The period T of a pendulum is given by:
[tex]T=2\pi\times\sqrt[]{\frac{L}{32ft/s^2}}[/tex]Where T is the period (s)
L the length (ft)
π is 3.14
A value of T=12.56 seconds is given. Let's find the length then:
[tex]\begin{gathered} \text{Divide both sides by 2}\pi \\ \frac{12.56s}{2\pi}=\frac{2\pi\times\sqrt[]{\frac{L}{32ft/s^2}}}{2\pi} \\ \text{Simplify} \\ \frac{12.56s}{2(3.14)}=\sqrt[]{\frac{L}{32ft/s^2}} \\ 2s=\sqrt[]{\frac{L}{32ft/s^2}} \\ \text{Apply square to both sides} \\ (2s)^2=\sqrt[]{\frac{L}{32ft/s^2}}^2 \\ 4s^2=\frac{L}{32ft/s^2} \\ \text{Multiply both sides by 32}ft/s^2 \\ 4s^2\times32ft/s^2=\frac{L}{32ft/s^2}\times32ft/s^2 \\ \text{Simplify} \\ 128ft=L \\ \text{And reorder terms} \\ L=128ft \end{gathered}[/tex]Thus, the length of a pendulum with a period of T=12.56s is equal to 128ft
