Final answer:
If the period of rotation is T2, the ratio of T1:T2 is 0.724: 0.314 which can be written as 2.31 approximately.
Step-by-step explanation:
The period of oscillation of a simple pendulum is given by the formula T = 2π√(L/g), where L is the length of the pendulum and g is the acceleration due to gravity. Substituting the given values, we get T1 = 2π√(1.3/10) = 0.724 s.
When the bob is pulled and released to move in a horizontal circle of radius 50.0 cm, it experiences a centripetal force that is provided by the tension in the string. The period of rotation T2 is given by the formula T2 = 2π√(r/g), where r is the radius of the circle. Substituting the given values, we get T2 = 2π√(0.5/10) = 0.314 s.
Therefore, the ratio of T1:T2 is T1/T2 = 0.724/0.314 = **2.31** (approx).