Question

21) The time T of oscillation of a simple pendulum of length
tage error in T, corresponding to an
Lis given by T = 2 . Find the percentage error in T, correspond
error of 2 % in the value of L.​

Answer 1

Answer:

this is answer by the formulae of simple pendulum

Answer 2

Case I: Percentage error when increase of 2% in l.

Given, T = 2π √(l/g)

Taking log on both sides,

log T  = log 2 + log π + 1/2 log l - 1/2 log g

Differentiating wrt l.

1/T dT/dl = 0 + 0 + 1/2l - 0

dT/T = 1/2 dl/l

so, dT/T*100 = 1/2(dl / l * 100) = 1/2 * (2) % = 1%

Hence, Percentage error in T = 1%.

Case II: Percentage error when decrease of 2% in l:

Similarly, as shown above, we will get percetage decrease in T as -1%.