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//4`
B. (b)4f
C. (c)2f
D. (d)`f//2`

Answer

C

Explanation

Here, ` K_(1) `and `k_(2)` are in parallel, so k=`k_(1)+k_(2)` The frequency `K_(1)` and `K_(2)` are made four times their original values,
K'=`4K_(1)+4K_(2)=4K_(2)=4(k_(1)+k_(2)) `
So the new frequency of oscillation is `f'=1/2pisqrt(4(k_(1)+k_(2))/m` `therefore ` `f'//f=sqrt(4)=2impliesf'=2f`