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Question
Obtain the expression for the period of a simple pendulum performing S.H.M.
Solution
Let m is Mass of the bob.
L is the Length of mass-less string.
A free-body diagram to the forces acting on the bob,
θ − 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 displacement, 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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