Differential Equations Bungee Jumping question

The general solution of harmonic oscillator equation is a combination of $sin omega t$ and $cos omega t$ where $omega = sqrt $. Since $x(0)=0$, we only have the sine. So, $x(t)=Asinomega t$. To find $A$, you can use the relation $x'(0)=Aomega$, where $x'(0)$ can be found from energy consideration. Indeed, jumping from height $x=-h$, the person acquires kinetic energy $mgh $ by the time that $x=0$. Hence, $x'(0)=sqrt$.

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Becky

Updated on August 01, 2022

Comments

So I was given a problem about bungee jumping,and it’s obviously important to know in advance how far the cord of unstretched length L will stretch for a given weight of person. Consider the cord to be a weak spring with constant k and the person is a point mass m. Air resistance and pendulum motion are ignored. Point x=0 would be the point where the person freely fell until the entire slack of the cord is extended to length L. After the person passes x=0, the cord is stretched x(t).
Simple enough right? Well I was having trouble:

find a model for x(t) defined only on the first interval of time 0=0. Solve for x(t) and then determine maximum elongation. I do know that I should use x(t)=A sin(ωt+ φ) but after than I’m stuck. Could anyone possibly help me?

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Maybe the The Physics Of Bungee Jumping will help? See equation 13 and the substitution they make for a. Regards

Differential Equations Bungee Jumping question

The general solution of harmonic oscillator equation is a combination of $sin omega t$ and $cos omega t$ where $omega = sqrt $. Since $x(0)=0$, we only have the sine. So, $x(t)=Asinomega t$. To find $A$, you can use the relation $x'(0)=Aomega$, where $x'(0)$ can be found from energy consideration. Indeed, jumping from height $x=-h$, the person acquires kinetic energy $mgh $ by the time that $x=0$. Hence, $x'(0)=sqrt$.

Related videos on Youtube

Differential Equations Application: Bungee Jumping

How to Solve a Bungee Jumping Problem (using Energy )

Hooke's Constant - Bungee Jumping Example 1

MINUTE PHYSICS: the physics of bungee jumping

Hooke's Constant - Bungee Jumping Example 3 (Part 1)

Becky

Updated on August 01, 2022

Comments

So I was given a problem about bungee jumping,and it’s obviously important to know in advance how far the cord of unstretched length L will stretch for a given weight of person. Consider the cord to be a weak spring with constant k and the person is a point mass m. Air resistance and pendulum motion are ignored. Point x=0 would be the point where the person freely fell until the entire slack of the cord is extended to length L. After the person passes x=0, the cord is stretched x(t).
Simple enough right? Well I was having trouble:

find a model for x(t) defined only on the first interval of time 0=0. Solve for x(t) and then determine maximum elongation. I do know that I should use x(t)=A sin(ωt+ φ) but after than I’m stuck. Could anyone possibly help me?

Maybe the The Physics Of Bungee Jumping will help? See equation 13 and the substitution they make for a. Regards

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Differential Equations Bungee Jumping question

The general solution of harmonic oscillator equation is a combination of $sin omega t$ and $cos omega t$ where $omega = sqrt $. Since $x(0)=0$, we only have the sine. So, $x(t)=Asinomega t$. To find $A$, you can use the relation $x'(0)=Aomega$, where $x'(0)$ can be found from energy consideration. Indeed, jumping from height $x=-h$, the person acquires kinetic energy $mgh $ by the time that $x=0$. Hence, $x'(0)=sqrt$.

Related videos on Youtube

Differential Equations Application: Bungee Jumping

How to Solve a Bungee Jumping Problem (using Energy )

Hooke's Constant - Bungee Jumping Example 1

MINUTE PHYSICS: the physics of bungee jumping

Hooke's Constant - Bungee Jumping Example 3 (Part 1)

Becky

Updated on August 01, 2022

Comments

So I was given a problem about bungee jumping,and it’s obviously important to know in advance how far the cord of unstretched length L will stretch for a given weight of person. Consider the cord to be a weak spring with constant k and the person is a point mass m. Air resistance and pendulum motion are ignored. Point x=0 would be the point where the person freely fell until the entire slack of the cord is extended to length L. After the person passes x=0, the cord is stretched x(t).
Simple enough right? Well I was having trouble:

find a model for x(t) defined only on the first interval of time 0=0. Solve for x(t) and then determine maximum elongation. I do know that I should use x(t)=A sin(ωt+ φ) but after than I’m stuck. Could anyone possibly help me?

Maybe the The Physics Of Bungee Jumping will help? See equation 13 and the substitution they make for a. Regards

Source https://9to5science.com/differential-equations-bungee-jumping-question

Source https://9to5science.com/differential-equations-bungee-jumping-question

Source https://9to5science.com/differential-equations-bungee-jumping-question

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