A simple pendulum, consists of an object suspended by a string. The object is assumed to be a particle. The string, with its top end fixed, has negligible mass and does not stretch. In the absence of air friction, the system oscillates by swinging back and forth in a vertical plane. If the string is 2.35 m long and makes an initial angle of 27.5° with the vertical, calculate the speed of the particle at the following positions.

Question

(a) at the lowest point in its trajectory
(b) when the angle is 15.0°

Answer 1

Energy conservation indicates that

M g L (1-cos A) + (1/2)M V^2
= M g L (1-cos Amax)
at any angle location A. M cancels out.

Your Amax angle is 27.5 degrees, so:

g L (1 - cosA) + V^2/2 = 0.11299 gL

Use that equation to solve for V at any angle A

Answer 2

Thank you..