A massless spring, having force constant k, oscillates with frequency n when a mass m is suspended from it. The spring is cut into two equal halves and a mass `2m` is suspended from one half. The frequency of oscillation will now be

Options

A. n
B. 2n
C. `(n)/(sqrt(2))`
D. `n(2)^(1//2)`

Answer

A

Explanation

We have, `n = (1)/(2 pi)sqrt((k)/(m)),`
`n'=(1)/(2pi)sqrt((k')/(2m))=(1)/(2pi)sqrt((2k)/(2m))=n" "(becausek'=2k)`