Maximizing acceleration using only mechanical energy - In Ohanian, Ch. 8.2, problem #25, the question asks to find the point at which a bungee jumper is experiencing maximum acceleration, and what that acceleration is. Anyone know the procedure for arriving at these answers? I have the equation for SHO derived, but x'' and x are both unknowns. Haven't taken Diff Eq. yet....
MIT 8.01 Physics I Classical Mechanics, Fall 1999
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Maximum acceleration is where F(x) is greatest. You see from the formula: F(x) = -dU/dx = -687 -150x = 0, that when x becomes more negative F(x) is increasing. This is obvious, because as the cord is stretched, the force it applies becomes higher. Thus maximum acceleration is at the the bottom, i.e. at the turn, where v=0 and x=-14.7m.
F(x) = -687 - 150(-14.7) = 1518N
a = F(x)/m = 1518N/70kg = 22 m/s2 (-2.2g)