Q. A mass $m$ is suspended by means of two coiled spring which have the same length in unstretched condition as in figure. Their force constant are $k_{1}$ and $k_{2}$ respectively. When set into vertical vibrations, the period will be
A) $2\pi\sqrt{\left( \frac{m}{k_{1}k_{2}} \right)}$
B) $2\pi\sqrt{m\left( \frac{k_{1}}{k_{2}} \right)}$
C) $2\pi\sqrt{\left( \frac{m}{k_{1} - k_{2}} \right)}$
D) $2\pi\sqrt{\left( \frac{m}{k_{1} + k_{2}} \right)}$
Explanation
(d) Given spring system has parallel combination, so
$K_{eq} = K_{1} + K_{2}$ and time period $T = 2\pi\sqrt{\frac{m}{(K_{1} + K_{2})}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.