Question

The time T, taken for a complete oscillation of a single pendulum with length l, is given by the equation T = `2pi sqrt(l/g)` where g is a constant. Find the approximate percentage error in the calculated value of T corresponding to an error of 2 percent in the value of l

Answer 1

Given T = `2pi sqrt(l/g)`

On taking log both sides, we get

log T = `log 2 + log pi + 1/2 log "l" - 1/2 log "g"`

On differentiating both sides w. r. to l, we get

`1/"T" "dT"/"dl" = 1/21`

`1/"T" "dT"/"dl" Deltal = 1/(2l) Deltal`  ......(Multiplying both sides `Deltal`)

∴ `Delta"T" = "dT"/"dl" Deltal`

`1/"T" Delta"T" = 1/(2l) Deltal`

`(Delta"T")/"T" xx 100 = 1/2 (Deltal)/l xx 100`

= `1/2 xx 2`

= 1%

So, the percentage error in T is 1%