Question

A boy on a carnival bungee ride is released from rest at the bottom of the ride and oscillates up and down with an amplitude of 8 meters and a period of 4s. How long after he is released does he first get to a height 10m above the place where he started? [Hint: Choose the top of the poles as y=0, and measure positive downward.]

Answer 1

If amplitude of oscillatory motion is 8 m/s, then at the time of releasing from rest, position of the boy

from equilibrium position is -8 m

Hence equation for displacement is written as , y = 8 sin [ t - ( ) ]

angular speed = 2/T = 2/4= /2 , where T is period of oscillation

If he is at 10 m from starting point, displacement from equilibrium position is (10-8) = 2 m

Hence time is obtained from the following equation

2 = 8 sin [ t - ( ) ]   or    [ t - ( ) ] = sin^-1 ( 0.25 ) = 0.253 rad

Hence t = [ 0.253 + ( /2 ) ] / ( /2 ) = 1.161 s