Question
Physics Question on simple harmonic motion
The period of oscillation of a simple pendulum is T=2πgL. Measured value of L is 20.0cm known to 1mm accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1s resolution. The accuracy in the determination of g is
A
3%
B
2%
C
1%
D
5%
Answer
3%
Explanation
Solution
TdT=21LdL−21gdg 10090+1001 21gdg=21LdL−TdT 21×200.1=90/1001/100=4001+901 21gdg=4001+901 gdg=(400×90490)×2 =(200×90490)=0.20272 =dg/g×100≈2.72%≈3%