Question
Physics Question on simple harmonic motion
The period of oscillation of a simple pendulum is T=2πgL. Measured value of 'L' is 1.0m from meter scale having a minimum division of 1mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01s resolution. The percentage error in the determination of 'g' will be :
A
1.13%
B
1.03%
C
1.33%
D
1.30%
Answer
1.13%
Explanation
Solution
T=2πgℓ g=T24π2ℓ gΔg=ℓΔℓ+T2ΔT gΔg=11×10−3+2×1.950.01 gΔg=0.0113 or 1.13%