A particle performs simple harmonic motion satisfying $\ddot{x} = -10gx$ . Taking $g = 9.81\,\mathrm{ms^{-2}}$ , find the period of the motion in seconds, to 2 decimal places.

Model answer

0.63

Also accepted: 0.63 s, 0.634, 0.6344

Explanation

For

\ddot{x} = -\omega^2 x

the period is

T = \dfrac{2\pi}{\omega}

. Here

\omega^2 = 10g = 98.1

, so

\omega = 9.9045

and

T = \dfrac{2\pi}{9.9045} = 0.6344

, that is

0.63

s. Taking

g = 9.8

would give

0.6347

s and

g = 10

would give

0.6283

s, so the value of

g

shows at the third decimal place.

Derived from ZIMSEC Additional Mathematics Paper 2, Specimen Paper, Q4

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