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Question
Show that, under certain conditions, simple pendulum performs the linear simple harmonic motion.
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Solution
Practical simple pendulum – In practice a small but heavy sphere can be regarded as point mass and a light string whose weight is negligible compared with weight of the bob can be taken as a weightless fibre.
Suppose that a simple pendulum of length ‘L’ is displaced through a small angle θ and released. It oscillates two sides of its equilibrium position. At displaced position, force acting on the bob are (1) its weight mg (2) the tension T in the string. Resolved ‘mg’ into two components ‘mg sinθ’ to ⊥ the string and ‘mg cos θ’ parallel to
the string. The component ‘mg cos’ is balanced by the tension in the string. The
component ‘mg sinθ’ is unbalanced. This acts as restoring force.
F = - mg sinθ -ve sign indicates that force is opposite.
But θ is very small , sinθ = θ
`F=-mg theta and theta =X/L`
`therefore F=-(mgX)/L`
`F=-((mg)/L)X`
But `F=ma_"cc"`
`therefore ma_"cc"=-((mg)/L)X`
`a_"cc" =-(g/L)X`
`a_"cc" alpha (-X)`
The motion of simple pendulum is linear S.H.M.
`a_" cc" =-(g/L)X`
Condition for simple pendulum: (1) Bob must be small but heavy sphere.
(2) It must be suspended by light string.
(3) It must be supported by rigid support.
(4) Amplitude must be very small.
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