The equivalent stiffness of two springs of stiffness S₁ and S₂ joined in series is:
Try it first: pick an option, then read the explanation.
Correct answer: A. S₁S₂/(S₁ + S₂)
Asked in SSC JE 2013
Detailed Explanation
In series, each spring carries the full load W, and the total deflection is the sum of the two deflections.
\[ \delta = \frac{W}{S_1} + \frac{W}{S_2} = \frac{W}{S} \;\Rightarrow\; \frac1S = \frac1{S_1} + \frac1{S_2} \;\Rightarrow\; S = \frac{S_1S_2}{S_1+S_2} \]
Why “S₁S₂/(S₁ + S₂)” Is Correct
The equivalent stiffness is S₁S₂/(S₁ + S₂) — always less than either spring.
Why the Other Options Are Wrong
- B. (S₁/S₂)(S₁ + S₂): Dimensionally fine but has no physical basis.
- C. S₁ + S₂: S₁ + S₂ is for springs in parallel.
- D. S₁S₂: S₁S₂ has the wrong units (stiffness²).
Key Point
Series: 1/S = Σ1/Si (softer). Parallel: S = ΣSi (stiffer).
Key Concepts for Students
- Two equal springs in series give half the stiffness.
- Cutting a spring in half doubles its stiffness.
- Like resistors in parallel (series springs).
Practice more: this is question 4 in Spring MCQs – Part 1. Try the whole set and check every answer instantly.