91Ó°ÊÓ

The period $T$of a simple pendulum of length $L$is given by:

$T=2\mathrm{Ï€}\sqrt{\frac{L}{g}}$

where $g=9.80$$\mathrm{m}/{\mathrm{s}}^2$, is the acceleration because of gravity.

Note that the period of a simple pendulum depends only on its length and the magnitude of its gravitational constant.

It does now no longer rely upon the mass of the item striking at its quit or the amplitude of vibration.

The clock ticks every time the pendulum exceeds the lowest point.

The pendulum is modeled as a simple pendulum.

You will be asked to determine the length of the pendulum so that the ticks occur once per second.

The pendulum passes the lowest point twice in each cycle, so it is the cycle of the pendulum$2.0\mathrm{s}$.

Solving Equation (*) for $L$, we obtain:

$L=\frac{T^{2}g}{4{\mathrm{Ï€}}^2}$

Substitute numerical values:

$L=\frac{(2.0\mathrm{s}{)}^2(9.80\mathrm{s}{)}^2}{4{\mathrm{Ï€}}^2}$

$=0.99\mathrm{m}$.