The eriodic time of a simple pendulum is given by T=2π(√(L/g)). if L=100±0.1cm(s.e) and the time s for 10 ossilations are: 48.2,48.2,48.7,48.5,48.4,48.2,48.4, 48.6,48.5,48.2. calcukate g and the standard error involved.


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Question Number 29260 by NECx last updated on 06/Feb/18

$T h e e r i o d i c t i m e o f a s i m p l e$ $p e n d u l u m i s g i v e n b y T = 2 π \sqrt{\frac{L}{g}} .$ $i f L = 100 \pm 0.1 c m (s . e) a n d t h e$ $t i m e s f o r 10 o s s i l a t i o n s a r e :$ $48.2, 48.2, 48.7, 48.5, 48.4, 48.2, 48.4,$ $48.6, 48.5, 48.2 .$ $c a l c u k a t e g a n d t h e s t a n d a r d$ $e r r o r i n v o l v e d .$
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