Four springs Oscillation

1. The problem statement, all variables and given/known data

Figure shows a particle of mass m attached with four identical springs each of length L0. Initial tension in each spring is F0.Neglect gravity calculate the period of small oscillations of the particle along a line perpendicular to the plane of figure. (Take F0 = (0.01 π2)N, m = 100 gm, L0 = 10cm)

2. Relevant equations

3. The attempt at a solution

In the equilibrium state length of the spring L = L0 + z ,where L0 is the original length and Z is the initial extension in the spring when in equilibrium.

Initial tension in each spring when in equilibrium = kz = F0

Suppose the mass is displaced by x amount upwards .In the attachment L’ represents the new length.


Force by a spring = k(L’-L0) = = k(L’-L+z) = k(L’-L)+kz

This force will have a a horizontal component and a a vertical component .The horizontal component will be cancelled by a similar component from the opposite string .

The net force from all four springs will be F = 4k(L’-L0 )sinθ = 4k(L’-L)sinθ + 4kzsinθ

sinθ = x/L’

So net force =4kx(L’-L)/L’ + 4kxz/L’

=4kx(1-L/L’) + 4kxz/L’

I somehow feel I am not approaching the problem correctly .

I would be grateful if somebody could help me with the problem.

Attached Images
File Type: gif four spring.GIF (3.7 KB)
File Type: gif spring1.GIF (2.3 KB)

Leave a Reply

Name *
Email *