The period T of a pendulum is given by:
[tex]T=2 \pi \sqrt{ \frac{L}{g} } [/tex]
where L is the length of the pendulum while [tex]g=9.81 m/s^2[/tex] is the gravitational acceleration.
In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is [tex]T=0.200 s[/tex]. Using this data, we can solve the previous formula to find L:
[tex]L=g ( \frac{T}{2\pi} )^2=(9.81 m/s^2)( \frac{0.2 s}{2 \pi} )^2=1 \cdot 10^{-3} m=1 mm[/tex]
