In v = √(T ÷ μ), what is μ?

the mass per unit length of the string. It is measured in kg/m.

The tension in a string is made four times bigger. What happens to the wave speed?

It doubles. Speed is proportional to the square root of the tension.

Which change raises the frequency of a vibrating string?

making it shorter. A shorter string means a shorter wavelength at the same wave speed.

The length of a string is doubled, with the same tension and the same string. What happens to the frequency of its simplest mode?

It halves. f = v ÷ 2L, and v hasn't changed.

Waves on a string · Edexcel IAL Physics WPH revision

In plain words

Tighten a guitar string and the note goes up; fit a heavier string and it goes down. Both follow from one fact: how fast a wave runs along a string depends on how tight the string is and how heavy it is.

Three things to know

Speed of a transverse wave on a string: v = √(T ÷ μ), where T is the tension in newtons and μ is the mass per unit length in kg/m.
For the simplest stationary wave on a string of length L, λ = 2L, so f = v ÷ 2L.
So the frequency goes up if the string is made shorter, tighter or lighter.

Worked example

A string 0.80 m long has a mass per unit length of 2.0 × 10⁻³ kg/m and is under a tension of 80 N. Find the frequency of its simplest mode.

v = √(T ÷ μ) = √(80 ÷ 2.0 × 10⁻³) = √40 000 = 200 m/s.
λ = 2L = 1.6 m.
f = v ÷ λ = 200 ÷ 1.6 = 125 Hz.

Tips and tricks

μ is the mass of the string divided by its length, in kg/m. Convert grams to kilograms first.
Frequency is proportional to √T, so doubling the frequency takes four times the tension.

5 questions, about 2 minutes.

It lands in your notebook with its questions as flashcards.

Waves on a string: 5 questions and answers

These are the quiz’s questions. Do the quiz first, then come back here for the ones that got you.

In v = √(T ÷ μ), what is μ?
- the mass per unit length of the string (the answer)
- the mass of the string
- the length of the string
- the density of the string
It is measured in kg/m.
The tension in a string is made four times bigger. What happens to the wave speed?
- It stays the same.
- It doubles. (the answer)
- It becomes four times bigger.
- It halves.
Speed is proportional to the square root of the tension.
Which change raises the frequency of a vibrating string?
- making it shorter (the answer)
- making it heavier
- making it slacker
- making it longer
A shorter string means a shorter wavelength at the same wave speed.
A string has a tension of 100 N and a mass per unit length of 0.010 kg/m. What is the wave speed?
- 1.0 m/s
- 10 m/s
- 100 m/s (the answer)
- 10 000 m/s
√(100 ÷ 0.010).
The length of a string is doubled, with the same tension and the same string. What happens to the frequency of its simplest mode?
- It halves. (the answer)
- It doubles.
- It stays the same.
- It falls to a quarter.
f = v ÷ 2L, and v hasn't changed.

5 questions

Tap an answer and you’ll see straight away whether it’s right, and why.

Worksheet

3 questions, 7 marks. Write your answers on paper, then check them.

Waves on a string

Edexcel International A Level Physics WPH · 7 marks · papermunch.org

Name ______________________________ Date ______________

Find the speed of waves on a string with a tension of 50 N and a mass per unit length of 5.0 × 10⁻³ kg/m.[2]
Show answerHide answer
100 m/s.
I got this one
A string 2.0 m long has a mass of 6.0 g. Find its mass per unit length.[2]
Show answerHide answer
3.0 × 10⁻³ kg/m.
I got this one
Describe how you would investigate how the frequency of a vibrating string depends on its tension.[3]
Show answerHide answer
Drive the string with a vibrator connected to a signal generator, with the string running over a pulley to hanging masses. For each tension (the weight of the masses), adjust the frequency until a single loop is seen, and record it. Keep the length and the string the same. A graph of frequency against √tension should be a straight line through the origin.
I got this one

Answers: Waves on a string

1. 100 m/s.
2. 3.0 × 10⁻³ kg/m.
3. Drive the string with a vibrator connected to a signal generator, with the string running over a pulley to hanging masses. For each tension (the weight of the masses), adjust the frequency until a single loop is seen, and record it. Keep the length and the string the same. A graph of frequency against √tension should be a straight line through the origin.