A 65.0 kg bungee jumper steps off abridge with a light bungee cord tied to her a

Question

A 65.0 kg bungee jumper steps off abridge with a light bungee cord tied to her and to the bridge. Theunstretched length of the cord is 11.0m. She reaches the bottom of her motion 36.0 m below the bridge before bouncing back. Hermotion can be separated into an 11.0 mfree-fall and a 25.0 m section of simple harmonic oscillation.

The solution of time taken to have the cord stretched by 25m is tfrom -8.68=16.3cos(1.06t), but I don't understand why this is theanswer. I thought t in that equation is just the time taken toreach the position -8.68.

Explanation / Answer

take the phase as zero at maxiumum down ward extension.
we find what the phase was 25m higher when x= -8.68m x = A cost -8.68= 16.3m cos(1.06t/s) 1.06t/s = cos-1(-8.68/16.3m) =-1220 = -1220/57.30 = -2.13 rad (1rad = 57.30) t = -2.13 rad/1.06 = -2.01s then +2.01s is the time over which the springstretches.

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