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A body of mass 200 g is in equibrium at ...

A body of mass `200 g` is in equibrium at `x = 0` under the influence of a force `F(x) = (- 100x + 10x^(2))N`.
(a) If the body is displacement a small distance from equilibrium, what is the period of its oscillations ?
(b) If the amplitude is `4.0 cm`, by how much do we eror in assuming that `F(x) = - kx` at the end points of the motion.

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Verified by Experts

The correct Answer is:
A, B, D

(a) For small displacement `x`, the term `x^(2)` can be neglected.
`:. F = - 100x`
Comparing with `F = - kx` we have,
`k = 100 N//m`
`T = 2pi sqrt ((m)/(k))`
` = 2pi sqrt ((0.2)/(100)) = 0.28 s`
(b) `|Delta F| = 10x^(2) = 10(0.04)^(2) = 0.016 N`
`|F| = 100 x = 100 (0.04) = 4 N`
`:. %` error `= (|Delta F|)/(|F|) xx 100 = 0.4%` .
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