The period of oscillation of a simple pendulum is T=2π√L/g. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillation of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is: (a). 3% (b). 1% (c). 2% (d). 5%
By BYJU'S Exam Prep
Updated on: September 13th, 2023
It is given that
ΔL = 1mm = 0.001 m
L = 20 cm
We know that
ΔT = least count/ number of oscillations
ΔT = 1/100 = 0.01 s
T = Total time/ Number of oscillations
T = 90/100 = 0.9 s
We can write g from the given equation as
g = 4π2L/T2
The relative error in g is written as
Δg/g = ΔL/L + 2 (ΔT/T)
ΔL/L = 0.001/0.20’
ΔT/T = 0.01/0.9
So we get
100 (Δg/g) = [(0.001/0.20) + 2 x (0.01/0.9)] x 100% = 2.7% = 3%
Therefore, accuracy in the determination of g is 3%.
Summary:
The period of oscillation of a simple pendulum is T=2π√L/g. The measured value of L is 20.0 cm known to have 1 mm accuracy and the time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is: (a). 3% (b). 1% (c). 2% (d). 5%
The period of oscillation of a simple pendulum is T=2π√L/g. The measured value of L is 20.0 cm known to have 1 mm accuracy and the time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. The accuracy in the determination of g is 3%.
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