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denpristay [2]
3 years ago
13

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

Mathematics
2 answers:
ivanzaharov [21]3 years ago
7 0

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\%

mel-nik [20]3 years ago
4 0

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.

-God Bless

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