# Question 47b8a

Dec 26, 2017

See the explanation below

#### Explanation:

The length of the pendulum is $= l$

The period of the pendulum is $= T$

The period of a simple pendulum is

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

Taking natural logarithms on both sides

$\ln T = \ln \left(2 \pi\right) + \frac{1}{2} \ln \left(l\right) - \frac{1}{2} \ln \left(g\right)$

Differentiating the equation

$\frac{\Delta T}{T} = 0 + \frac{1}{2} \frac{\Delta l}{l} - \frac{1}{2} \frac{\Delta g}{g}$

You cannot substract errors

Therefore,

$| \frac{\Delta T}{T} | = \frac{1}{2} | \frac{\Delta l}{l} | + \frac{1}{2} | \frac{\Delta g}{g} |$

The error in measuring the length is (Deltal)/l=2%

The error in measuring the acceleration due to gravity is (Deltag)/g=2%

So,

(DeltaT)/T=1/2*2%+1/2*2%=2%#