Two springs of force constants `K_(1) " and " K_(2)` are connected to a mass m as shown I the figure.

If both `K_(1) " and " K_(2)` are made four times their original values, the frequency of oscillation becomes,

Options

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

Answer

A

Explanation

The two springs are in parallel.
In the first case, the equivalent spring constant `K=K_(1)+K_(2)` and in the second case, the equivalent spring constant
`K=4(K_(1)+K_(2))`
`therefore` The frequency of oscillation
`f=(1)/(2pi) sqrt((K_(1)+K_(2))/(m))`
and `f'=(1)/(2pi) sqrt((4(K_(1)+K_(2)))/(m))`
`therefore (f')/(f)=sqrt((4(K_(1)+K_(2)))/(K_(1)+K_(2)))=2`
`therefore f'=2f`