Physics · Simulation and virtual lab Free
Waves on a string: standing waves, harmonics and wave speed
Vibrate a string, find the frequencies that make steady loops, and discover how tension controls wave speed.
- You will learn to
- Produce standing waves and identify nodes, antinodes and harmonics.
- Find wavelength from the loops and calculate v = fλ.
- Show that wave speed is proportional to √T and find the mass per unit length.
- You will measure
- Frequency for each number of loops (Hz), tension (N), wavelength (m), wave speed (m/s).
- Time needed
- About 35–45 min
- Equipment
- Nothing extra. Real version: signal generator and vibrator, string, pulley, slotted masses, metre rule.
- Price
- Free: no payment, no account.
Virtual lab
Follow the steps with the simulation above (go back to the simulation ↑). Each time you tap Record, a row is added to the table and a point to the graph.
Aim
To produce standing waves on a stretched string, to relate the number of loops to the frequency, and to investigate how the wave speed depends on the tension and on the mass per unit length of the string.
What to do
- Keep 200 g, 1.0 m, thin thread. Slowly increase the frequency from 5 Hz until one big loop appears. Record.
- Find the frequencies for 2, 3 and 4 loops (or use Tune to the nearest resonance). What pattern do the frequencies follow?
- Calculate λ = 2L ÷ n and v = fλ for each. Is v the same each time?
- Change the hanging mass to 100 g, 300 g, 400 g and 500 g. Find the one-loop frequency each time and record.
- Look at the graph of v against √T. Use the gradient to find the mass per metre of the thread.
- Repeat with the thick string. Explain why it gives lower notes.
Results
Questions
What is a standing (stationary) wave?
A wave pattern formed when two waves of the same frequency travel in opposite directions and superpose; it has fixed nodes (no motion) and antinodes (maximum motion) and does not move along the string.
What is the distance between two neighbouring nodes?
Half a wavelength (λ ÷ 2).
A 1.0 m string vibrates in 3 loops. What is the wavelength?
λ = 2L ÷ n = 2 × 1.0 ÷ 3 = 0.67 m.
The fundamental frequency of a string is 30 Hz. What are the next two harmonics?
60 Hz and 90 Hz (2f and 3f).
How does the wave speed depend on the tension?
v = √(T ÷ μ): the speed is proportional to the square root of the tension; four times the tension doubles the speed.
Why does tightening a guitar string raise its note?
Higher tension gives a higher wave speed; for the same length (same λ) the frequency f = v ÷ λ is higher.
For teachers: expected results and how the model works
Model: the string is driven at one end and fixed at the pulley; the amplitude is A ∝ sin k(L − x) ÷ √(sin² kL + δ²), so large steady loops appear only near f = n v ÷ 2L. Tension T = mg with g = 9.78 m/s². Check values for L = 1.0 m: thin thread μ = 0.50 g/m; 200 g gives T = 1.96 N, v = 62.5 m/s, f₁ = 31.3 Hz (harmonics 62.5, 93.8, 125 Hz); 100 g gives v = 44.2 m/s, f₁ = 22.1 Hz; 400 g gives v = 88.5 m/s, f₁ = 44.2 Hz. Thick string μ = 2.0 g/m at 200 g: v = 31.3 m/s, f₁ = 15.6 Hz. Strobe view shows the string moving slowly; blur view shows what the eye sees. Nodes are marked N, antinodes A.
Exam practice Exam set
National-exam style questions on this simulation (14 marks). Try each one on paper first, then open the marking scheme.
A string 1.2 m long has a mass per unit length of 0.50 g/m and is under a tension of 4.5 N. Calculate (a) the speed of waves on it, (b) its fundamental frequency, (c) the frequency of the third harmonic. (6 marks)
Marking scheme: (a) v = √(4.5 ÷ 0.0005) (1) = 94.9 m/s (1). (b) λ = 2L = 2.4 m (1); f = 94.9 ÷ 2.4 = 39.5 Hz (1). (c) 3 × 39.5 = 119 Hz (2).
Explain how a standing wave is formed on a string fixed at both ends. (3 marks)
Marking scheme: Waves travel along the string and are reflected at the fixed ends (1); the incident and reflected waves superpose (1); at certain frequencies they form fixed nodes and antinodes (resonance) (1).
In Melde’s experiment the frequency is kept at 50 Hz. With a tension of 1.0 N the string shows 4 loops. How many loops appear when the tension is increased to 4.0 N? Explain. (3 marks)
Marking scheme: Four times the tension doubles the wave speed (1), so the wavelength doubles at the same frequency (1), giving half as many loops: 2 loops (1).
Give two differences between a progressive wave and a standing wave. (2 marks)
Marking scheme: A progressive wave transfers energy along the medium, a standing wave does not (1); in a standing wave the amplitude varies from zero at nodes to maximum at antinodes, while in a progressive wave all points have the same amplitude (1).
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Waves on a string: standing waves, harmonics and wave speed
Aim: To produce standing waves on a stretched string, to relate the number of loops to the frequency, and to investigate how the wave speed depends on the tension and on the mass per unit length of the string.
Apparatus (real laboratory): signal generator and vibrator, string, pulley, slotted masses, metre rule.
What to do (with the simulation on www.donatstem.co.rw)
- Keep 200 g, 1.0 m, thin thread. Slowly increase the frequency from 5 Hz until one big loop appears. Record.
- Find the frequencies for 2, 3 and 4 loops (or use Tune to the nearest resonance). What pattern do the frequencies follow?
- Calculate λ = 2L ÷ n and v = fλ for each. Is v the same each time?
- Change the hanging mass to 100 g, 300 g, 400 g and 500 g. Find the one-loop frequency each time and record.
- Look at the graph of v against √T. Use the gradient to find the mass per metre of the thread.
- Repeat with the thick string. Explain why it gives lower notes.
Results
| # | Mass (g) | Tension T (N) | Loops n | Frequency f (Hz) | λ = 2L ÷ n (m) | v = fλ (m/s) | √T |
|---|
| 1 | | | | | | | |
| 2 | | | | | | | |
| 3 | | | | | | | |
| 4 | | | | | | | |
| 5 | | | | | | | |
| 6 | | | | | | | |
| 7 | | | | | | | |
| 8 | | | | | | | |
| 9 | | | | | | | |
| 10 | | | | | | | |
Graph
Plot v (m/s) on the vertical axis against √T. Draw the best line through the origin and find μ from the gradient (gradient = 1 ÷ √μ).
Questions
- What is a standing (stationary) wave? [2]
- What is the distance between two neighbouring nodes? [2]
- A 1.0 m string vibrates in 3 loops. What is the wavelength? [2]
- The fundamental frequency of a string is 30 Hz. What are the next two harmonics? [2]
- How does the wave speed depend on the tension? [2]
- Why does tightening a guitar string raise its note? [2]
- A string 1.2 m long has a mass per unit length of 0.50 g/m and is under a tension of 4.5 N. Calculate (a) the speed of waves on it, (b) its fundamental frequency, (c) the frequency of the third harmonic. [6]
- Explain how a standing wave is formed on a string fixed at both ends. [3]
- In Melde’s experiment the frequency is kept at 50 Hz. With a tension of 1.0 N the string shows 4 loops. How many loops appear when the tension is increased to 4.0 N? Explain. [3]
- Give two differences between a progressive wave and a standing wave. [2]
Donat Sciences and MathsPhysics · Virtual lab worksheet
School: ........................................ Name: ................................ Class: ........ Date: ............
Marking scheme: Waves on a string: standing waves, harmonics and wave speed
Expected results
Model: the string is driven at one end and fixed at the pulley; the amplitude is A ∝ sin k(L − x) ÷ √(sin² kL + δ²), so large steady loops appear only near f = n v ÷ 2L. Tension T = mg with g = 9.78 m/s². Check values for L = 1.0 m: thin thread μ = 0.50 g/m; 200 g gives T = 1.96 N, v = 62.5 m/s, f₁ = 31.3 Hz (harmonics 62.5, 93.8, 125 Hz); 100 g gives v = 44.2 m/s, f₁ = 22.1 Hz; 400 g gives v = 88.5 m/s, f₁ = 44.2 Hz. Thick string μ = 2.0 g/m at 200 g: v = 31.3 m/s, f₁ = 15.6 Hz. Strobe view shows the string moving slowly; blur view shows what the eye sees. Nodes are marked N, antinodes A.
Answers
- What is a standing (stationary) wave? [2]
A wave pattern formed when two waves of the same frequency travel in opposite directions and superpose; it has fixed nodes (no motion) and antinodes (maximum motion) and does not move along the string.
- What is the distance between two neighbouring nodes? [2]
Half a wavelength (λ ÷ 2).
- A 1.0 m string vibrates in 3 loops. What is the wavelength? [2]
λ = 2L ÷ n = 2 × 1.0 ÷ 3 = 0.67 m.
- The fundamental frequency of a string is 30 Hz. What are the next two harmonics? [2]
60 Hz and 90 Hz (2f and 3f).
- How does the wave speed depend on the tension? [2]
v = √(T ÷ μ): the speed is proportional to the square root of the tension; four times the tension doubles the speed.
- Why does tightening a guitar string raise its note? [2]
Higher tension gives a higher wave speed; for the same length (same λ) the frequency f = v ÷ λ is higher.
- A string 1.2 m long has a mass per unit length of 0.50 g/m and is under a tension of 4.5 N. Calculate (a) the speed of waves on it, (b) its fundamental frequency, (c) the frequency of the third harmonic. [6]
(a) v = √(4.5 ÷ 0.0005) (1) = 94.9 m/s (1). (b) λ = 2L = 2.4 m (1); f = 94.9 ÷ 2.4 = 39.5 Hz (1). (c) 3 × 39.5 = 119 Hz (2).
- Explain how a standing wave is formed on a string fixed at both ends. [3]
Waves travel along the string and are reflected at the fixed ends (1); the incident and reflected waves superpose (1); at certain frequencies they form fixed nodes and antinodes (resonance) (1).
- In Melde’s experiment the frequency is kept at 50 Hz. With a tension of 1.0 N the string shows 4 loops. How many loops appear when the tension is increased to 4.0 N? Explain. [3]
Four times the tension doubles the wave speed (1), so the wavelength doubles at the same frequency (1), giving half as many loops: 2 loops (1).
- Give two differences between a progressive wave and a standing wave. [2]
A progressive wave transfers energy along the medium, a standing wave does not (1); in a standing wave the amplitude varies from zero at nodes to maximum at antinodes, while in a progressive wave all points have the same amplitude (1).