Some springs are combined in series and parallel arrangement as shown in the figure and a mass m is suspended from them. The ratio of their frequencies will be

Options

A. `1:1`
B. `2:1`
C. `sqrt(3):2`
D. `4:1`

Answer

C

Explanation

For `1^(st)` condition `k_(eff) = k/2`
For `2^(nd)` condition `k_(eff) = ((2k)(k))/(2k+k) = 2/3k`
`:. T = 2pisqrt((m)/(k)) :. f = 1/T = 1/2pi sqrt((k)/(m))`
`(f_(1))/(f_(2)) = sqrt((k//2)/(2k//3)) = (sqrt(3))/(2)`