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I am given the formula for the period of a pendulum: $T=2\pi\sqrt{\frac{L}{g}}$ along with inputs $T=1$ and $g=32$.

I am to find L, the textbook provides the answer as 10 (to the nearest whole unit). I arrive at 0.8.

I'm not sure I approached this the right way initially. How can I arrive at 10?

$1=2\pi\sqrt{\frac{L}{4\sqrt{2}}}$

$1^2=4\pi^2\frac{L}{32}$ # square both sides to get rid of radical

$\frac{1}{1}=\frac{4\pi^2L}{32}$ # writing the left side a a fraction just makes me see clearer when working with fraction equations

$32=4\pi^2 L$ # multiple out denominator

$32=39.5L$ # $4*(3.14159^22)$ ~ 39.5

$L = 32/39.5=0.8$

How can I arrive at 10?

Here's a screen shot from the textbook in case I've missed anything. It's the second question starting "If the gravity is 32..." enter image description here

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You got an answer of $0.8$. But since the question gave gravity in feet, your final answer is in feet! Since 1 foot is 12 inches, your final answer would be $$0.8\times 12=9.6\,\text{inches}$$

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If you square both sides you get $T^2=4 \pi^2 \frac{L}{g} \rightarrow L = \frac{T^2 g}{4\pi^2} = \frac{1^2 \cdot 32}{4\cdot \pi^2} \approx 0.81$ which is what you get. However, the problem is in ft, not in inches. So you get $0.81$ feet which is 9.72 inches, which is about 10 inches.

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$L= g ({ T \over 2 \pi})^2 = 32{1 \over 4 \pi^2}= { 8 \over \pi^2} \approx {8 \over 9} \approx 1$.

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  • $\begingroup$ Why the downvote? The part about inches was added after my answer. $\endgroup$ – copper.hat May 27 '19 at 23:29

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