Two springs of force constants `k_(1)` and `k_(2)`, are connected to a mass m as shown. The frequency of oscillation of the mass is `f`. If both `k_(1)` and `k_(2)` are made four times their original values, the frequency of oscillation becomes

Options

A. `f//2`
B. `f//4`
C. `4f`
D. `2f`

Answer

D

Explanation

The orginal frequency of osillation
`f = (1)/(2pi) sqrt((k_(1) + k_(2))/(m))`
On increasing the `k_(1)` and `k_(2)` by `4` times, then `f`
`f' = (1)/(2pi) sqrt((4(k_(1)k_(2))/(m)) = 2f`