Two spring have force constants `k_(1)` and `k_(2)` respectively. They are attached to a mass m and two fixed supports as shwon in figure. If the surface is frictionless, fing the time period of oscillations. What is the spring factore of this combination.

Explanation

If the mass m is displaced a little through distance x towards right hand side, the spring `k_(1)` gets compressed and spring `k_(2)` gets stretched. Due to it, the restoring froces `F_(1)` and `F_(2)` developed in two springs will be towards left, i.e., in the same direction.
Since `k_(1)` and `k_(2)` are the spring constants fo the two springs. hence,
`F_(1)=-k_(1)x` and `F_(2)=-k_(2)x`
Total restoring force,
`F=F_(1)+F_(2)=(-k_(1x)+(-k(2)x)`
`=(k_(1)+k_(2))x` ...(i)
i.e., `Fprop x`. This force F is directed towards equilibrium position of th ebody. If body is left free, it will execute linear SHM. If k is the force constant of a spring which is equivalent to the combination of the two spring constants as given above, then
`F=-ky` ...(ii)
Form (i) and (ii), `k=k_(1)+k_(2)`
Here, spring factor`=k_(1)+k_(2)`
Inertia factore `=` mass of the body`=m`
`:.` Time period , `T=2pisqrt(("inertia factor")/("spring factor"))`
`=2pisqrt((m)/(k_(1)+k_(2)))`