Oscillations on a horizontal string or spring

One spring

At extension x the tension is λx/l towards the fixed point, so m ẍ = -λx/l, that is ẍ = -(λ/(ml))x. This is SHM with ω² = λ/(ml) about the natural length position.

A string goes slack

An elastic string only pulls. If the particle passes the natural length position, the string goes slack and the particle moves with constant speed (on a smooth table) until the string becomes taut again. So the motion is SHM only while the string is taut; add the times of each stage.

Two strings

A particle P between two fixed points A and B, joined to each by a string: at displacement x from equilibrium one extension increases by x and the other decreases by x. Write both tensions at the general position, which is where Pearson's examiners saw errors such as an extension of 0.5L (a constant tension) or (0.5 - x) with the wrong dimensions. SHM continues while both strings stay taut.

Worked example

A particle of mass 0.5 kg on a smooth horizontal table is attached to a spring of natural length 0.8 m and modulus 10 N. It is pulled out to an extension of 0.2 m and released. Find the period and greatest speed.

1. m ẍ = -10x/0.8, so ẍ = -25x: SHM with ω = 5.
2. T = 2π/5 = 1.26 s.
3. Amplitude 0.2 m, greatest speed aω = 1 m/s.

Answer: T = 1.26 s, greatest speed 1 m/s

Check yourself

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

   1.26 s. T = 2π/5 = 1.26 s.

2. In the worked example, find the greatest speed.
   Show answer

   1 m/s. aω = 0.2 × 5 = 1 m/s.

3. Fixed points A and B are 4 m apart on a smooth horizontal table. A particle P of mass 0.6 kg is joined to A and to B by elastic strings, each of natural length 1.5 m and modulus 9 N. P is displaced along AB from the midpoint and released. Find the period of the oscillations, to 3 significant figures.
   Show answer

   1.40 s. At displacement x from the midpoint, extensions are 0.5 + x and 0.5 - x. Net force = -9(0.5 + x)/1.5 + 9(0.5 - x)/1.5 = -12x, so 0.6ẍ = -12x, ω² = 20, T = 2π/sqrt 20 = 1.40 s.

4. A particle of mass 0.2 kg on a smooth horizontal table is attached to a fixed point O by an elastic string of natural length 0.5 m and modulus 10 N. It is released from rest with the string stretched by 0.1 m. Find the time until the string becomes slack, to 3 significant figures.
   Show answer

   0.157 s. ω² = 10/(0.2 × 0.5) = 100, ω = 10. The string becomes slack after a quarter period: π/20 = 0.157 s.

5. For the same particle, find the total time from release until it reaches O, to 3 significant figures.
   Show answer

   0.657 s. At slack the speed is aω = 0.1 × 10 = 1 m/s; it then travels the natural length 0.5 m at constant speed in 0.5 s. Total = π/20 + 0.5 = 0.657 s.

6. In the two-string set-up, the particle oscillates about the midpoint where each string has extension 0.5 m. When is the whole motion SHM?
   • For any amplitude
   • When the amplitude is at most 0.5 m
   • Only when the amplitude is more than 0.5 m
   • Only when the strings have different moduli
   Show answer

   When the amplitude is at most 0.5 m. If the amplitude exceeds 0.5 m one string goes slack and the equation of motion changes.

7. P is at the midpoint of AB (AB = 4 m) and joined to A by a string of natural length 1.5 m. At displacement x from the midpoint towards B, what is the extension of string AP?
   • 0.5
   • 0.5 - x
   • 2 + x
   • 0.5 + x
   Show answer

   0.5 + x. AP = 2 + x, so the extension is 2 + x - 1.5 = 0.5 + x.

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Sources

• Pearson Edexcel International Advanced Level in Mathematics specification (issue 3), Unit M3
• Pearson Edexcel IAL Mechanics M3 (WME03) question paper, June 2025: front-cover instruction on g and accuracy
• Pearson Edexcel IAL Mechanics M3 (WME03) examiners' report, June 2025

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