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Show That, Under Certain Conditions, Simple Pendulum Performs the Linear Simple Harmonic Motion.

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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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2012-2013 (March)

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