Stationary waves test with answers for A Level physics (free PDF download)

Quick answer: A stationary wave forms when two identical waves travel in opposite directions and superpose, giving fixed nodes and antinodes. This Physics 11-12 test on stationary waves has 8 questions worth 20 marks, to do in 45 minutes: violin string, closed pipe resonances and reading the loops. The mark scheme is on a separate page; free printable PDF.

This test lasts 45 minutes and is worth 20 marks. Use v = 343 m/s for sound in air, show all your working and give answers with units. Draw clear sketches where they help you.

Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB C.4

Time: 45 minutes · 8 questions · Total: 20 marks · Calculator allowed

Contents

Question 1 Violin string (3 marks)

A string of length 0.90 m is fixed at both ends and carries waves of speed 180 m/s.

Calculate the wavelength of the fundamental. (1 mark)
Calculate the fundamental frequency. (1 mark)
State the frequency of the third harmonic and the number of antinodes it has. (1 mark)

Question 2 Closed pipe resonances (3 marks)

A pipe closed at one end has length 0.40 m (v = 343 m/s).

Calculate the fundamental frequency. (1 mark)
Calculate the next two resonant frequencies. (2 marks)

Question 3 Reading the loops (3 marks)

The diagram shows a stationary wave on a string driven at 200 Hz.

Stationary wave with four loops on a string of length 1.20 m.

State the harmonic number. (1 mark)
Calculate the wavelength. (1 mark)
Calculate the wave speed. (1 mark)

Question 4 Two kinds of wave (3 marks)

State three differences between a stationary wave and a progressive wave.

Question 5 Odd harmonics (2 marks)

Explain why a pipe closed at one end produces only odd harmonics.

Question 6 Doubling the tension (2 marks)

A string has a fundamental of 220 Hz. Calculate the new fundamental if the tension is doubled with length and mass per unit length unchanged.

Question 7 Fretting a string (2 marks)

An open guitar string of length 0.64 m has a fundamental of 110 Hz. A finger presses the string so its vibrating length is 0.48 m. Calculate the new fundamental frequency.

Question 8 End correction idea (2 marks)

Explain what is meant by the end correction of a pipe and how it affects the calculated frequency if ignored.

Distance between adjacent nodes? — λ/2

String harmonics (fixed ends)? — f = nv/2L

Wave speed on a string? — v = √(T/μ)

Open pipe harmonics? — f = nv/2L, all harmonics

Closed pipe harmonics? — f = (2n − 1)v/4L, odd only

What is at a closed end? — A node

What is at an open end? — An antinode

End correction for a closed pipe? — Add about 0.6r to the length

Effect of doubled tension on f? — f rises by a factor of √2

Do stationary waves carry energy along? — No net energy transfer along the medium