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Vertical oscillations on a string or spring

Edexcel IAL Mathematics · WME03 Mechanics 3

A particle hanging on a spring or string bounces up and down. The motion is SHM about the equilibrium position, not about the natural length position, and a string may go slack if the oscillation is too big.

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Equilibrium first

In equilibrium the extension e satisfies mg = λe/l. Measure x downwards from the equilibrium position.

The proof

At a general position the extension is e + x, so

m ẍ = mg - λ(e + x)/l = mg - λe/l - λx/l = -λx/l.

So ẍ = -(λ/(ml))x: SHM about the equilibrium position with ω2 = λ/(ml) = g/e.

When a string goes slack

If the particle rises more than e above the equilibrium position, the string becomes slack and the particle moves freely under gravity until it falls back. So if the amplitude is bigger than e, find the speed when x = -e with v2 = ω2(a2 - x2), then use suvat with acceleration g. A spring does not go slack: it pushes.

Worked example

A particle of mass 0.4 kg hangs from an elastic string of natural length 0.5 m and modulus 19.6 N. It is pulled down 0.05 m below its equilibrium position and released. Find the equilibrium extension, the period and the greatest speed.

  1. 0.4g = 19.6e/0.5, so e = 0.1 m.
  2. ω2 = g/e = 98, so T = 2π/sqrt 98 = 0.635 s.
  3. Amplitude 0.05 m (less than e, so the string stays taut). Greatest speed = 0.05 × sqrt 98 = 0.495 m/s.

Answer: e = 0.1 m, T = 0.635 s, greatest speed 0.495 m/s

Key idea: Measure x from the equilibrium position: ω2 = λ/(ml) = g/e. If the amplitude exceeds e, a string goes slack and gravity takes over.

Check yourself

  1. A particle of mass 0.3 kg hangs in equilibrium from a spring of natural length 0.6 m and modulus 29.4 N. Find the extension.
    Show answer

    0.06 m. 0.3g = 29.4e/0.6, e = 0.06 m.

  2. In the worked example, find the period, to 3 significant figures.
    Show answer

    0.635 s. ω = sqrt 98, T = 0.635 s.

  3. In the worked example, find the greatest speed, to 3 significant figures.
    Show answer

    0.495 m/s. aω = 0.05 × sqrt 98 = 0.495 m/s.

  4. In the worked example the particle is instead pulled down 0.15 m below its equilibrium position and released. Find its speed when the string becomes slack, to 3 significant figures.
    Show answer

    1.11 m/s. The string goes slack at the natural length, x = -0.1 m. v2 = 98(0.152 - 0.12) = 1.225, v = 1.11 m/s.

  5. Continuing, find how much further the particle rises after the string becomes slack, to 3 significant figures.
    Show answer

    0.0625 m. Free motion under gravity: h = v2/(2g) = 1.225/19.6 = 0.0625 m.

  6. A particle hangs on a spring with equilibrium extension e. What is ω2 for vertical oscillations?
    • g/e
    • e/g
    • g
    • λ/m
    Show answer

    g/e. ω2 = λ/(ml) and λ/l = mg/e, so ω2 = g/e.

  7. About which point does a particle on a vertical spring oscillate?
    • The point where the spring has its natural length
    • The fixed end of the spring
    • Its equilibrium position
    • The lowest point of the motion
    Show answer

    Its equilibrium position. The proof shows ẍ = -ω2x with x measured from equilibrium.

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Last updated 8 November 2026. Spotted a mistake? Tell us.