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A clock with metallic pendulum is 5 sec fast every day at a temperature of 20°C and 10 sec slow at a temp of 35°C. Find coefficient of linear expansion.

I am unable to understand the application of formula here .

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  • $\begingroup$ Pendulum equation, and changes in the length of the pendulum $\endgroup$ – user56903 Jan 31 '16 at 18:33
  • $\begingroup$ Hint: The duration of the swing of a pendulum clock depends only on its effective length. $\endgroup$ – Lewis Miller Jan 31 '16 at 18:34
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Recall that the period of small oscillations of a pendulum depends on the length. This length is influenced by thermal expansion. At hotter temperatures (e.g. $35^\circ$ celcius) the pendulum will expand, and its period will increase, causing the clock to count off fewer seconds. At colder temperatures, the pendulum contracts and the period of oscillation decreases.

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You may remember that the period of a pendulum is given by

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

When the clock is 5 seconds fast, it is counting 5 seconds too many per day, or 86405 seconds. When it is 20 seconds slow, it is counting only 86380 seconds.

Regardless of the number of swings of the pendulum (does it swing once per second? more? less?), you can use this to find the ratio of the lengths for the two situations:

$$\frac{\ell_1}{\ell_2} = \left(\frac{86405}{86380}\right)^2$$

From this you can find the ratio of lengths; and that should allow you to find the thermal expansion coefficient.

Now I am an old-fashioned physicist, and I would urge you NOT to "plug the numbers into a calculator". Instead, you can rewrite the above equation:

$$1 + \frac{\Delta \ell}{\ell_2} = \left(1 + \frac{25}{86380}\right)^2 \approx 1 + \frac{2*25}{86380}$$

That will allow you to deal with "small changes" directly, instead of having to carry five digits of precision in your calculation. I hope that going from here to thermal expansion equations is straightforward.

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  • $\begingroup$ Hi! Don't you think it's a homework question? If so, then, IMO, you've provided a complete solution which I don't deem aptly. $\endgroup$ – user36790 Jan 31 '16 at 19:31
  • $\begingroup$ Well - I think that the thermal expansion part of the question is left to the OP. This would not rate as a complete solution in my books. But we can have different opinions about that. I thought that the "small differences" approach is a bit of physics that a lot of people unfortunately don't learn, or appreciate - and that's the crux of my answer. $\endgroup$ – Floris Jan 31 '16 at 19:35
  • $\begingroup$ I always appreciate your answers but IMO you actually solved the problem : / $\endgroup$ – user36790 Jan 31 '16 at 19:38

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