Question and Answers Forum

All Questions      Topic List

Mechanics Questions

Previous in All Question      Next in All Question      

Previous in Mechanics      Next in Mechanics      

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

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.

$${The}\:{eriodic}\:{time}\:{of}\:{a}\:{simple} \\ $$$${pendulum}\:{is}\:{given}\:{by}\:{T}=\mathrm{2}\pi\sqrt{\frac{{L}}{{g}}}. \\ $$$${if}\:{L}=\mathrm{100}\pm\mathrm{0}.\mathrm{1}{cm}\left({s}.{e}\right)\:{and}\:{the} \\ $$$${time}\:{s}\:{for}\:\mathrm{10}\:{ossilations}\:{are}: \\ $$$$\mathrm{48}.\mathrm{2},\mathrm{48}.\mathrm{2},\mathrm{48}.\mathrm{7},\mathrm{48}.\mathrm{5},\mathrm{48}.\mathrm{4},\mathrm{48}.\mathrm{2},\mathrm{48}.\mathrm{4}, \\ $$$$\mathrm{48}.\mathrm{6},\mathrm{48}.\mathrm{5},\mathrm{48}.\mathrm{2}. \\ $$$${calcukate}\:{g}\:{and}\:{the}\:{standard} \\ $$$${error}\:{involved}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com