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I came across this equation: $T_m= \sqrt{6}T_e $.

Can anyone tell me how this equation is derived? This is how I tried to, but I got stuck after some time.

So the time period of a simple pendulum on the earth= $2\pi\sqrt{l/g}$

The time period of a pendulum on the moon is $2\pi\sqrt{l/(g/6)}$

Now how do I create an equation which shows the time period of a pendulum on the moon with respect to the time period of a pendulum on the earth.

And please be as detailed as possible!

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    $\begingroup$ The first time period is $T_e$ and the second is $T_m$, so what problem are you having? $\endgroup$ – G. Smith Jul 10 at 20:20
  • $\begingroup$ Just divide one expression by the other, keeping in mind that $\sqrt {ab} =\sqrt{a}\sqrt{b} $ $\endgroup$ – Agnius Vasiliauskas Jul 10 at 20:24
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    $\begingroup$ The 6 is approximate, not exact. Writing $\sqrt 6$ wrongly suggests that it is exact. $\endgroup$ – G. Smith Jul 10 at 20:33
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You can modify the second formula:

$$T_m = 2 \pi \sqrt{\frac{l}{\frac{g}{6}}} = 2 \pi \sqrt{\frac{6l}{g}} = 2 \pi \sqrt{6\frac{l}{g}} = \sqrt{6} \left(2 \pi \sqrt{\frac{l}{g}}\right) = \sqrt{6} T_e$$

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  • $\begingroup$ ahh this is what I was looking for! Thanks so much! $\endgroup$ – Viradeus Jul 10 at 20:20
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You are almost there!

\begin{align*} T_e&=2\pi\sqrt{\frac lg}\\ T_m&=2\pi\sqrt{\frac l{\frac g6}}=\sqrt6\cdot2\pi\sqrt{\frac lg}=\sqrt6\cdot T_e\\ \end{align*}

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