# Question 70d85

Jul 31, 2016

The equation of simple pendulum

$T = 2 \pi \sqrt{\frac{L}{g}}$

$\implies g = 4 {\pi}^{2} \cdot \frac{L}{T} ^ 2$

Taking logarithm on both sides

$\ln g = \ln \left(4 {\pi}^{2}\right) + \ln L - 2 \ln T$

Taking differential

$\frac{\Delta g}{g} = \frac{\Delta L}{L} - \frac{2 \Delta T}{T} \ldots . \left(1\right)$

Now for calculation of maximum error of g every error term should be taken positive.

Hence equation (1) becomes

$\frac{\Delta g}{g} = \frac{\Delta L}{L} + \frac{2 \Delta T}{T}$

In percent form it becomes

$\frac{\Delta g}{g} \times 100 = \frac{\Delta L}{L} \times 100 + \frac{2 \Delta T}{T} \times 100. . \left(2\right)$

Now it is given that in measuring $L = 20 c m$ the minimum length measurable by scale being 1mm or 0.1cm the error $\Delta L = 0.1 c m$

$\therefore \frac{\Delta L}{L} = \frac{0.1}{20}$

Again the resolution of watch used for the measurement of T being 1s (the smallest graduation on the face of the watch),error for the measurement of time for 100 oscillations will be 1s i.e.$\Delta T = \frac{1}{100} s = 0.01 s$

Now time measured for 100 oscillations is 90s. So time period $T = \frac{90}{100} = 0.9 s$

Hence $\frac{\Delta T}{T} = \frac{0.01}{0.9}$

Now the percent error in measurement of g by equation(2)

$\frac{\Delta g}{g} \times 100 = \frac{\Delta L}{L} \times 100 + \frac{2 \Delta T}{T} \times 100$

$\implies \frac{\Delta g}{g} \times 100 = \frac{0.1}{20} \times 100 + \frac{2 \times 0.01}{0.9} \times 100$

~~(0.5+2.2)%=2.7%~~3%#