Chapter 6: Review

The chapter problem, solved

The technician is checking a hall before a concert. The loudspeakers are 1.20 m apart and are driven in phase at 1.00 kHz. The speed of sound is $340\,\text{m}\,\text{s}^{-1}$ . The laser has a wavelength of 633 nm and is used with a double slit 0.25 mm apart and a screen 2.50 m away, and with a grating of 300 lines per millimetre. The guitar string is 0.650 m long, has a mass per unit length of $4.0 \times 10^{-4}\,\text{kg}\,\text{m}^{-1}$ , and a tension of 81 N. It is meant to sound E₄, 330 Hz.

1. The wavelength of the sound.

\lambda = \frac{v}{f} = \frac{340}{1.00 \times 10^{3}} = 0.340\,\text{m}

2. The first loud band. The speakers are coherent, so the maxima lie where $s\sin\theta = n\lambda$ . For the first maximum, $n = 1$ .

\begin{aligned} \sin\theta &= \frac{\lambda}{s} = \frac{0.340}{1.20} = 0.2833 \\ \theta &= 16.5^\circ \end{aligned}

A listener hears the first loud band at $16.5^\circ$ to the centre line. On the centre line the two paths are equal, so the path difference is zero and the sound is loud there too.

3. The spacing of the light fringes.

w = \frac{\lambda D}{s} = \frac{633 \times 10^{-9} \times 2.50}{0.25 \times 10^{-3}} = 6.33 \times 10^{-3}\,\text{m}

The fringes are 6.33 mm apart.

4. The grating spacing. The grating has 300 lines per millimetre, which is $3.00 \times 10^{5}$ lines per metre.

d = \frac{1}{N} = \frac{1}{3.00 \times 10^{5}} = 3.33 \times 10^{-6}\,\text{m}

5. The first-order angle.

\begin{aligned} \sin\theta &= \frac{\lambda}{d} = \frac{633 \times 10^{-9}}{3.33 \times 10^{-6}} = 0.190 \\ \theta &= 10.9^\circ \end{aligned}

The first-order maximum is at $10.9^\circ$ from the straight-through direction. Because the first order is well clear of the centre, the audience sees several bright spots.

6. The highest order the grating can give. The order must satisfy $n\lambda \le d$ .

\frac{d}{\lambda} = \frac{3.33 \times 10^{-6}}{633 \times 10^{-9}} = 5.26

The highest whole number is 5, so the grating gives orders 0 to 5, with five maxima on each side of the centre.

7. The speed of a wave on the guitar string.

\begin{aligned} v &= \sqrt{\frac{T}{\mu}} = \sqrt{\frac{81}{4.0 \times 10^{-4}}} \\ &= \sqrt{2.025 \times 10^{5}} = 450\,\text{m}\,\text{s}^{-1} \end{aligned}

8. The fundamental frequency of the string.

f_1 = \frac{v}{2L} = \frac{450}{2 \times 0.650} = 346\,\text{Hz}

9. The polarised light. The analyser axis is at $60^\circ$ to the plane of polarisation, so the fraction of the intensity that passes is

\frac{I}{I_0} = \cos^2 60^\circ = 0.25

One quarter of the light passes through, so the spot is dimmer but still visible.

10. The judgement on the string. The string sounds 346 Hz, and the note E₄ is 330 Hz. The measured frequency is too high, so the string is too tight. The technician must slacken it. The tension needed is

\begin{aligned} T_2 &= T_1\left(\frac{f_2}{f_1}\right)^2 \\ &= 81 \times \left(\frac{330}{346}\right)^2 = 73.6\,\text{N} \end{aligned}

The tension should be reduced from 81 N to about 74 N. This final judgement is a reasoned answer. It is not the kind of claim a computer can check, so compare your reasoning with the evidence above.

11. The first prediction. The loud and quiet places move when the note changes. Their positions depend on the path difference measured in wavelengths, and a higher note has a shorter wavelength. The same path difference is then a different number of wavelengths, so the places of reinforcement and cancellation are in new positions.

Mixed practice — choose the method first

For each question, say which rule or equation you will use before you calculate.

The loudspeakers are driven at 1.00 kHz and the speed of sound is 340 m/s. Find the wavelength of the sound.
The speakers are 1.20 m apart and the sound has wavelength 0.340 m. Find the angle of the first loud band to the side of the centre line.
The laser has wavelength 633 nm, the double slit is 0.25 mm wide and the screen is 2.50 m away. Find the fringe spacing.
The diffraction grating has 300 lines per millimetre. Find the spacing between neighbouring slits, in metres.
The grating spacing is 3.33e-6 m and the light has wavelength 633 nm. Find the angle of the first-order maximum.
For the grating spacing 3.33e-6 m and the wavelength 633 nm, find the highest order the grating can give.
The guitar string has tension 81 N, a mass per unit length of 4.0e-4 kg/m, and a length of 0.650 m. Find the speed of a wave on it and the frequency of its fundamental note.
Plane-polarised laser light passes through an analyser whose axis is at 60 degrees to the plane of polarisation. Find the fraction of the intensity that passes.
The string sounds 346 Hz at a tension of 81 N. Find the tension needed for 330 Hz.
The grating spacing is 3.33e-6 m. Write this length in micrometres.
A sound wave has a frequency of 500 Hz with a percentage uncertainty of 0.4%, and a wavelength of 0.680 m with a percentage uncertainty of 1.5%. Find its speed and its absolute uncertainty.
A tension of 12.0 N acts along a string at 35 degrees to the horizontal. Find the horizontal and vertical components of the tension.
A mass of 0.10 kg hangs on a spring of force constant 40 N/m. Find the period of the oscillation.
You must measure the wavelength of a laser. Say which method you would choose, a double slit or a diffraction grating, and explain why. State the measurements you need.
Compare a stationary wave with a progressive wave, in terms of the amplitude at different points and the transfer of energy.

Answers and concise solutions

1. The loudspeakers are driven at 1.00 kHz and the speed of sound is 340 m/s. Find the wavelength of the sound.

Answer: $\lambda = \dfrac{v}{f} = \dfrac{340}{1.00 \times 10^{3}} = 0.340\,\text{m}$ .

2. The speakers are 1.20 m apart and the sound has wavelength 0.340 m. Find the angle of the first loud band to the side of the centre line.

Answer: $\sin\theta = \dfrac{\lambda}{s} = \dfrac{0.340}{1.20} = 0.2833$ , so $\theta = 16.5^\circ$ .

3. The laser has wavelength 633 nm, the double slit is 0.25 mm wide and the screen is 2.50 m away. Find the fringe spacing.

Answer: $w = \dfrac{\lambda D}{s} = \dfrac{633 \times 10^{-9} \times 2.50}{0.25 \times 10^{-3}} = 6.33 \times 10^{-3}\,\text{m}$ , or $6.33\,\text{mm}$ .

4. The diffraction grating has 300 lines per millimetre. Find the spacing between neighbouring slits, in metres.

Answer: $300\,\text{lines mm}^{-1} = 3.00 \times 10^5\,\text{lines m}^{-1}$ , so $d = \dfrac{1}{3.00 \times 10^5} = 3.33 \times 10^{-6}\,\text{m}$ .

5. The grating spacing is 3.33e-6 m and the light has wavelength 633 nm. Find the angle of the first-order maximum.

Answer: $\sin\theta = \dfrac{633 \times 10^{-9}}{3.33 \times 10^{-6}} = 0.190$ , so $\theta = 10.9^\circ$ .

6. For the grating spacing 3.33e-6 m and the wavelength 633 nm, find the highest order the grating can give.

Answer: $\dfrac{d}{\lambda} = \dfrac{3.33 \times 10^{-6}}{633 \times 10^{-9}} = 5.26$ . The highest whole number with $n\lambda \le d$ is 5.

7. The wave speed and the fundamental frequency

Answer: $v = \sqrt{\dfrac{T}{\mu}} = 450\,\text{m}\,\text{s}^{-1}$ , and $f_1 = \dfrac{v}{2L} = \dfrac{450}{2 \times 0.650} = 346\,\text{Hz}$ .

8. Plane-polarised laser light passes through an analyser whose axis is at 60 degrees to the plane of polarisation. Find the fraction of the intensity that passes.

Answer: $\dfrac{I}{I_0} = \cos^2 60^\circ = 0.25$ , so one quarter of the light passes.

9. The tension

Answer: $T_2 = T_1\left(\dfrac{f_2}{f_1}\right)^2 = 81 \times \left(\dfrac{330}{346}\right)^2 = 73.6\,\text{N}$ . The string is too tight and must be slackened.

10. The grating spacing is 3.33e-6 m. Write this length in micrometres.

Answer: $3.33 \times 10^{-6}\,\text{m} = 3.33\,\mu\text{m}$ . One micrometre is $10^{-6}\,\text{m}$ .

11. The speed and its uncertainty

Answer: The speed is $v = f\lambda = 500 \times 0.680 = 340\,\text{m}\,\text{s}^{-1}$ . The percentage uncertainties add: $0.4\% + 1.5\% = 1.9\%$ . The absolute uncertainty is $340 \times \dfrac{1.9}{100} = 6.46\,\text{m}\,\text{s}^{-1}$ , or about $6\,\text{m}\,\text{s}^{-1}$ .

12. The two components

Answer: The horizontal component is $12.0\cos 35^\circ = 9.83\,\text{N}$ . The vertical component is $12.0\sin 35^\circ = 6.88\,\text{N}$ .

13. The period

Answer: $T = 2\pi\sqrt{\dfrac{m}{k}} = 2\pi\sqrt{\dfrac{0.10}{40}} = 2\pi \times 0.0500 = 0.314\,\text{s}$ .

14. The method and the measurements

Use the diffraction grating, because its maxima are narrow and sharp, so the angle can be measured more precisely than a broad fringe position.
Measure the angle theta to a maximum of known order n, and find the slit spacing d from the printed number of lines per millimetre.
Use $d\\\\sin\\\\theta = n\\\\lambda$ , rearranged to give $\\\\lambda = \\\\dfrac{d\\\\sin\\\\theta}{n}$ . If a double slit is used instead, measure the fringe spacing w, the slit separation s and the screen distance D, then use $\\\\lambda = \\\\dfrac{ws}{D}$ .

Answer: Use the grating: its sharp maxima give a more precise angle than a broad fringe. Measure the angle to a maximum of known order and find the slit spacing from the lines per millimetre, then use $\lambda = \dfrac{d\sin\theta}{n}$ .

15. The comparison

On a stationary wave the amplitude varies from zero at each node to a maximum at each antinode. On a progressive wave every point has the same amplitude.
The nodes of a stationary wave never move, while every point of a progressive wave moves by the same amount.
A stationary wave transfers no net energy along the wave. Energy is stored in the vibration.
A progressive wave carries energy from one place to another as it travels.

What you can now do

describe a progressive wave, and tell a transverse wave from a longitudinal one;
use $v = f\lambda$ and $T = \dfrac{1}{f}$ ;
find a phase difference and a path difference;
state the principle of superposition, and decide whether waves reinforce or cancel;
explain what makes two sources coherent;
use the fringe formula for two slits;
describe diffraction, and explain how the gap size changes the pattern;
use the grating equation, and find the highest order a grating can give;
explain polarisation, and work out the light through an analyser;
explain how a stationary wave forms;
find the harmonics of a string and of an air column;
use $v = \sqrt{\dfrac{T}{\mu}}$ for a wave on a string;
check whether an instrument is tuned to the note it should sound.

Your relevance reflection

Return to the prompt from the chapter opener: think of a time you noticed a wave effect, such as sound changing as you moved, or colours in a soap film, an oil slick or a CD. Write one more example now that you have finished the chapter. Has the chapter changed what you notice about sound, light or the strings and pipes that make music?