Q. A mass $m$ is suspended from a spring of length $l$ and force constant $K.$ The frequency of vibration of the mass is $f_{1}.$ The spring is cut into two equal parts and the same mass is suspended from one of the parts. The new frequency of vibration of mass is $f_{2}.$ Which of the following relations between the frequencies is correct
A) $f_{1} = \sqrt{2}f_{2}$
B) $f_{1} = f_{2}$
C) $f_{1} = 2f_{2}$
D) $f_{2} = \sqrt{2}f_{1}$
Explanation
(d) When spring is cut into two equal parts then spring constant of each part will be $2K$ and so using $n \propto \sqrt{K}$, new frequency will be $\sqrt{2}$ times, $i.e.\ f_{2} = \sqrt{2}\ f_{1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.