Question

The measurement value of the 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 1 s resolution. Calculate the percentage accuracy in the determination of acceleration due to gravity ‘g’ from the above measurement.

Answer 1

Given,

Length of simple pendulum (l) = 20 cm

absolute error in length (∆l) = 2 mm = 0.2 cm

Time taken for 50 oscillation (t) = 40 s

error in time ∆T = 1 s

Solution: Time period for one oscillation (T)

= `t/n = 40/50`s

∴ ∆T = `(∆t)/n`

So, `(∆T)/T = ((∆t)/n)/(t/n) = (∆t)/t`

∆l, ∆T are least count errors,

`(∆g)/g = (∆l)/l + 2(∆T)/T = 0.2/20 + 2(1/40) = 1.2/20 = 0.06`

Hence, the percentage error in g is

`((∆g)/g) xx 100 = ((∆l)/l + 2(∆T)/T) xx 100% = 0.06 xx 100% = 6%`