# Find L in $T=2\pi\sqrt{\frac{L}{g}}$

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..."

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 $$0.8\times 12=9.6\,\text{inches}$$
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.
$$L= g ({ T \over 2 \pi})^2 = 32{1 \over 4 \pi^2}= { 8 \over \pi^2} \approx {8 \over 9} \approx 1$$.