A compound pendulum oscillates though a small angle θ about its equilibrium position such that 10a((dθ/dt))^2 = 4g cos θ , a >0 . its period is [A] 2π(√(((5a)/(4g)) )) [B] ((2π)/5)(√(a/g)) [C] 2π(√(((2g)/(5a)) )) [D] 2π(√((5a)/g))


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Question Number 84231 by Rio Michael last updated on 10/Mar/20

$A compound pendulum oscillates though a$ $small angle θ about its equilibrium position$ $such that$ $10 a {(\frac{d θ}{d t})}^{2} = 4 g \cos θ, a > 0 . its period is$ $[A] 2 π \sqrt{\frac{5 a}{4 g}} [B] \frac{2 π}{5} \sqrt{\frac{a}{g}} [C] 2 π \sqrt{\frac{2 g}{5 a}} [D] 2 π \sqrt{\frac{5 a}{g}}$
	Answered by TANMAY PANACEA last updated on 10/Mar/20

	$\frac{d θ}{d t} = w = \frac{2 π}{T} \to T = \frac{2 π}{w}$ $w^{2} = {(\frac{d θ}{d t})}^{2} = \frac{4 g}{10 a} c o s θ \approx \frac{4 g}{10 a} \times 1$ $w = \sqrt{\frac{2 g}{5 a}} \to T = \frac{2 π}{\sqrt{\frac{2 g}{5 a}}}$ $T = 2 π \times \sqrt{\frac{5 a}{2 g}} \to i t h i n k s o$
	Commented byRio Michael last updated on 10/Mar/20

	$thanks, i^{'} ve gotten the idea$
	Commented bymr W last updated on 11/Mar/20

	$a n s w e r i s w r o n g s i r!$ $ω h e r e i s n o t t h e a n g u l a r v e l o c i t y, i . e .$ $ω \neq \frac{d θ}{d t}$ $ω i s t h e a n g u l a r f r e q u e n c y .$ $w e k n o w t h e a n g u l a r v e l o c i t y \frac{d θ}{d t} i s$ $n o t c o n s t a n t, o t h e r w i s e t h e r e i s n o$ $o s c i l l a t i o n! b u t ω i s a c o n s t a n t!$ $i n f a c t ω h e r e i s d e f i n e d a s \sqrt{\frac{e l a s t i c i t y}{m a s s}} .$ $p l e a s e r e f e r t o y o u r p h y s i c s b o o k .$ $c o r r e c t a n s w e r :$ $10 a {(\frac{d θ}{d t})}^{2} = 4 g \cos θ$ $\Rightarrow \frac{d θ}{d t} = \sqrt{\frac{2 g}{5 a}} \times \sqrt{\cos θ}$ $\Rightarrow \frac{d^{2} θ}{{d t}^{2}} = \sqrt{\frac{2 g}{5 a}} \times \frac{- \sin θ}{2 \sqrt{\cos θ}} \times \frac{d θ}{d t}$ $\Rightarrow \frac{d^{2} θ}{{d t}^{2}} = \sqrt{\frac{2 g}{5 a}} \times \frac{- \sin θ}{2 \sqrt{\cos θ}} \times \sqrt{\frac{2 g}{5 a}} \times \sqrt{\cos θ}$ $\Rightarrow \frac{d^{2} θ}{{d t}^{2}} = - \frac{g \sin θ}{5 a} \approx - \frac{g}{5 a} θ$ $\Rightarrow \frac{d^{2} θ}{{d t}^{2}} + \frac{g}{5 a} θ = 0$ $\Rightarrow \frac{d^{2} θ}{{d t}^{2}} + ω^{2} θ = 0 w i t h ω = \sqrt{\frac{g}{5 a}}$ $p e r i o d T = \frac{2 π}{ω} = 2 π \sqrt{\frac{5 a}{g}}$ $\Rightarrow a n s w e r [D] i s c o r r e c t .$
	Commented byTANMAY PANACEA last updated on 11/Mar/20

	$y e s s i r y o u a r e r i g h t . . . s o p l s i g n o r e m y p o s t$
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