If the length and time period of an oscillating pendulum have a error of 1%and3% respectively then the error in measurement is an person due to gravity is
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I'm assuming you're asking error in acceleration due to gravity,
answer will be 7%.
[tex]T= 2\pi \sqrt{L/g
T^{2}=4 \pi^{2} L/g
g = 4 \pi^{2} L/g
4 and \pi [/tex] are constant values.
Error in g = L + T^2
= 1% + (3)^2 %
= 1%+6%
= 7%
answer will be 7%.
[tex]T= 2\pi \sqrt{L/g
T^{2}=4 \pi^{2} L/g
g = 4 \pi^{2} L/g
4 and \pi [/tex] are constant values.
Error in g = L + T^2
= 1% + (3)^2 %
= 1%+6%
= 7%
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