The measurement value of length of a simple pendulum is 20 cm known with 2 mm accuracy. The time for 50 oscillations was measured to be 40 s within 1s resolution. Calculate the percentage accuracy in the determination of acceleration due to gravity g from the above measurement.

The measurement value of length of a simple pendulum is 20 cm known wi

1.	The measurement value of length of a simple pendulum is 20 cm known with 2 mm accuracy. The time for 50 oscillations was measured to be 40 s within 1s resolution. Calculate the percentage accuracy in the determination of acceleration due to gravity g from the above measurement.
Answer» SOLUTION :Time period for ONE oscillation (T) `= t/n = (40)/(50) s` `:. DeltaT = (Deltat)/(n)` So, `(DeltaT)/(T) = (Deltat//n)/(t//n) = (Deltat)/(t)` `Deltal, DeltaT` are LEAST COUNT errors `(Deltag)(g) = (Deltal)/(l) + 2 (DeltaT)/(T) = (0.2)/(20) +2 ((1)/(40)) = (1.2)/(20) = 0.06` Hence, the percentage error in g is `((Deltag)/(g)) xx 100 = ((Deltal)/(l) +2 (DeltaT)/(T)) xx 100% = 0.06 xx 100% = 6%`