The Weighing Machine - Isaac physics

Does the bit at the bottom help
https://phys.libretexts.org/Bookshelves/University_Physics/Book%3A_Introductory_Physics_-_Building_Models_to_Describe_Our_World_(Martin_Neary_Rinaldo_and_Woodman)/13%3A_Simple_Harmonic_Motion/13.02%3A_Vertical_spring-mass_system

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.

Sure, I hope it's easy to read. Thank you so much.

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

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:


Thank you so much for your help I really appreciate it

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:


Thank you so much for your help I really appreciate it

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.

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 Thanks!

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.

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