Answer:
[tex]L=6.21m[/tex]
Explanation:
For the simple pendulum problem we need to remember that:
[tex]\frac{d^{2}\theta}{dt^{2}}+\frac{g}{L}sin(\theta)=0[/tex],
where [tex]\theta[/tex] is the angular position, t is time, g is the gravity, and L is the length of the pendulum. We also need to remember that there is a relationship between the angular frequency and the length of the pendulum:
[tex]\omega^{2}=\frac{g}{L}[/tex],
where [tex]\omega[/tex] is the angular frequency.
There is also an equation that relates the oscillation period and the angular frequeny:
[tex]\omega=\frac{2\pi}{T}[/tex],
where T is the oscillation period. Now, we can easily solve for L:
[tex](\frac{2\pi}{T})^{2}=\frac{g}{L}\\\\L=g(\frac{T}{2\pi})^{2}\\\\L=9.8(\frac{5}{2\pi})^{2}\\\\L=6.21m[/tex]
