Question 12
Writing the summation of force = ma for shm is similar to circular motion where the positive direction is always taken to point towards the equilibrium (instead of center of circle). For 12 b) the 2 forces are T and mg . At the highest point , point positive as down , hence mg -T = ma. At the lowest point , point positive up , hence T - mg = ma.
Do the same for question 14 b) which is at the lowest point , hence point positive up , T - mg = ma where T = ke where e is the extension of the spring.
Question 14 requires a good handling of the sine equation . Be fully aware of displacement of all the points first. Each spacing is 8 cm, amplitude is 20 cm . Work out the displacement of A, B, C ,D ,E and F . Displacement of C, D , B , E and A and F are 4cm , -4cm , 12 cm , -12 cm , 20 cm and -20 cm respecitively.
To find time taken to reach C from equilibrium , substitute into x = xosinwt taking x to be 4cm and xo = 20cm, t is 0.32s. Repeating the same for B and A gives 1.02 s and 2.5s respectively. Hence all the timings can be found:
1. AB : (2.5 - 1.02) x2 s = 2.96 s
2. BC : (1.02 - 0.32) x2 = 1.4 s
3. CD : (0.32 ) x 4 = 1.28 s
