# Thread: Algebra II

1. ## Algebra II

1.) The period for the pendulum in a clock is givin by: T(L)=2(3.14)√L/32, where T(L) is the period in seconds and L is the length in ft.
a) Find the period of a pendulum that is 18 inches long

b) find the period of a Foucault pendulum that is 72 ft. long

c) How long is the pendulum that has a period of 7.5 seconds?

Thank you :]

2. Originally Posted by ep78
1.) The period for the pendulum in a clock is givin by: T(L)=2(3.14)√L/32, where T(L) is the period in seconds and L is the length in ft.
a) Find the period of a pendulum that is 18 inches long

b) find the period of a Foucault pendulum that is 72 ft. long

c) How long is the pendulum that has a period of 7.5 seconds?

Thank you :]
For the first two, substitute the lengths into the equation.

a) Note that we need to convert the length from inches to feet: $18\text{ in}=1.5\text{ ft}$

Thus, $T(1.5)=2\pi\sqrt{\frac{1.5}{32}}$

Just plug this into a calculator, where $\pi\approx 3.14$

b) $T(72)=2\pi\sqrt{\frac{72}{32}}$

Just plug this into a calculator, where $\pi\approx 3.14$

c) Here, we know that $T=7.5\text{ sec}$, but we want to find L:

$T=2\pi\sqrt{\frac{L}{32}}\implies \frac{T}{2\pi}=\sqrt{\frac{L}{32}}\implies \bigg(\frac{T}{2\pi}\bigg)^2=\frac{L}{32}\implies L=32\bigg(\frac{T}{2\pi}\bigg)^2$

Now plug in T=7.5 and plug this into your calculator:

$L=32\bigg(\frac{7.5}{2\pi}\bigg)^2\approx ......$, where $\pi\approx 3.14$

Does this make sense?

--Chris

3. very much so..thanks so much!