The time T of oscillation of a simple pendulu | vedclass

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(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.Taking the natural logarithm on both sides,we get $\ln T = \ln(2 \

Vedclass · Accepted Answer

$-\frac{5}{18}$

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(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{L}{g}}$.Taking the natural logarithm on both sides,we get $\ln T = \ln(2 \pi) + \frac{1}{2} \ln L - \frac{1}{2} \ln g$.Differentiating both sides,we get $\frac{dT}{T} = \frac{1}{2} \frac{dL}{L}$.Here,the clock loses $2$ minutes per day,so $\Delta T = -2$ minutes.The total time in a day is $24 	imes 60 = 1440$ minutes.Thus,the fractional change in time is $\frac{\Delta T}{T} = \frac{-2}{1440} = -\frac{1}{720}$.Using the relation $\frac{\Delta L}{L} = 2 \frac{\Delta T}{T}$,we have $\frac{\Delta L}{L} = 2 	imes \left( -\frac{1}{720} ight) = -\frac{1}{360}$.To express this as a percentage change,we multiply by $100$:$\frac{\Delta L}{L} \% = -\frac{1}{360} 	imes 100 = -\frac{10}{36} = -\frac{5}{18} \%$.Therefore,the length must be changed by $-\frac{5}{18} \%$.

The time $T$ of oscillation of a simple pendulum of length $L$ is governed by $T=2 \pi \sqrt{\frac{L}{g}}$,where $g$ is constant. The percentage by which the length must be changed in order to correct an error of loss equal to $2$ minutes of time per day is

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