High school Physics - Simple Harmonic Motion

[bigbosswilly]

Homework Statement

Penny (m=31kg) is learning to fly. She is attached to a 9.5m long bungee cord that has a spring constant of 72N/m. She jumps off the platform with an initial velocity of 3.5m/s [up]. Calculate the maximum distance below the platform that penny will reach (before bouncing back up).

Relevant Equations

Energy conservation, Forces equations, Energy equations

I started off by finding when Fg=Fx:

(72)(x)=(31)(9.8)
x=4.2193m

After this I'm stuck and have a few things I'm confused about:

When the penguin's jumping, is there elastic energy? (because the rope's getting compressed? Or maybe not). Also, I know you can use energy conservation, but there's too many variables and I can't solve it.

For example, I tried:

Et1= Et2
Eg + Ek = Ee (when the penguin's jumping and at the maximum distance below ledge with reference level set there as well, so no Eg)
(9.8)(31)(h) + (0.5)(31)(-3.5)^2 = (0.5)(72)(x^2)

I don't have the answer to this problem, sorry, but I was still hoping someone can help guide me through.

 

[ehild]

I started off by finding when Fg=Fx:

(72)(x)=(31)(9.8)
x=4.2193m

This would be true in equilibrium. But Penny moves up and down at the end of the cord.

After this I'm stuck and have a few things I'm confused about:

When the penguin's jumping, is there elastic energy? (because the rope's getting compressed? Or maybe not).

No, the cord can not be compressed. Have you seen bungee-jumping? the cord is slack initially. See

Also, I know you can use energy conservation, but there's too many variables and I can't solve it.

For example, I tried:

Et1= Et2
Eg + Ek = Ee (when the penguin's jumping and at the maximum distance below ledge with reference level set there as well, so no Eg)
(9.8)(31)(h) + (0.5)(31)(-3.5)^2 = (0.5)(72)(x^2)

I don't have the answer to this problem, sorry, but I was still hoping someone can help guide me through.

Energy conservation is a good start, but what is Penny's state at the start of the jump, and what is it at the deepest point? What energies it has? And please explain your notations, x and h.

 

[bigbosswilly]

This would be true in equilibrium. But Penny moves up and down at the end of the cord.

No, the cord can not be compressed. Have you seen bungee-jumping? the cord is slack initially. See

Energy conservation is a good start, but what is Penny's state at the start of the jump, and what is it at the deepest point? What energies it has? And please explain your notations, x and h.

Sorry, I'm not exactly sure how to use this site yet and the quoting functions so I hope this simple reply format is ok.

For the x=4.1m, that was me trying to find the amplitude of the system (thinking back on it however, it wasn't helpful). The fact that there's no elastic energy makes a LOT more sense haha (great video too!). For the "h" and "x" variables, I set the reference point at the ledge so "h" represents the height Penny jumps above the ledge and "x" represented the stretching distance at the maximum distance at the bottom.

So I've been thinking a bit more, and managed to find an answer this way but am not sure if it is correct:

I first used kinematics to find the velocity of Penny at x=0

d= 9.5m
vi= -3.5m/s
a= 9.8m/s^2
vf= ?

I got final velocity equal to approximately 13.64 m/s [down]

I then set my reference point at maximum distance below the ledge, and set the energy of Penny when jumping equal to the energy of Penny at the maximum distance below the ledge:

Et1 = Et2
Eg + Ek = Ee
(31)(9.8)(x) + (0.5)(31)(13.64)^2 = (0.5)(72)(x^2)
303.8x + 2886.1 = 36x^2
0 = 36x^2 - 303.8x - 2886.1

I solved the quadratic and got two answers, 14m and -5m, the latter which I rejected. The maximum distance would be 9.5m + 14m, therefore the max distance would be 23.6m

Would this be correct?

 

[Master1022]

Sorry, I'm not exactly sure how to use this site yet and the quoting functions so I hope this simple reply format is ok.

For the x=4.1m, that was me trying to find the amplitude of the system (thinking back on it however, it wasn't helpful). The fact that there's no elastic energy makes a LOT more sense haha (great video too!). For the "h" and "x" variables, I set the reference point at the ledge so "h" represents the height Penny jumps above the ledge and "x" represented the stretching distance at the maximum distance at the bottom.

So I've been thinking a bit more, and managed to find an answer this way but am not sure if it is correct:

I first used kinematics to find the velocity of Penny at x=0

d= 9.5m
vi= -3.5m/s
a= 9.8m/s^2
vf= ?

I got final velocity equal to approximately 13.64 m/s [down]

I then set my reference point at x=0, and set the energy of Penny when jumping equal to the energy of Penny at the maximum distance below the ledge:

Et1 = Et2
Eg + Ek = Ee
(31)(9.8)(x) + (0.5)(31)(13.64)^2 = (0.5)(72)(x^2)
303.8x + 2886.1 = 36x^2
0 = 36x^2 - 303.8x - 2886.1

I solved the quadratic and got two answers, 14m and -5m, the latter which I rejected. The maximum distance would be 9.5m + 14m, therefore the max distance would be 23.6m

Would this be correct?

Right method, but you should think about your EPE term. More specifically, does it start storing energy as soon as she drops (i.e. is it at all slack during the descent)? Hint: strings store EPE when they are stretched beyond their initial length (I am assuming you do indeed mean a string rather than a spring)? You should be able to make some quick adjustments and solve from there.

 

[bigbosswilly]

Right method, but you should think about your EPE term. More specifically, does it start storing energy as soon as she drops (i.e. is it at all slack during the descent)? Hint: strings store EPE when they are stretched beyond their initial length (I am assuming you do indeed mean a string rather than a spring)? You should be able to make some quick adjustments and solve from there.

I think it does not start storing energy as she drops? I assumed that the string would not be compressed or stretched for 9.5m down as it is the length of the string.

 

[Master1022]

I think it does not start storing energy as she drops? I assumed that the string would not be compressed or stretched for 9.5m down as it is the length of the string.

Oh my apologies, I see you have gone through and set x at equilibrium. In that case, your answer should be correct. However, I think it would be much easier to go straight from the top to the bottom (initial KE + GPE loss ---> EPE gain) so you wouldn't need to worry about this extra middle calculation. That is often the great thing about energy arguments- you can go from beginning to end and not necessarily worry about what happened in between.

 

[bigbosswilly]

Oh my apologies, I see you have gone through and set x at equilibrium. In that case, your answer should be correct. However, I think it would be much easier to go straight from the top to the bottom (initial KE + GPE loss ---> EPE gain) so you wouldn't need to worry about this extra middle calculation. That is often the great thing about energy arguments- you can go from beginning to end and not necessarily worry about what happened in between.

Thank you very much for your help!