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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. |
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Answer» 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)))` |
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