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qid 11794 · engineering

Question: A steel helical spring of index 10 has 12 active coils. The spring is made of oil-tempered no. 7 gauge wire. Determine the fundamental frequency of the spring if (1) one end is fixed and one end is free. (2) both ends are fixed. For the first case, check for the safety of the spring against failure due to surging when the operating load frequency due to external excitation source is 120 cycles per minute. The specific weight of steel is y = 0.285 lb/in3 and acceleration due to gravity g = 386 in/sec. Modulus of rigidity for steel G = 11.5 × 10^6 lb/in^2.

  1. 29 cycles/sec and 58 cycles/sec
  2. 28 cycles/sec and 56 cycles/sec
  3. 31 cycles/sec and 62 cycles/sec
  4. 34 cycles/sec and 68 cycles/sec
  5. 30 cycles/sec and 60 cycles/sec
  6. 33 cycles/sec and 66 cycles/sec
  7. 37 cycles/sec and 74 cycles/sec
  8. 35 cycles/sec and 70 cycles/sec
  9. 32 cycles/sec and 64 cycles/sec
  10. 36 cycles/sec and 72 cycles/sec

Our answer: F. 33 cycles/sec and 66 cycles/sec Source quote machine-checked (exact quote)

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How it was answered

Multi-step solver (maze), replayed by code

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Current source

RoyMech, "Spring Surge and Natural Frequency Explained"

https://www.roymech.co.uk/Useful_Tables/Springs/Springs_Surge.html

“Spring surge occurs when a helical spring operates near its natural frequency, causing large internal vibrations and stresses. To avoid failure, the natural frequency should typically be at least 15–20 times higher than the operating frequency.”

Source quote machine-checked (exact quote)

retrieved 2026-09-18T02:26:51.768Z

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addendum maze_qid11794_b4_maze_input_ADDENDUM_source_rs20260918T022651Z · sha256 42262725f4b4f47f… · replaces the version below, addendum sha256 88eb7102fb62f814…

Earlier version (superseded)

No public source has been found for this card yet (3 places checked internally).

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