt(g/l)

Time period T =
= 2π / ω = 2π .rt(l/g)

Frequency n = 1
/ 2 π . rt(g/l)

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