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Obtain the expression for the period of a simple pendulum performing S.H.M.

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Question

Obtain the expression for the period of a simple pendulum performing S.H.M.

Derivation
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Solution

Let m is Mass of the bob.

L is the Length of the massless string.

A free-body diagram of the forces acting on the bob.

θ is the angle made by the string with the vertical.

T is tension along the string.

g is the acceleration due to gravity.

∴ Restoring force, F = −mg sin θ   ...(i)

As θ is very small (θ < 10°), we can write 

sin θ ≅ θ° ∴ F ≅ mgθ

From the figure, the small angle `theta = x/L`

∴ `F = -mgx/L`   ...(ii)

As m, g and L are constants, F ∝ −x

Thus, for small displacements, the restoring force is directly proportional to the displacement and is oppositely directed.

Hence, the bob of a simple pendulum performs linear S.H.M. for small amplitudes. The period T of oscillation of a pendulum from can be given as,

= `(2pi)/omega`

= `(2pi)/sqrt("acceleration per unit displacement")`

Using eq. (ii), `F = -mgx/L`

∴ `a = -gx/L`

∴ `a/x = -g/L = g/L`   ...(in magnitude)

Substituting in the expression for T, we get,

`T = 2pi sqrt((L)/(g))`   ...(iii)

The equation (iii) gives the expression for the time period of a simple pendulum.

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Chapter 5: Oscillations - Exercises [Page 129]

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Balbharati Physics [English] Standard 12 Maharashtra State Board
Chapter 5 Oscillations
Exercises | Q 2.3 | Page 129

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