6.3 Stationary waves — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check. Draw the shape of the wave before you calculate: the nodes are at the fixed ends or at a closed end.

Warm-up: remember

A wave has a frequency of 0.50 kHz. Find its period.
A spring of force constant $200\,\text{N}\,\text{m}^{-1}$ is stretched by a force of 10 N. Find the extension.
Write $4.0 \times 10^{-4}\,\text{kg}\,\text{m}^{-1}$ in grams per metre.

Answers

$T = \dfrac{1}{f} = \dfrac{1}{500} = 2.0 \times 10^{-3}\,\text{s}$ .
$F = k\Delta L$ , so $\Delta L = \dfrac{F}{k} = \dfrac{10}{200} = 0.050\,\text{m}$ .
$4.0 \times 10^{-4}\,\text{kg}\,\text{m}^{-1} = 0.40\,\text{g}\,\text{m}^{-1}$ .

If question 1 goes wrong, change the kilohertz to hertz before dividing. If question 2 goes wrong, use Hooke's law from Section 4.2. If question 3 goes wrong, one kilogram is 1000 grams.

Practise with help

1. A string has a tension of 100 N and a mass per unit length of $2.5 \times 10^{-4}\,\text{kg}\,\text{m}^{-1}$ . Find the speed of a wave on the string.

Hint

The speed depends on the square root of the tension divided by the mass per unit length.

Next step

Work out $\dfrac{100}{2.5 \times 10^{-4}}$ first, then take its square root.

Full working

\begin{aligned} v &= \sqrt{\frac{T}{\mu}} = \sqrt{\frac{100}{2.5 \times 10^{-4}}} \\ &= \sqrt{4.0 \times 10^{5}} = 632\,\text{m}\,\text{s}^{-1} \end{aligned}

The wave travels at $632\,\text{m}\,\text{s}^{-1}$ , to three significant figures.

2. A string 0.500 m long is fixed at both ends. The speed of a wave on it is $400\,\text{m}\,\text{s}^{-1}$ . Find the frequency of its fundamental note.

Hint

The fundamental note has one antinode, so the string holds half a wavelength.

Next step

Use $f = \dfrac{v}{2L}$ with $v = 400$ and $L = 0.500$ .

Full working

f_1 = \frac{v}{2L} = \frac{400}{2 \times 0.500} = 400\,\text{Hz}

The fundamental frequency is 400 Hz.

3. A string 0.600 m long has a wave speed of $300\,\text{m}\,\text{s}^{-1}$ . Find the frequency of its third harmonic.

Hint

The $n$ th harmonic has $n$ half wavelengths on the string, and $n$ times the fundamental frequency.

Next step

Find the fundamental first, then multiply by 3.

Full working

\begin{aligned} f_1 &= \frac{v}{2L} = \frac{300}{2 \times 0.600} = 250\,\text{Hz} \\ f_3 &= 3f_1 = 3 \times 250 = 750\,\text{Hz} \end{aligned}

The third harmonic is 750 Hz. Its shape has three antinodes and four nodes, counting the two ends.

4. A string sounds a note of 200 Hz when its tension is 100 N. Find the tension needed for a note of 300 Hz.

Hint

The frequency is proportional to the square root of the tension.

Next step

Square the ratio of the frequencies, then multiply by the original tension.

Full working

\begin{aligned} T_2 &= T_1\left(\frac{f_2}{f_1}\right)^2 \\ &= 100 \times \left(\frac{300}{200}\right)^2 = 100 \times 2.25 = 225\,\text{N} \end{aligned}

The tension must be 225 N.

5. A pipe is open at both ends and is 0.500 m long. The speed of sound is $340\,\text{m}\,\text{s}^{-1}$ . Find the frequency of its fundamental note and of its second harmonic.

Hint

An open pipe behaves like a string fixed at both ends: both ends are antinodes, and all harmonics are present.

Next step

Use $f_n = \dfrac{nv}{2L}$ with $n = 1$ and $n = 2$ .

Full working

\begin{aligned} f_1 &= \frac{340}{2 \times 0.500} = 340\,\text{Hz} \\ f_2 &= \frac{2 \times 340}{1.00} = 680\,\text{Hz} \end{aligned}

The fundamental is 340 Hz and the second harmonic is 680 Hz.

6. A pipe of the same length, 0.500 m, is closed at one end. Find the frequency of its fundamental note and of the next possible harmonic, and explain why some harmonics are missing.

Hint

A closed end is a node and an open end is an antinode. There is a node at the closed end and an antinode at the open end.

Next step

For a closed pipe, $L = (2n - 1)\dfrac{\lambda}{4}$ , so use $f_n = \dfrac{(2n - 1)v}{4L}$ with $n = 1$ and $n = 2$ .

Full working

\begin{aligned} f_1 &= \frac{v}{4L} = \frac{340}{4 \times 0.500} = 170\,\text{Hz} \\ f_3 &= \frac{3v}{4L} = \frac{3 \times 340}{2.00} = 510\,\text{Hz} \end{aligned}

The fundamental is 170 Hz and the next possible harmonic is the third, at 510 Hz. The even harmonics are missing because a closed end must be a node. An even harmonic would need a whole number of half wavelengths with an antinode at the closed end, where the air cannot move.