Core practical 5: frequency of a vibrating string

Apparatus and method

• Stretch a string or wire over two bridges (a sonometer), with a mass hanger over a pulley to set the tension T = mg.
• Drive it with a vibration generator connected to a signal generator.
• Adjust the frequency until the string vibrates in its fundamental mode (one loop of maximum amplitude); a small paper rider at the antinode helps judge this.
• Change one variable at a time: length L (move a bridge), tension (change the mass), or μ (use different wires, finding μ by weighing a measured length).

What theory predicts

Combining v = sqrt(T/μ) with λ = 2L gives

f = (1 / 2L) sqrt(T / μ)

• f ∝ 1/L (fixed T and μ): plot f against 1/L for a straight line through the origin.
• f ∝ sqrt(T): plot f² against T for a straight line.
• f ∝ 1/sqrt(μ): plot f² against 1/μ.

Good practice

• Measure L between the bridges with a metre rule.
• Find resonance from both sides (increasing and decreasing frequency) and average.
• Keep masses low enough not to permanently stretch the wire; place a soft surface below the masses in case they fall.

Worked example

A wire of length 0.80 m and μ = 1.2 x 10^-3 kg/m has a 2.0 kg mass hung from it. Find the fundamental frequency.

1. T = mg = 2.0 x 9.81 = 19.6 N
2. v = sqrt(19.6 / 1.2 x 10^-3) = 128 m/s
3. f = v / 2L = 128 / 1.60

Answer: f = 79.9 Hz

Check yourself

1. Which graph should give a straight line through the origin when L is varied with T and μ fixed?
   • f against L
   • f² against L
   • f against 1/L
   • f against L²
   Show answer

   f against 1/L. f = (1/2L) sqrt(T/μ) shows f ∝ 1/L.

2. A 0.50 kg mass hangs from the end of the wire. Find the tension in N, to 3 s.f.
   Show answer

   4.91 N. T = mg = 0.50 x 9.81 = 4.91 N.

3. A string has a fundamental frequency of 120 Hz. Its vibrating length is halved, with T and μ unchanged. Find the new fundamental frequency in Hz.
   Show answer

   240 Hz. f ∝ 1/L, so halving L doubles f: 240 Hz.

4. A string has a fundamental frequency of 150 Hz. The tension is made 4 times larger. Find the new fundamental frequency in Hz.
   Show answer

   300 Hz. f ∝ sqrt(T), so f becomes 150 x 2 = 300 Hz.

5. A wire of length 0.60 m has tension 30 N and μ = 1.5 x 10^-3 kg/m. Find its fundamental frequency in Hz, to 3 s.f.
   Show answer

   118 Hz. v = sqrt(20000) = 141 m/s; f = 141 / 1.20 = 118 Hz.

6. A graph of f² against T is a straight line through the origin. What is its gradient?
   • 1 / (2Lμ)
   • 1 / (4L² μ)
   • 4L² μ
   • sqrt(μ) / 2L
   Show answer

   1 / (4L² μ). Squaring f = (1/2L) sqrt(T/μ) gives f² = T / (4L² μ).

7. How can you tell the string is vibrating in its fundamental mode?
   • The string stops moving
   • There are two loops
   • The signal generator gets warm
   • There is a single loop and its amplitude is at a maximum
   Show answer

   There is a single loop and its amplitude is at a maximum. At resonance in the fundamental, one loop forms with the largest amplitude.

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Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
• Pearson qualifications: International Advanced Levels Physics (2018)
• Cognito: Edexcel International A level Physics specification and core practicals
• Pearson Edexcel IAL Physics Unit 1 (WPH11) question paper, January 2023: list of data (g = 9.81 m/s²)

Last updated 7 November 2026. Spotted a mistake? Tell us.