Q. Two identical springs of constant $K$ are connected in series and parallel as shown in figure. A mass $m$ is suspended from them. The ratio of their frequencies of vertical oscillations will be
A) 2 :1
B) 1 :1
C) 1 :2
D) 4 :1
Explanation
(c) $n = \frac{1}{2\pi}\sqrt{\frac{K}{m}} \Rightarrow \frac{n_{S}}{n_{P}} = \sqrt{\frac{K_{S}}{K_{P}}} \Rightarrow \frac{n_{s}}{n_{p}} = \sqrt{\frac{\left( \frac{K}{2} \right)}{2K}} = \frac{1}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.