A simple spring mass vibrating system has a natural frequency of f_n. If the spring stiffness is halved and mass is double, then the natural frequency will become

Explanation:

The natural frequency of a spring-mass system is given by,

\(f_n = \frac {\omega_n}{2\pi}\)   and  \(\omega_n = \sqrt {\frac km} \)

where k = spring stiffness and m = mass 

If spring stiffness is halved and mass is doubled

then the natural frequency is given by

\(\omega_n' = \sqrt {\frac {0.5k}{2m}}\)

\(\omega_n'= \sqrt {\frac {k}{4m}}\)

\(\omega_n'= \frac 12 \sqrt {\frac {k}{m}}\)

\(\omega_n'=\frac {w_n}{2} \)

\(f_n' = \frac {f_n}{2}\)

Hence, the natural frequency will become half.