Question 237: Units , Dimensions and Error Analysis

Question

The acceleration due to gravity is found upto an accuracy of 4% on a planet. The energy supplied to a simple pendulum to known mass 'm' to undertake oscillations of time period T is being estimated. If time period is measured to an accuracy of 3%, the accuracy to which E is known as ..............%

Accepted Answer

$$T = 2\pi \sqrt {{l \over g}}  \Rightarrow l = {{{T^2}g} \over {4{\pi ^2}}}$$
$$E = mgl{{{	heta ^2}} \over 2} = m{g^2}{{{T^2}{	heta ^2}} \over {8{\pi ^2}}}$$
$${{dE} \over E} = 2\left( {{{dg} \over g} + {{dT} \over T}} ight)$$
$$ = (4 + 3) = 14\% $$

Units , Dimensions and Error Analysis

The acceleration due to gravity is found upto an accuracy of 4% on a planet. The energy supplied to a simple pendulum to known mass 'm' to undertake oscillations of time period T is being estimated. If time period is measured to an accuracy of 3%, the accuracy to which E is known as ..............%

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