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The Weighing Machine - Isaac physics

Im really stuck on part D of this question. Could someone please help?
https://isaacphysics.org/questions/weigh_machine?board=0c11bd1d-f8f3-4dbf-a972-e980cd6f1e4b

I've done the first three parts: a)omega =(g/alphaM)^0.5
b) 2M c)pi(alphaM/g)^0.5

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Reply 3
Original post by amell1234
I've read that and tried the question again but I'm still not getting the right answer?

Can you upload what you did for the earlier parts? I'll have a look at it later this afternoon if it's not been sorted by someone.
Reply 4
Original post by mqb2766
Can you upload what you did for the earlier parts? I'll have a look at it later this afternoon if it's not been sorted by someone.

Sure, I hope it's easy to read. Thank you so much.
3DADCA69-4948-4051-AF96-2C65CD50C6A5_1_201_a.jpeg
Reply 5
Original post by amell1234
Sure, I hope it's easy to read. Thank you so much.
3DADCA69-4948-4051-AF96-2C65CD50C6A5_1_201_a.jpeg

Working it through at the moment, and it seems like they want you to split the motion up into two regions
* x < M0 equilibrium position
* x >= M0 equilibrium position
So effectively you'll have two half periods added together. The spring constant will be different in each region.
(edited 3 years ago)
Reply 6
Original post by mqb2766
Working it through at the moment, and it seems like they want you to split the motion up into two regions
* x < M0 equilibrium position
* x >= M0 equilibrium position
So effectively you'll have two half periods added together. The spring constant will be different in each region.

Thank you so much. That is what I have tried but I will try again and see if I can find any mistakes
Reply 7
Original post by amell1234
Thank you so much. That is what I have tried but I will try again and see if I can find any mistakes

Again, can you upload what you've tried? I'm fairly tired today and plodding ...
Reply 8
Original post by mqb2766
Again, can you upload what you've tried? I'm fairly tired today and plodding ...

I found this video on youtube https://www.youtube.com/watch?v=ZpmCZLypUxQ&ab_channel=IsaacPhysics
So i did what they tried but i carried on getting that k1=k2=g/alpha1

So then I thought it would have one half of the motion where there is just k1 and then the other half where there is k1+k2 and added the two half periods together but that didn't work.
Here are my workings:

Screenshot 2020-11-19 at 19.30.23.jpg
Thank you so much for your help I really appreciate it :smile:
Reply 9
Original post by amell1234
I found this video on youtube https://www.youtube.com/watch?v=ZpmCZLypUxQ&ab_channel=IsaacPhysics
So i did what they tried but i carried on getting that k1=k2=g/alpha1

So then I thought it would have one half of the motion where there is just k1 and then the other half where there is k1+k2 and added the two half periods together but that didn't work.
Here are my workings:

Screenshot 2020-11-19 at 19.30.23.jpg
Thank you so much for your help I really appreciate it :smile:

Cheers. I'm zonked for the evening, will get it sorted in the morning. The approach is about right, just need to get the second spring sorted properly.
Original post by amell1234
I found this video on youtube https://www.youtube.com/watch?v=ZpmCZLypUxQ&ab_channel=IsaacPhysics
So i did what they tried but i carried on getting that k1=k2=g/alpha1

So then I thought it would have one half of the motion where there is just k1 and then the other half where there is k1+k2 and added the two half periods together but that didn't work.
Here are my workings:

Screenshot 2020-11-19 at 19.30.23.jpg
Thank you so much for your help I really appreciate it :smile:

As a bit of a bump, it looks like the second spring, K2, is shorter than the one in parts A-C and that's part of the solution.
It looks like it's easier to consider the displacement from the top of the second spring so for the first spring, the displacement is
x+X
Where X is the difference in spring length = alpha_1*M_0 and x is the subsequent displacement for both springs. For a mass M > M_0, you should get for the equilibrium displacement
x = alpha_2*(M - M_0)

Ok with that? There's an elegant simplification when you solve for K2 and it's related to
https://www.johndcook.com/blog/2009/03/26/springs-and-resistors-harmonic-mean/
(edited 3 years ago)
Reply 11
Original post by mqb2766
As a bit of a bump, it looks like the second spring, K2, is shorter than the one in parts A-C and that's part of the solution.
It looks like it's easier to consider the displacement from the top of the second spring so for the first spring, the displacement is
x+X
Where X is the difference in spring length = alpha_1*M_0 and x is the subsequent displacement for both springs. For a mass M > M_0, you should get for the equilibrium displacement
x = alpha_2*(M - M_0)

Ok with that? There's an elegant simplification when you solve for K2 and it's related to
https://www.johndcook.com/blog/2009/03/26/springs-and-resistors-harmonic-mean/

OMG thank you so much. I used that to find k1+K2=g/alpha2 which gave me the right answer. :smile:
Original post by amell1234
OMG thank you so much. I used that to find k1+K2=g/alpha2 which gave me the right answer. :smile:

It's a bit of a slog to solve for K2, but a good exercise as it's ~ the "harmonic difference" of the two alphas.
Hi there, sorry to jump on this thread, but I am stuck on the final part, E. So far, I drew the force - extension graph, and came to the conclusion that M_0 + 1/a_2 (A - a_1 * M_0) is the maximum reading. I then used conservation of energy at the amplitude of oscillation:

M_0 * g * A = 1/2 K_1 * A^2 + 1/2 K_2 * (A-a_1M_0)

I then finally came to a position where I could complete the square to find A - a_1 * M_0 to be sqrt ( M_0 * a_2 ). However I am not getting the correct answer... Hopefully I made an error in my working, not my initial equations! Any help appreciated and I can upload my working if that's helpful :smile: Thanks!
Original post by tande33
Hi there, sorry to jump on this thread, but I am stuck on the final part, E. So far, I drew the force - extension graph, and came to the conclusion that M_0 + 1/a_2 (A - a_1 * M_0) is the maximum reading. I then used conservation of energy at the amplitude of oscillation:

M_0 * g * A = 1/2 K_1 * A^2 + 1/2 K_2 * (A-a_1M_0)

I then finally came to a position where I could complete the square to find A - a_1 * M_0 to be sqrt ( M_0 * a_2 ). However I am not getting the correct answer... Hopefully I made an error in my working, not my initial equations! Any help appreciated and I can upload my working if that's helpful :smile: Thanks!

Do upload what you did for d) and e)
Not sure how you did part b) but a very simple argument says the max displacement is twice the equilibrium position.
Sorry for the mess, I re wrote out my working so it was clearer, and then came to the right answer. I also included how I got K_2 and K_1 but didn't include my whole answer to d as it was clear once you had K_1 and K_2. Thanks again!
Original post by tande33
Sorry for the mess, I re wrote out my working so it was clearer, and then came to the right answer. I also included how I got K_2 and K_1 but didn't include my whole answer to d as it was clear once you had K_1 and K_2. Thanks again!

So you're sorted? Looks like it, but haven't checked line by line.
Original post by mqb2766
So you're sorted? Looks like it, but haven't checked line by line.

Yes I believe so! Thanks again, Are you a maths teacher btw? You are extremely helpful and very knowledgable!
Original post by tande33
Yes I believe so! Thanks again, Are you a maths teacher btw? You are extremely helpful and very knowledgable!

Flattery will get you far :smile:
You realize I didn't do anything, you solved your own problem.

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