Question

The period of oscillation of a simple pendulum is T=2√


. Measured value of L is 20.0 m
known to 1mm accuracy and time for 100 oscillations is found to be 90s using a stop
watch of 2 s resolution. What is the accuracy in determining the accuracy in “g”?
(Please notice there is 2s in resolution and good luck I'm waiting for the answer let's see who is the brainiest )

Answer 1

Answer:

Step-by-step explanation:

he relative error in the determination of g is

Δg/g = ΔL/L + 2 ΔT/T

where L is length and T is time period.

Now ΔL/L = 0.1/20 = 0.005 and ΔT/T = (1/90)/100 = .01/90 = 0.0001

So, Δg/g =0.005 + 2 * 0.0001 = 0.0052 or 0.52%.