A particle of mass `m` is attached to three identical springs of spring constant `k` as shwon in figure. The time period of vertical oscillation of the particle is

Options

A. `2pisqrt((m)/(k))`
B. `2pisqrt((m)/(2k))`
C. `2pisqrt((m)/(3k))`
D. `pisqrt((m)/(k))`

Answer

B

Explanation

When the particle of mass `m` is pushed by distance `y` in the direction of `A`,the spring `A` will be compressed by `y`, while spring `B` and `C` each will be stretched by `y^(')(=cos 45^(@))` , Total restoring force on the mass `m` along `AO` is
`-F=F_(A)+F_(B)cos 45^(@)+F_(C)cos 45^(@)`
`=ky+ky^(')cos45^(@)+ky^(')cos45^(@)`
`=ky+2ky^(')cos45^(@)`
`=ky+2k(y cos 45^(@))cos 45^(@)=2ky`
or `F=-2ky` ...(i)
As `Fprop y` and `F` is directed towards the equilibrium position `O`. Hence mass `m` if left free, will execute lkinear SHM. Comparing (i) with `F=-Ky`, we find that
Spring factor, `K=2k`
Here, inertia factor `=` mass of particle `=m`
`:. ` Time period, `T=2pisqrt((i n ertia fact o r (m) )/(spri ng fact o r(K)))`
`=2pisqrt((m)/(2k))`