Time period of Mass attached to 3 springs

[Smarty7]

Homework Statement

There are three springs A, B and C with equal force constants as k. Springs are placed such as the angle between two consecutive springs is 120 degrees. Calculate the time period of oscillation if the mass is slightly pushed towards C and released.

Homework Equations

Force from spring A = -kx
Force from spring B = -kx
Force from spring C = -kx

Vector sum of forces from spring A and B = kx
As the direction of resultant force vector is on same line as force from spring C.
Net Force = 2kx

Which lands me up on wrong answer.

The Attempt at a Solution

I think all springs would have the force -kx in the direction stated in the diagram above. I know that it is also the point I'm making mistake.(As vector sum of B and A springs comes out to be kx, which means total force would be 2kx)
My doubts are,
1) Would be helpful if you correct me there with where I'm wrong with forces from A and B, i can solve it after that.
2) Too silly to ask but when i do the vector sum and magnitude comes out to be \sqrt{k^{2}x^{2}}, should i always take it as kx or there is a possibility of -kx also? I know the negative sign implies the direction in single dimension and when dealing with vectors we get the direction from tan \alpha. Just wanted to clear the small doubt.

 

[zorro]

A small displacement 'x' in the direction of 'C' does not cause the same displacements for springs A and B resp. Try to find the component of 'x' in the direction of A and B.

 

[Abhishekdas]

I had a similar doubt so i am joining in the thread...Anyway Abdul Quadeer...can u just check what i did...I am getting extention of either of spring A or B to be xcos 60 ie x/2 (please refer to my attachment diagram)... So two forces each of magnitude kx/2 at 120degree to each other gives the net force due to these two to be kx/2 in the direction symetrical to A and B...So net force on mass is 3/2*kx...So noe i have the net spring constant so i can get the time period...Am i correct?

 

[zorro]

Yeah you are right.

 

[Abhishekdas]

Ok...thanks...