Q. A mass $m$ is suspended from the two coupled springs connected in series. The force constant for springs are $K_{1}$ and $K_{2}.$ The time period of the suspended mass will be
A) $T = 2\pi\sqrt{\left( \frac{m}{K_{1} + K_{2}} \right)}$
B) $T = 2\pi\sqrt{\left( \frac{2m}{K_{1} + K_{2}} \right)}$
C) $T = 2\pi\sqrt{\left( \frac{m(K_{1} + K_{2})}{K_{1}K_{2}} \right)}$
D) $T = 2\pi\sqrt{\left( \frac{mK_{1}K_{2}}{K_{1} + K_{2}} \right)}$
Explanation
(c) In series $k_{eq} = \frac{K_{1}K_{2}}{K_{1} + K_{2}}$ so time period $T = 2\pi\sqrt{\frac{m(K_{1} + K_{2})}{K_{1}K_{2}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.