Answer to Question #116985 in Mechanics | Relativity for Riya

Question #116985
The following observations we're made during an experiment to find the value of g using simple pendulum l=90.0 cm
Time t for 20 vibrations =36 sec
Find percentage error in measurement of g given that T=2π√l/g length is being measured to an accuracy of 0.1 cm and time 0.2 sec
1
Expert's answer
2020-05-21T13:09:13-0400

We know that

T=2πlgg=4π2lT2.T = 2\pi\sqrt{\dfrac{l}{g}} \Longrightarrow g = \dfrac{4\pi^2l}{T^2}.

T is equal to 36:20 = 1.8 seconds, so

g=4π20.91.8211.0m/s2.g = \dfrac{4\pi^2 \cdot0.9}{1.8^2} \approx 11.0\,\mathrm{m/s^2}. The uncertainty of period will be 0.220=0.01s.\dfrac{0.2}{20} = 0.01\,\mathrm{s}.

Next, we should obtain the formula for calculating the percentage error (see part 5 in http://lectureonline.cl.msu.edu/~mmp/labs/error/e2.htm)

δg=(gTδT)2+(glδl)2=(8π2lT3δT)2+(4π2T2δl)2=(8π20.91.830.01)2+(4π21.820.001)20.12m/s2.\delta g = \sqrt{\left(\dfrac{\partial g}{\partial T}\delta T \right)^2 + \left(\dfrac{\partial g}{\partial l}\delta l \right)^2} = \sqrt{\left(-\dfrac{8\pi^2l}{T^3}\cdot {\delta T} \right)^2 + \left(\dfrac{4\pi^2}{T^2}\cdot{\delta l} \right)^2} = \sqrt{\left(-\dfrac{8\pi^2\cdot0.9}{1.8^3}\cdot{0.01} \right)^2 + \left(\dfrac{4\pi^2}{1.8^2}\cdot{0.001} \right)^2} \approx 0.12\,\mathrm{m/s^2}.

The percentage error will be 0.1211.01.1%.\dfrac{0.12}{11.0} \approx 1.1\%.


The simplier way to calculate the percentage error is to use the formula

δll+2δTT=0.0010.9+20.011.8=1.2%.\dfrac{\delta l}{l} + 2\dfrac{\delta T}{T} = \dfrac{0.001}{0.9} + 2\dfrac{0.01}{1.8} = 1.2\%.


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