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A helical compression spring of stiffness k is cut into two pieces, each having equal number of turns and kept side by side under compression. The equivalent spring stiffness of this new arrangement is equal to
  • a)
    4k
  • b)
    2k
  • c)
    k
  • d)
    0.5k
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A helical compression spring of stiffness k is cut into two pieces, e...
Understanding Spring Stiffness
When a helical compression spring is cut into two equal pieces, each piece retains some properties of the original spring.
Initial Stiffness
- The original spring has a stiffness denoted by k.
- Stiffness (k) indicates the force required to compress the spring by a unit distance.
Cutting the Spring
- When the spring is cut into two equal pieces, each piece has half of the original number of turns.
- The stiffness of each half-spring becomes 2k because stiffness is inversely proportional to the number of turns.
Combining the Pieces
- When these two halves are placed side by side under compression, they act in parallel.
- For springs in parallel, the equivalent stiffness (K_eq) is the sum of the individual stiffnesses.
Calculating Equivalent Stiffness
- If each half has a stiffness of 2k, the equivalent stiffness can be calculated as follows:
K_eq = k1 + k2
K_eq = 2k + 2k
K_eq = 4k
Conclusion
- Therefore, when the two pieces are compressed together, the equivalent spring stiffness of this new arrangement is equal to 4k.
- The correct answer is option 'A'.
This process illustrates how dividing a spring affects its stiffness and how combining springs influences the overall stiffness in a parallel configuration.
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A helical compression spring of stiffness k is cut into two pieces, each having equal number of turns and kept side by side under compression. The equivalent spring stiffness of this new arrangement is equal toa) 4kb) 2kc) kd) 0.5kCorrect answer is option 'A'. Can you explain this answer?
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