Question

The time T of a complete oscillation of a simple pendulum of length 'l' is given by the equation T = 2 π √ 1/g where g is constant. What is the percentage error in T when 'l' is increased by 1% ?

Answer 1

Hi,

Here is your answer,

              T = 2π√1/g

→ Δl/l × 100 = 1 (Given)         Here we are using Percentage error formula                                                              dy/y × 100 (OR) dx/x × 100                   
→ dl/l × 100 = 1

→ T = 2π√1/g

→ T = 2π(l/g)¹/²

Now, Apply Log on Both sides

→ logT = log2π(l/g)¹/²

→ logT - log2π + log(l/g)¹/²

→ logT = log2π + 1/2log(l/g)

→ logT = log2π + 1/2[log l - log g ]

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

→ 1/T × dt = 0 + 1/2 × 1/l - 1/2 × (0)

→ 1/T × dt = 1/2 dl/l

→ dT/T = 1/2 dl/l

→ dT/T × 100 = 1/2 dl/l × 100   { Apply Percentage error formula }

→ dT/T × 100 = 1/2 %        { FINAL ANSWER }

Answer 2

Hey!!

Plz refer to the attachment for the answer...