Question

Writing g for the acceleration due to gravity, the period,T, of a pendulum of length l is given by

T = 2π√l/g.
(a) Show that if the length of the pendulum changes byΔl, the change in the period, ΔT, is given by
ΔT ˜ T/2l Δl
(b) If the length of the pendulum increases by 2% by whatpercent does the period change?

Answer 1

Here, the Taylor approximation for a square root was applied, and O(x) stands for all negligible terms of Taylor's sum with respect to variable x.

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Answer 2

Explanation:

`T=2\pi\sqrt{(l)/(g)}`

`T+\Delta T=2\pi\sqrt{((l+\Delta l))/(g)} =2\pi\sqrt{(l)/(g)}\sqrt{1+(\Delta l)/(l)}=2\pi\sqrt{(l)/(g)}(1+(1)/(2)(\Delta l)/(l)+0((\Delta l)/(l)))=T(1+(1)/(2)(\Delta l)/(l)+0((\Delta l)/(l)))`

Here, the Taylor approximation for a square root was applied, and O(x) stands for all negligible terms of Taylor's sum with respect to variable x.

So,
`\Delta T=T(1)/(2)(\Delta l)/(l)`

b. For an increase of 2%, that is:

`(\Delta l)/(l)=0.02`

`(\Delta T)/(T)=(1)/(2)0.02=0.01=1\%`