A particle of mass m is attached to four springs with spring constant k , k , 2 k and 2 k as shown in the figure. Four springs are attached to the four corners of a square and a particle is placed at the center. If the particle is pushed slightly towards any side of the square and released, the period of oscillation will be

Step-by-step solution

when the mass is displaced in the upward direction then the forces acting ar as shown in the diagram, F 1 = - k x = F 2   &   F 3 = - 2 k x = F 4   Referring to second diagram, F R = F 1 + F 3 2 + F 2 + F 4 2 = 3 2 k x   But since the particle performing SHM, ∴     F R = m ω 2 x ∴       m ω 2 x = 3 2 k x ∴       ω = 3 2 k m 2 π T = 3 2 k m ⇒ T = 2 π m 3 2 k