The time t, of a complete oscillation of a simple pendulum of length ‘l’ is given by the equation T=2πlg where ‘g’ is a gravitational constant find approximate percentage error in ‘T’ when the percentage of error in ‘l’ is 1%.

Given time period of simple pendulumt=2πlg,δll×100=1log⁡t=log⁡2πgl=log⁡(2π)g+12log⁡lTake differential on both sides Δtt≅12Δll ⇒Δtt×100=12Δll×100=12(1)≅12 ∴Δtt×100≅12

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The time $t$ , of a complete oscillation of a simple pendulum of length ‘ $l$ ’ is given by the equation $T = 2 π \sqrt{\frac{l}{g}}$ where ‘ $g$ ’ is a gravitational constant find approximate percentage error in ‘ $T$ ’ when the percentage of error in ‘ $l$ ’ is $1 %$ .

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Detailed Solution

Given time period of simple pendulum

$t = 2 π \sqrt{\frac{l}{g}}, \frac{δ l}{l} \times 100 = 1$

$\log t = \log \frac{2 π}{\sqrt{g}} \sqrt{l} = \log \frac{(2 π)}{\sqrt{g}} + \frac{1}{2} \log l$

Take differential on both sides

$\frac{Δ t}{t} ≅ \frac{1}{2} \frac{Δ l}{l} \Rightarrow \frac{Δ t}{t} \times 100$

$= \frac{1}{2} (\frac{Δ l}{l} \times 100) = \frac{1}{2} (1) ≅ \frac{1}{2} ∴ \frac{Δ t}{t} \times 100 ≅ \frac{1}{2}$

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