✔ 最佳答案
(a) Angular frequency of the oscillation, w = sqrt[g/2] = sqrt(10/2) s^-1 = 2.236 s^-1[where sqrt=square-root, and g is the acceleration due to gravity, taken as 10 m/s2 for simplicity].Using y = (yo).sin(wt) as the equation describing the oscillation
where y is the angular displacementat time t
yo is the angular amplitude of oscillation (= 15.pi/180 radinas, where pi = 3.14159...)Hence, 5.pi/180 = (15.pi/180).sin(w.t1)
where t1 is the time taken for the pendulum bob to travel from the equilibrium position to angular amplitude of 5 degrees.
solve for t1 gives t1 = 0.152 s
Similary, 10.pi/180 = (15.pi/180).sin(w.t2)
where t2 is the time to travel from equilibrium position to the 10-degree position.
solve for t2 gives t2 = 0.326 s
Hence, time required to travel from 5-degree position to 10-degree position
= (0.326 - 0.152) s = 0.174 s
(b) At the 5-degree position.angular velocity = (15.pi/180).wcos(w.t1) = (15.pi/180).(2.236.)cos(2.236x0.152) S^-1 = 0.552 s^-1
hence, linear velocity = 0.552 x 2 m/s = 1.104 m/s
angular acceleration = w^2.[5.pi/180] s^-2 = (2.236)^2 x (5.pi/180) s^-2 = 0.436 s^-1
hence, acceleration = 0.436 x 2 m/s2 = 0.872 m/s2
You could use similar steps to calculate the velocity and acceleration at the 10-degree position, just be substituting t1 by t2, which takes the value of 0.326 s, as from part (a).
