Question

Plz answer its very urgent.........
Find the Time Period of the given SHM.
I will mark the best answer as THE BRAINLIEST

Answer 1

   This is a Spring mass SHM system. Here there are totally 8 springs. Four springs are joined into one long spring.  There are free horizontal supports at the joints of the springs. Assume that they are light (massless). Assume that they remain horizontal.

   Assume that each of the 8 springs has a spring constant = k.

Case (1)

   Let us first take the case of four springs connected in a Rhombus shape like shown in the figure. Let the mass m be connected to the lower two springs. Let the springs make an angle Ф with the vertical.

   Let there be a restoration force T in each spring.  Due to T each spring extends by x units from its equilibrium length. Then T = k x.  Then the mass m moves down vertically by 2 x CosФ.

Vertical displacement = y = 2 x CosФ

Equation of motion:    m ÿ = m d²y/dt² = - 2 T CosФ + m g
                                   m ÿ = - 2 k x CosФ + m g
                                          = - k y + m g
                                      ÿ = - k/m * y + g

Hence  ω² = k/m  and  Time period = 2π √(m/k)

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case 2)   Let us take all 8 springs as shown in the first figure.

Here  the restoration force in each spring = T.  Each spring extends by x from its equilibrium position.

          T = k x

The vertical displacement of the mass m be y.
           y = 4 x Cos Ф

Equation of motion:    m ÿ = - 2 T CosФ + m g
                                   m ÿ =  - 2 k x CosФ + m g
                                   m ÿ =  - k y/2  + m g
                                       ÿ = - k/(2m) y + g

Hence the angular frequency :  ω² = k/2m
     Time period T = 2π √(2m/k)