Given the number of oscillations and the given acceleration due to gravity, the length of the pendulum is 0.692m.
Given the data in the question;
Time period; [tex]T = 25.0s\ for \ 25cycle = \frac{25}{15} = 1.67s[/tex]
Using the simple pendulum equation of time:
[tex]T = 2\pi \sqrt\frac{L}{g} }[/tex]
Where T is the period of time, L is the length of the pendulum and g is the acceleration due to gravity( [tex]9.8m/s^2[/tex] )
We make "[tex]L[/tex]" the subject of the formula
[tex]\frac{L}{g} = \frac{T^2}{4\pi ^2}\\\\L = \frac{T^2}{4\pi ^2}\ *\ g[/tex]
We substitute our given values into the equation
[tex]L = \frac{(1.67s)^2}{4*\pi ^2} \ *\ 9.8m/s^2\\\\L = \frac{2.7889s^2}{4*\pi ^2} \ *\ 9.8m/s^2\\\\L = 0.07064s^2\ *\ 9.8m/s^2\\\\L = 0.692m[/tex]
Therefore, given the number of oscillations and the given acceleration due to gravity, the length of the pendulum is 0.692m.
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