Question

ow long must a simple pendulum be if it is to make exactly ten swings per second? (That is, one complete vibration takes exactly 0.200 s.)

Answer 1

The period T of a pendulum is given by:
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the length of the pendulum while $g=9.81 m/s^2$ is the gravitational acceleration.

In the pendulum of the problem, one complete vibration takes exactly 0.200 s, this means its period is $T=0.200 s$. Using this data, we can solve the previous formula to find L:
$L=g ( \frac{T}{2\pi} )^2=(9.81 m/s^2)( \frac{0.2 s}{2 \pi} )^2=1 \cdot 10^{-3} m=1 mm$