Question

Two identical springs of spring constant k are attached to a block of mass m and to fixed supports as shwon in figure . Show that when the mass is displaced from its equilibrium position on either side, it executes a simple harmonic motion. Find th eperiod of oscillation.

Answer 1

Let the mass m be displaced by a small distance x to the right as shown in figure. Due to it, the spring on the left hand side gets stretched by length x and the spring on the right hand side gets compressed by length x. The forces acting on the mass due to springs are
`F_(1)=-kx` towards left hand side
`F_(2)=-kx` towards left hand side
Therefore, total restoring force on mass m is
`F=F(1)+F_(2)=-kx+(-kx)=-2kx` ...(i) Here `-ve` sign showns that force F is directed towards the equilibrium position O and `Fpropx. ` Therefore, if the mass m i s left free, it will execute linear SHM.
Comparing (i) with the relation,
`F=-Kx, we have `
Spring factor, `K=2k`
Here, inertia factor`=` mass of the block`=` m
As time period, `T=2pisqrt((I n e r tia fact o r)/(spri ngfact o r))`
`=2pisqrt((m)/(2k))`