Question

T=2π√l\g,if l=50.0+\-0.2cm and the time for 15 oscillations are:24.2,24.2,24.7,24.5,24.2,24.2,24.4,24.5,24.6,24.7,24.5,24.6,24.5,24.3, determine the value of g and the standard error in g

Answer 1

$l=(0.50\pm0.02)m$

For the period after statistical calculations

$T=(1.6\pm0.1)s$

$g=4\pi^2\frac{l}{T^2}=4\cdot3.14^2\cdot\frac{0.5}{1.6^2}=7.71(m/s^2)$

$\frac{\Delta g}{g}=\sqrt{(\frac{\Delta l}{l})^2+(2\frac{\Delta T}{T})^2}=$

$=\sqrt{(\frac{0.02}{0.50})^2+(2\cdot\frac{0.1}{1.6})^2}=0.131$

$\Delta g=7.71\cdot0.131=1.01(m/s^2)$

$g=(7.7\pm1.0)m/s^2$