Q. A weightless spring which has a 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 2 m is suspended from one part of spring. The frequency of oscillation will now become
A) N
B) 2n
C) $\frac{n}{\sqrt{2}}$
D) $n\ (2)^{1/2}$
Explanation
(a) $n = \frac{1}{2\pi}\sqrt{\frac{k}{m}};$
$n^{'} = \frac{1}{2\pi}\sqrt{\frac{k^{'}}{2m}} = \frac{1}{2\pi}\sqrt{\frac{2k}{2m}}\ \ \ \ \ \lbrack\because k^{'} = 2k\rbrack$
$\therefore\ \ \ n^{'} = n$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.