Question

Derivation of time period of oscillation of compound pendulum

Answer 1

Simple pendulum consists of a point mass suspended by inextensible weightless string in a uniform gravitational field.

Simple pendulum can be set into oscillatory motion by pulling it to one side of equilibrium position and then releasing it.

In case of simple pendulum path ot the bob is an arc of a circle of radius l, where l is the length of the string.

We know that for SHM F=-kx and here x is the distance measured along the arc as shown in the figure below.


When bob of the simple pendulum is displaced from its equilibrium position O and is then released it begins to oscillate.

Suppose it is at P at any instant of time during oscillations and θ be the angle subtended by the string with the vertical.

mg is the force acting on the bob at point P in vertically downward direction.

Its component mgcosθ is balanced by the tension in the string and its tangential component mgsinθ directs in the direction opposite to increasing θ .

Thus restoring force is given by
     F=-mgsinθ               (19)

The restoring force is proportional to sinθ not to The restoring force is proportional to sinθ, so equation 19 does not represent SHM.

If the angle θ is small such that sinθ very narly equals θ then above equation 19 becomes
     F=-mgθ
since x=lθ then,
     F=-(mgx)/l
where x is the displacement OP along the arc. Thus,
     F=-(mg/l)x               (20)

From above equation 20 we see that restoring forcr is proportional to coordinate for small displacement x , and the constant (mg/l) is the force constant k.

Time period of a simple pendulum for small amplitudes is 


Corresponding frequency relations are

and angular frequency
     ω=√(g/l)          (23)

Notice that the period of oscillations is independent of the mass m of the pendulumand for small oscillations pperiod of pendulum for given value of g is entirely determined by its length.

Answer 2

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