Asked by please help
An 73-kg bungee jumper is enjoying an afternoon of jumps. The jumper's first oscillation has an amplitude of 14 m and a period of 6 s. sprint constant is 80.1 n/m and max speed is 14.7 m/s. Treating the bungee cord as a spring with no damping, calculate each of the following:
the time for the amplitude to decrease to 2 m (with air resistance providing the damping of the oscillations at 8.3 kg/s
the time for the amplitude to decrease to 2 m (with air resistance providing the damping of the oscillations at 8.3 kg/s
Answers
Answered by
drwls
Use the formula for an underdamped oscillator than can be found at:
http://hyperphysics.phy-astr.gsu.edu/hbase/oscda.html
What you call the "sprint constant" is the spring constant of the cord.
http://hyperphysics.phy-astr.gsu.edu/hbase/oscda.html
What you call the "sprint constant" is the spring constant of the cord.
Answered by
please help
there are two unknowns in that equation..how do i solve it?
Answered by
drwls
The question seems to contradict itself. They provide a damping constant but tell you to ignore "spring" damping. Damping is what the problem is all about.
You want the amplitude to decrease by a factor of 1/7, from 14 to 2. The exponential factor in front of the sine term will tell you how long that takes, in terms of the damping constant.
You want the amplitude to decrease by a factor of 1/7, from 14 to 2. The exponential factor in front of the sine term will tell you how long that takes, in terms of the damping constant.
Answered by
please help
do you know what would be the value of x?
Answered by
Elena
The equation of damped oscillations is
(d2x/dt2) +2•β• (dx/dt) +(ωo)^2•x = 0, where
β =r/2•m = 8.3/2•73 = 0.194
14/2= 7 = Ao/A=Ao/Ao•e^(- β•t) = e^ (β•t),
β•t = ln7
t= ln7/ β = 1.94/0.194 = 10 s.
(d2x/dt2) +2•β• (dx/dt) +(ωo)^2•x = 0, where
β =r/2•m = 8.3/2•73 = 0.194
14/2= 7 = Ao/A=Ao/Ao•e^(- β•t) = e^ (β•t),
β•t = ln7
t= ln7/ β = 1.94/0.194 = 10 s.
Answered by
no idea
Thank you Elena
Answered by
Adri
your B is wrong, 8.3/2*73=0.05685
which would make your time
t=ln7/0.05685=34.23sec
which would make your time
t=ln7/0.05685=34.23sec
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