Suppose that a piston is moving straight up and down and that its position at time t seconds iss = Acos(2πbt)with A and b positive. The value of A is the amplitude of the motion and b is the frequency(the number of times the piston moves up and down each second). What effect does doubling the frequency have on the piston's velocity and jerk?Follow up question, why does machinery break when you run it too fast?
f = 1/P, where f is the frequency and P is the period P = 2π/(2πb) = 1/b, so P = b Let v = velocity v = ds/dt = -A sin (2πbt) ? 2πb = -2πAb sin (2πbt) Let f* = 2/b, so P* = 2π/(2/b) = bπ v* = -2πAb sin (πbt) The amplitude of the new velocity is 2πb times the original amplitude. Let a = acceleration a = dv/dt = -4π^2Ab^2 cos (2πbt) Let j = jerk j = da/dt = 8π^3Ab^3 sin (2πbt) a* = 8π^3Ab^3 sin (πbt) The amplitude of the new jerk is 8π^3b^3 times the original amplitude. An increased jerk means the machinery is subjected to greater stress, which causes it to break down.
Q1 === If you double the frequency, this means that you are doubling the number of times that the cylinder goes up, hits it's highest point, turns around and comes back down again in a second. You must be doubling the distance the piston travels. The time is one second, so effectively you are doubling the speed. Q2 === Jerk in this sense is one of those wonderful descriptive words that works in the real world, but is hard to put into a formula. I wonder if you have to do that. Certainly the force increases because the turn around time is much smaller when the frequency is doubled. That means time in this formula F = m*(vf - vi)/t is much smaller and that makes the force larger. Q3 == So machinery must break down because the force on moving the piston up and down has become much larger.
