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A-compound-pendulum-ocsillates-through-angles-about-its-equilibrium-position-such-that-8a-2-9g-cos-a-gt-0-its-period-is-A-2pi-8a-9g-B-3pi-8-a-g-C-2pi-9g-8a-D-8pi-




Question Number 76817 by Rio Michael last updated on 30/Dec/19
A compound pendulum ocsillates  through angles θ about its equilibrium  position such that   8aθ^2  = 9g cosθ, a>0. its period is   A. 2π(√((8a)/(9g)))  B. ((3π)/8)(√(a/g))  C. 2π(√((9g)/(8a)))  D. ((8π)/3)(√(a/g))
$$\mathrm{A}\:\mathrm{compound}\:\mathrm{pendulum}\:\mathrm{ocsillates} \\ $$$$\mathrm{through}\:\mathrm{angles}\:\theta\:\mathrm{about}\:\mathrm{its}\:\mathrm{equilibrium} \\ $$$$\mathrm{position}\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\mathrm{8}{a}\theta^{\mathrm{2}} \:=\:\mathrm{9}{g}\:{cos}\theta,\:{a}>\mathrm{0}.\:\mathrm{its}\:\mathrm{period}\:\mathrm{is}\: \\ $$$$\mathrm{A}.\:\mathrm{2}\pi\sqrt{\frac{\mathrm{8}{a}}{\mathrm{9}{g}}} \\ $$$$\mathrm{B}.\:\frac{\mathrm{3}\pi}{\mathrm{8}}\sqrt{\frac{{a}}{{g}}} \\ $$$$\mathrm{C}.\:\mathrm{2}\pi\sqrt{\frac{\mathrm{9}{g}}{\mathrm{8}{a}}} \\ $$$$\mathrm{D}.\:\frac{\mathrm{8}\pi}{\mathrm{3}}\sqrt{\frac{{a}}{{g}}} \\ $$
Commented by mr W last updated on 31/Dec/19
8aθ^2  = 9g cosθ makes no sense.
$$\mathrm{8}{a}\theta^{\mathrm{2}} \:=\:\mathrm{9}{g}\:{cos}\theta\:{makes}\:{no}\:{sense}. \\ $$
Commented by Rio Michael last updated on 31/Dec/19
really i was damn confuse when i saw that
$${really}\:{i}\:{was}\:{damn}\:{confuse}\:{when}\:{i}\:{saw}\:{that} \\ $$

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