Question
A pendulum clock gives correct time. What is the error in time per day if the length increases by 0.05%?
✓
Answer
$\text{T}=2\pi\sqrt{\frac{\text{L}}{\text{g}}}$ and $\text{T}'=2\pi\sqrt{\frac{\big(\text{L}+\frac{0.05}{100}\text{L}\big)}{\text{g}}}$
Dividing, $\frac{\text{T}'}{\text{T}}=\sqrt{\frac{\big(\text{L}+\frac{0.05}{100}\text{L}\big)}{\text{g}}}$
$=\Big[1+\frac{0.05}{100}\Big]^{\frac{1}2{}}$
Applying binomial theorem, and neglecting squares and higher powers, we get
$\frac{\text{T}'}{\text{T}}=1+\frac{1}{2}\times0.0005$
$\frac{\text{T}'}{\text{T}}-1=0.00025$
$\frac{\text{T}'-\text{T}}{\text{T}}=0.00025$
$\therefore$ Loss of time pre second $=0.00025\text{s}$
Loss of time per day
$=0.00025\times24\times60\times60\text{s}$
$=21.6\text{s}$
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