Q. A spring of spring constant $k$ is cut into two equal parts. A block of mass $m$ is attached with one part of spring. What is the frequency of the system if $\alpha$ is frequency of block with original spring?
A) $\sqrt{2}\ \alpha$
B) $\frac{\alpha}{2}$
C) $2\alpha$
D) $\alpha$
Explanation
(a) Spring constant of each part, $k^{'} = 2k$
Original frequency of system
$\alpha = \frac{1}{2\pi}\ \sqrt{\frac{k}{m}}$
where $m$ is the mass of the block.
New frequency of system
$\alpha' = \frac{1}{2\pi}\ \sqrt{\frac{k'}{m}}$
or $\alpha' = \frac{1}{2\pi}\ \sqrt{\frac{2k}{m}}$
$\therefore$ $\alpha^{'} = \sqrt{2}\alpha$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.