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I'm trying to solve the following question:

The period of the pendulum of a grandfather clock is $T = 2\pi \sqrt{\frac{L}{g}}$ , where $L$ is the length (in meters) of the pendulum, $T$ is the period (in seconds), and $g$ is the acceleration due to gravity ($9.8\frac{\text{m}}{\text{s}^2}$ ). Suppose an increase in temperature increases the length $L$ of the pendulum by $1\%$. What is the corresponding percentage error in the period?


Attempt at solving the question:

I know I have to find $\frac{dT}{T}= \frac{\frac{\pi \sqrt{\frac{L}{g}}}{L}dL}{2\pi \sqrt{\frac{L}{g}}} = \frac{1}{2L}dL = \frac{1}{2L}\times 0.01$. Is this the correct way to do it. I thought the answer should be a number but I can't see how to remove the final $L$. Thanks for the help.

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  • $\begingroup$ Note that a $1\%$ change in $L$ means that $\Delta L = 0.01\cdot L$ not $\Delta L = 0.01$. Otherwise it's fine. $\endgroup$ – Winther Nov 16 '15 at 1:00
  • $\begingroup$ @Winther I see, so my final answer should be 0.005. Thank you for your help $\endgroup$ – user290479 Nov 16 '15 at 1:04
  • $\begingroup$ Yes thats right, or $0.5\%$ if you will. btw a small detail: I would not call it the 'error in the period' it's rather the change in the period you are calculating. $\endgroup$ – Winther Nov 16 '15 at 1:04

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