Question

Calculate percentage error in determination of time period of a pendulum. T = 2pi (sqrt(l))/(g) where, l and g are measured with +-1% and +-2%.

Answer 1

Maximum percentage error for the time period of pendulum with $T = 2\pi \sqrt{\frac{l}{g} }$ is

(ΔT/T) = $\frac{1}{2}$ (Δl/l) + $\frac{1}{2\\}$ (Δg/g)

Given (Δl/l) = 1% and (Δg/g) = 2%

So, Maximum percentage error of time period= (ΔT/T) = 0.5 * ( 1% + 2% ) = 1.5%

Answer 2

The Time period of a simple pendulum is given by,

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

Maximum relative error is given by,

$\frac{ \Delta \:T }{T } = \frac{1}{2} (\frac{ \Delta \: l}{l} + \frac{ \Delta \: g}{g} )$

Maximum percentage error is given by,

$\frac{ \Delta \:T }{T }\% = \frac{1}{2} (\frac{ \Delta \: l}{l} \% + \frac{ \Delta \: g}{g}\% )$

According to the given question

$\frac{ \Delta \: l}{l} \% = \pm \: 1\%\\ \\ \frac{ \Delta \: g}{g}\% = \pm \: 2\%$

So, Percentage error in the Time period of a simple pendulum is,

$\frac{ \Delta \:T }{T }\% = \frac{1}{2} (\frac{ \Delta \: l}{l} \% + \frac{ \Delta \: g}{g}\% ) \\ \\ \frac{ \Delta \:T }{T }\% = \frac{1}{2} (1 + 2)\% = \frac{3}{2} \% = 1.5\%$

Therefore, The maximum percentage error in the Time period of a simple pendulum is 1.5 %