- #1
OrisAble
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Well as above I've completely hit a blank for some uni work I've got.
If anyone could give me an idea of what i actually need to do here that would be great. I don't need someone do this for me. Just clear up what it is they want me to do.
"...Bungee jumping situation
Choose x= x(t) (positive donwards) to represent the location of the jumper's feet below the launching platform. Assume the bungee cord is of length L = 50m. Hooke's Law is to be used to represent the restoring force when the cable is stretched beyond length L, that is
Restoring force =
{0 x<=L
{k(x-L) x > L
Assume further, that their is an air resistance term opposing the motion of magnitude
cx(with the dot on top)|x(with the dot on top)|
Use Newton's equation of motion to determine the DE that describes the motion at times t>=0."
Thats what i don't understand. What does that actually want?[/quote]
Newton's equation is "force= mass times acceleration" or, for constant mass,
[tex]m\frac{d^2x}{dt^2}= F[/tex]
Here you are told that F is the sum of two separate forces: the restoring force which is 0 for x< L and -k(x-L) for x>L (the "-" is because the force is back toward x= 0) and the air resistance force which is -c dx/dt (again negative because the air resistance is always opposite to the direction of motion). Put those together.
Would be great if anyone could give me some direction..btw, ill be replying with a different account as i had forgotten my password and can't check hotmail accounts at uni ;).
If anyone could give me an idea of what i actually need to do here that would be great. I don't need someone do this for me. Just clear up what it is they want me to do.
"...Bungee jumping situation
Choose x= x(t) (positive donwards) to represent the location of the jumper's feet below the launching platform. Assume the bungee cord is of length L = 50m. Hooke's Law is to be used to represent the restoring force when the cable is stretched beyond length L, that is
Restoring force =
{0 x<=L
{k(x-L) x > L
Assume further, that their is an air resistance term opposing the motion of magnitude
cx(with the dot on top)|x(with the dot on top)|
Use Newton's equation of motion to determine the DE that describes the motion at times t>=0."
Thats what i don't understand. What does that actually want?[/quote]
Newton's equation is "force= mass times acceleration" or, for constant mass,
[tex]m\frac{d^2x}{dt^2}= F[/tex]
Here you are told that F is the sum of two separate forces: the restoring force which is 0 for x< L and -k(x-L) for x>L (the "-" is because the force is back toward x= 0) and the air resistance force which is -c dx/dt (again negative because the air resistance is always opposite to the direction of motion). Put those together.
Would be great if anyone could give me some direction..btw, ill be replying with a different account as i had forgotten my password and can't check hotmail accounts at uni ;).
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