The measurement value of length of a simple pendulum is 20 cm with 2 m

1.	The measurement value of length of a simple pendulum is 20 cm with 2 mm accuracy. The time taken for 50 oscillations was measured to be 40 s within 1 s resolution. Calculate the percentage accuracy in the determination of acceleration due to gravity from the above measurement
Answer» CONSIDER, t is time and n is NUMBER of revolutions. Therefore, T = t/n. Hence, ∆T = ∆t/n; n is a constant term. On simplifying we get, ∆T/T = ∆t/t, let this be equation 1. We know that, T = 2π√(L/g) By squaring both the sides, T² = 4π²l/g ⇒ g = 4π²l/T² Therefore, ∆g/g = ∆l/l + 2∆T/T This implies that, ∆g/g = ∆l/l + 2∆t/t……. from equation 1. Therefore, percentage error in g = ( ∆l/l + 2∆t/t) X 100 We have, ∆l = 1mm = 0.1 CM , l = 20cm ∆t = 1 s and t = 90 s Therefore, % error in g = (0.1/20 + 2 × 1/90) X 100 = ( 0.005 + 0.022) 100 = (0.027) 100 = 2.7%

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

Answer»

CONSIDER, t is time and n is NUMBER of revolutions.

Therefore, T = t/n.

Hence, ∆T = ∆t/n; n is a constant term.

On simplifying we get, ∆T/T = ∆t/t, let this be equation 1.

We know that, T = 2π√(L/g)

By squaring both the sides,

T² = 4π²l/g ⇒ g = 4π²l/T²

Therefore, ∆g/g = ∆l/l + 2∆T/T

This implies that, ∆g/g = ∆l/l + 2∆t/t……. from equation 1.

Therefore, percentage error in g = ( ∆l/l + 2∆t/t) X 100

We have, ∆l = 1mm = 0.1 CM , l = 20cm

∆t = 1 s and t = 90 s

Therefore, % error in g = (0.1/20 + 2 × 1/90) X 100 = ( 0.005 + 0.022) 100 = (0.027) 100 = 2.7%