A spring of force constant `k` is cut into two equal parts, and mass 'm' is suspended from parallel combination of the springs. Then frequency of small oscillations is

Options

A. `f = (1)/(2pi) sqrt((K)/(m))`
B. `f = (1)/(2pi) sqrt((K)/(2m))`
C. `f = (1)/(2pi) sqrt((4K)/(m))`
D. `f = (1)/(2pi) sqrt((2K)/(m))`

Answer

C

Explanation

When the spring of force constant .K. is cut into two equal parts `rArr k_(2) = 2k`
In parallel combination `K^(1) = 2K + 2K = 4K`
`f = (1)/(2pi) sqrt((4k)/(m))`