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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
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Solution
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%
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