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Question
Obtain an expression for a time period of the simple pendulum.
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Solution
Expression for a time period of the simple pendulum:
- Let ‘m’ be the mass of the bob and T′ be the tension in the string. The pendulum remains in equilibrium in the position OA, with the center of gravity of the bob, vertically below the point of suspension O.
- If now the pendulum is displaced through a small angle θ, called angular amplitude, and released, it begins to oscillate on either side of the mean (equilibrium) position in a single vertical plane.
Simple pendulum - In the displaced position (extreme position), two forces are acting on the bob.
a. Force T′ due to tension in the string, directed along the string, towards the support.
b. Weight mg, in the vertically downward direction. - At the extreme positions, there should not be any net force along the string.
- The component of mg can only balance the force due to tension. Thus, weight mg is resolved into two components;
a. The component mg cosθ along the string, which is balanced by the tension T′.
b. The component mg sinθ perpendicular to the string is the restoring force acting on mass m tending to return it to the equilibrium position.
∴ Restoring force, F = –mg sinθ - As θ is very small (θ <10°),
sinθ ≈ θc
∴ F ≈ –mgθ
From the figure,
For small angle, θ = `"x"/"L"`
∴ F = -mg`"x"/"L"` ........(1)
As m, g and L are constant, 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 is given by,
T = `(2pi)/ω = (2pi)/sqrt("acceleration per unit displacement")`
Using equation (1),
F = -mg`"x"/"L"`
∴ ma = -mg`"x"/"L"` .............(∵ F = ma)
∴ a = -g`"x"/"L"`
∴ `"a"/"x" = -"g"/"L" = "g"/"L"` (in magnitude)
Substituting in the expression for T,
T = `2pisqrt("L"/"g")`
This gives the expression for the time period of a simple pendulum.
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