Vibration of a mass–spring system (vertical helical spring): One end of a helical spring is fixed and the other end carries a load W. The load executes simple harmonic motion vertically. If the static deflection under W is δ, what is the frequency f of oscillation?

Introduction / Context:Small vertical vibrations of a mass hung from a spring are modeled as simple harmonic motion (SHM). Designers often use the static deflection δ under weight W to estimate natural frequency without directly computing spring stiffness and mass separately.

Given Data / Assumptions:

• Load W causes a static deflection δ of the spring.
• Linear spring behavior; small oscillations about equilibrium.
• Gravity g acts downward; mass of spring neglected compared with load mass.

Concept / Approach:Let spring stiffness be k. At static equilibrium, W = k * δ, so k = W / δ. With W = m * g, we have k = m * g / δ. The circular natural frequency is ω = sqrt(k / m). Frequency in hertz is f = ω / (2π).

Step-by-Step Solution:From static deflection: k = W / δ = (m * g) / δ.Compute natural frequency: ω = sqrt(k / m) = sqrt((m * g / δ) / m) = sqrt(g / δ).Convert to hertz: f = ω / (2π) = (1 / 2π) * sqrt(g / δ).

Verification / Alternative check:Unit check: g/δ has units (m/s^2)/m = 1/s^2, so sqrt(g/δ) is 1/s, matching angular frequency; dividing by 2π gives cycles per second (Hz).

Why Other Options Are Wrong:

• (1/2π)*sqrt(δ/g): Inverted; would have units s, not 1/s.
• (1/2π)*sqrt(W/δ): Dimensionally inconsistent for frequency without dividing by sqrt(m).
• (1/2π)*sqrt(δ/W): Also dimensionally incorrect.

Common Pitfalls:

• Forgetting to use static deflection to eliminate k and m cleanly.
• Confusing angular frequency ω with frequency f; ω = 2π f.

Final Answer:f = (1 / 2π) * sqrt(g / δ)