91Ó°ÊÓ

The period $T$of an easy 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 that the acceleration of gravity .

Note that the amount of a straightforward pendulum depends only on its length and the magnitude of itsconstant of gravitation.

Itdoesn't rely upon the mass ofthe item hanging fromthe sting or the amplitude of vibration.

The mass of the ball is:$m=200$$g$

The ball is pulled to an angle $8.00°$.

The ball makes $10$oscillations in $12.0\mathrm{s}$.

To find the length of the string.

Model the ball string system as a straightforward pendulum.

The ball makes 10 oscillations in $12.0\mathrm{s}$,so the beat cycle is as follows:

$T=\frac{12.0\mathrm{s}}{10}$

$=1.20s$

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

$L=\frac{T^{2}g}{4{\mathrm{Ï€}}^2}$

Substitute numerical values:

$L=\frac{(1.20\mathrm{s}{)}^2\left(9.80\mathrm{m}/{\mathrm{s}}^2\right)}{4{\mathrm{Ï€}}^2}$

=0.357m.