In figure `S_(1)` and`S_(1)` are identical springs. The oscillation frequency of the mass `m` is `f`. if one spring is removed, the frequency will become

Options

A. `fxxsqrt(2)`
B. `(f)/(sqrt(2))`
C. `fxx2`
D. f

Answer

B

Explanation

Originally the springs are connected in parallel.
`therefore` Their equivalent spring constant =2K
`therefore` The oscillation frequency (f) of the mass m is given by
`f=(1)/(2pi) sqrt((2K)/(m))`
When one spring is removed, the spring constant will be only `K and f '=(1)/(2pi) sqrt((K)/(m))`
`therefore (f')/(f)= sqrt((K)/(m)xx(m)/(2K))=(1)/(sqrt(2)) therefore f'=(f)/(sqrt(2))`