Q. A mass $M$ is suspended by two springs of force constants $K_{1}$ and $K_{2}$ respectively as shown in the diagram. The total elongation (stretch) of the two springs is
A) $\frac{Mg}{K_{1} + K_{2}}$
B) $\frac{Mg(K_{1} + K_{2})}{K_{1}K_{2}}$
C) $\frac{MgK_{1}K_{2}}{K_{1} + K_{2}}$
D) $\frac{K_{1} + K_{2}}{K_{1}K_{2}Mg}$
Explanation
(b) For series combination $K_{eq} = \frac{K_{1}K_{2}}{K_{1} + K_{2}}$
$F = K_{eq}x \Rightarrow mg = \left( \frac{K_{1}K_{2}}{K_{1} + K_{2}} \right)x \Rightarrow x = \frac{mg(K_{1} + K_{2})}{K_{1}K_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.