SHM in a horizontal plane: 4 springs

mmh37
Messages
56
Reaction score
0
a point mass is constrained to move in a horizontal plane and is attached to four fixed pegs by 4 light springs. The four pegs are arranged at the corners of a square of side root2*a. Each spring has a natural length of a/2 and a spring constant a. If the mass is displaced towards one corner through a small distance d<<a and is then released from rest, what is the angular frequencyof this SHM?

So far I have tried to resolve forced in x and y direction but that didn't work. This seems to be quite tough. Thanks for any hints.
 
Physics news on Phys.org
Do you have a diagram of the system at all?
 
well, from the question, I think this is how it looks like (note that the system is arranged in the horizontal plane):

Can anyone please try to help - this is really important (exams start in one week!). Thanks so much!
 

Attachments

  • federn.jpg
    federn.jpg
    11.8 KB · Views: 459
mmh37 said:
Each spring has a natural length of a/2 and a spring constant a.
I assume that's a typo and you meant a spring constant "k" (or something), not "a" (which is a length).

I also assume that if you only had one spring, with spring constant k, you would know how to find the angular frequency.

For this problem, what you have to do is find the effective spring constant. Hint: As the mass is displaced towards one corner, what is the restoring force on it?
 
Doc Al said:
I assume that's a typo and you meant a spring constant "k" (or something), not "a" (which is a length).[

That threw me also Doc, I didn't think that it could be a typo :rolleyes:

~H
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top