A ball tied to the end of a string of length 10cm is rotated with constant speed of 2m/s in a circle about the origin. Find the equation of projection of the ball on the x-axis. At t=0, the ball makes an angle of 45° with the x-axis & is travelling in anticlockwise direction.(a) 0.1sin(20t + 3π/4)(b) 0.1cos(20t + π/4)(c) 10sin(20t + π/4)(d) 10cos(20t – 3π/4)This question was posed to me during an interview.This interesting question is from Oscillations topic in division Oscillations of Physics – Class 11

A ball tied to the end of a string of length 10cm is rotated with cons

1.	A ball tied to the end of a string of length 10cm is rotated with constant speed of 2m/s in a circle about the origin. Find the equation of projection of the ball on the x-axis. At t=0, the ball makes an angle of 45° with the x-axis & is travelling in anticlockwise direction.(a) 0.1sin(20t + 3π/4)(b) 0.1cos(20t + π/4)(c) 10sin(20t + π/4)(d) 10cos(20t – 3π/4)This question was posed to me during an interview.This interesting question is from Oscillations topic in division Oscillations of Physics – Class 11
Answer» Correct OPTION is (a) 0.1sin(20t + 3π/4) Easy explanation: Let the equation of motion be: x = A sin(WT + a).The amplitude is 0.1m,w = 2/0.1 = 20s^-1.a will be either π/4 or 3π/4 since at t=0, x = A/√2.The ball is travelling in anticlockwise direction so its VELOCITY of projection at t=0 should be towards the origin and therefore negative.v = Aw cos(wt + a). At t=0, v is negative, so a will be 3π/4.Therefore EQN is: x = 0.1sin(20t + 3π/4)

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