Question

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

Answer 1

Step-by-step explanation:

$T=2\pi\sqrt{\frac{l}{g}}$

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

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, $\Delta T=T\frac{1}{2}\frac{\Delta l}{l}$

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

$\frac{\Delta l}{l}=0.02$

$\frac{\Delta T}{T}=\frac{1}{2}0.02=0.01=1\%$

Answer 2

Answer:

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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