6.3 Stationary waves — Practise

Work through these on your own. Write the unit of every answer, and draw the shape of the stationary wave before you use the length. In the questions that ask for a reason, write your answer in full sentences.

The questions and the answers below come from a set that is checked by computer every time the book is built.

A string has a tension of 64 N and a mass per unit length of 4.0e-4 kg/m. Find the speed of a wave on the string.
A string 0.800 m long is fixed at both ends and has a wave speed of 320 m/s. Find the frequency of its fundamental note.
Find the frequency of the second harmonic of the string in question 2.
A string sounds a note of 150 Hz at a tension of 90 N. Find the tension needed for a note of 300 Hz.
An organ pipe is open at both ends and is 0.400 m long. The speed of sound is 340 m/s. Find the frequency of its fundamental note.
A pipe closed at one end is 0.400 m long. The speed of sound is 340 m/s. Find the frequency of its fundamental note.
The distance between two neighbouring nodes on a stationary wave is 0.15 m. Find the wavelength.
On a stationary sound wave the distance from a node to the next antinode is 0.20 m. Find the wavelength.
Explain how a stationary wave is formed on a guitar string, and say why it transfers no energy along the string.
Explain why a pipe closed at one end sounds only the odd harmonics.
A spring of force constant 250 N/m is stretched by 0.080 m. Find the energy stored in it.
A tension of 500 N acts along a string at 30 degrees to the horizontal. Find the horizontal and vertical components of the tension.

Answers and concise solutions

1. The wave speed

Answer: $v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{64}{4.0 \times 10^{-4}}} = 400\,\text{m}\,\text{s}^{-1}$ .

2. The fundamental frequency

Answer: $f_1 = \dfrac{v}{2L} = \dfrac{320}{2 \times 0.800} = 200\,\text{Hz}$ .

3. The second harmonic

Answer: $f_2 = 2f_1 = 2 \times 200 = 400\,\text{Hz}$ .

4. The tension

Answer: $T_2 = T_1\left(\dfrac{f_2}{f_1}\right)^2 = 90 \times 4 = 360\,\text{N}$ .

5. The fundamental frequency

Answer: $f_1 = \dfrac{v}{2L} = \dfrac{340}{2 \times 0.400} = 425\,\text{Hz}$ .

6. The fundamental frequency

Answer: $f_1 = \dfrac{v}{4L} = \dfrac{340}{4 \times 0.400} = 213\,\text{Hz}$ , to three significant figures.

7. The wavelength

Answer: Neighbouring nodes are half a wavelength apart, so $\lambda = 2 \times 0.15 = 0.30\,\text{m}$ .

8. The wavelength

Answer: A node to the next antinode is a quarter of a wavelength, so $\lambda = 4 \times 0.20 = 0.80\,\text{m}$ .

9. Explain how a stationary wave is formed on a guitar string, and say why it transfers no energy along the string.

A wave travels along the string and reflects from each fixed end.
The reflected wave and the incoming wave have the same frequency and travel in opposite directions.
They superpose, so their displacements add at every point.
The result has fixed nodes, where the displacement is always zero, and antinodes, where the amplitude is largest.
Each point oscillates in place, and no energy is carried from one end of the string to the other. Energy is stored in the vibration.

10. Explain why a pipe closed at one end sounds only the odd harmonics.

The closed end must be a node, because the air there cannot move.
The open end must be an antinode, because the air there is free to move.
The shortest wave that fits has a node at one end and an antinode at the other. That is a quarter of a wavelength, so the length is an odd number of quarter wavelengths.
An even harmonic would need a whole number of half wavelengths with an antinode at the closed end, which is impossible. So only the odd harmonics are present.

11. The energy stored

Answer: $E = \tfrac{1}{2}k(\Delta L)^2 = \tfrac{1}{2} \times 250 \times 0.080^2 = 0.80\,\text{J}$ .

12. The two components

Answer: The horizontal component is $500\cos 30^\circ = 433\,\text{N}$ . The vertical component is $500\sin 30^\circ = 250\,\text{N}$ .