Q. A weightless spring of length $60\ cm$ and force constant $200\ N/m$ is kept straight and unstretched on a smooth horizontal table and its ends are rigidly fixed. A mass of $0.25\ kg$ is attached at the middle of the spring and is slightly displaced along the length. The time period of the oscillation of the mass is
A) $\frac{\pi}{20}s$
B) $\frac{\pi}{10}s$
C) $\frac{\pi}{5}s$
D) $\frac{\pi}{\sqrt{200}}s$
Explanation
(a) System is equivalent to parallel combination of springs
$\therefore K_{eq} = K_{1} + K_{2} = 400\ $and
$T = 2\pi\sqrt{\frac{m}{K_{eq}}} = 2\pi\sqrt{\frac{0.25}{400}} = \frac{\pi}{20}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.