# 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 Shivank Goel|Updated : August 12th, 2022

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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