The given 2.0 seconds period, T, of the small angle oscillation gives the radius of the hoop as approximately 0.5 meters.
The small angle oscillation approximation for the period is presented as follows;
[tex] T_{0} = 2 \cdot \pi \cdot \sqrt{ \frac{l}{g} } [/tex]
Where;
T0 = The period of oscillation = 2.0 s
l = The length of the oscillating object = The diameter of the hoop
g = Acceleration due to gravity
Therefore;
l = The diameter of the hoop = 2 × The radius of the hoop, r.
Which gives;
[tex] T_{0} = 2 \cdot \pi \cdot \sqrt{ \frac{2 \cdot r}{g} } [/tex]
Solving gives;
[tex] 2 \approx 2 \times \pi \cdot \sqrt{ \frac{2 \cdot r}{9.81} } [/tex]
[tex] r \approx \frac{9.81}{2}\times \left(\frac{1}{π} \right)^2 \approx 0.5 [/tex]
The radius of the hoop, r ≈ 0.5 m
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