Sound Waves

Sound is a mechanical wave (it needs a medium) and it is longitudinal: the air molecules oscillate back and forth along the direction the wave travels, creating compressions and rarefactions.

Sound is a mechanical wave and needs a medium of particles to travel through. Space is a vacuum, so there is nothing to carry the vibration and no sound is heard.

A higher frequency is heard as a higher pitch. Amplitude sets the loudness, and the speed is fixed by the medium, not the note.

Loudness depends on amplitude (how far the air molecules swing, which sets the energy carried). Same pitch means the same frequency and wavelength, so only the amplitude differs.

Where molecules crowd together the air is momentarily at high pressure – a compression. Where they spread apart is a rarefaction (low pressure). These travel outward as the longitudinal sound wave.

The more tightly bound the particles, the faster a vibration passes between them, so sound is fastest in solids, slower in liquids and slowest in gases. It cannot travel at all in a vacuum.

$v = f\lambda = (170)(2.0) = 340\text{ m/s}$, the everyday speed of sound in air.

$\lambda = \dfrac{v}{f} = \dfrac{340}{220} = 1.55\text{ m}$. (Dividing the other way, $f/v$, gives the $0.65$ trap.)

The beat frequency is the difference of the two frequencies: $f_\text{beat} = |260 - 256| = 4\text{ Hz}$, so 4 beats every second.

Intensity from a point source obeys the inverse-square law, $I \propto \dfrac{1}{r^2}$. Doubling the distance multiplies the intensity by $\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$, so the intensity becomes $\tfrac{1}{4}I_0$. (Halving it, as if $I \propto 1/r$, is the trap.)

In $0.40\text{ s}$ the sound travels $d = vt = (1500)(0.40) = 600\text{ m}$, but that is down and back, so the depth is half: $300\text{ m}$. (Forgetting to halve gives the $600\text{ m}$ trap.)

Frequency is fixed by the source. The string drives the air at its own $330\text{ Hz}$, so the sound has that frequency. Only the speed and wavelength differ between the string and the air. (Thinking the medium sets the frequency is a common error – it sets the speed.)

$f = \dfrac{v}{2L} = \dfrac{340}{2(0.50)} = \dfrac{340}{1.0} = 340\text{ Hz}$. An open pipe has a displacement antinode at each end, so its fundamental fits half a wavelength in the length.

A closed pipe fits only a quarter of a wavelength in its length, so $f = \dfrac{v}{4L} = \dfrac{340}{4(0.50)} = \dfrac{340}{2.0} = 170\text{ Hz}$ – an octave below the open pipe of the same length.

A closed pipe must have a node at the closed end and an antinode at the open end. Only odd multiples of the fundamental fit that condition, so a closed pipe sounds the 1st, 3rd, 5th, ... harmonics – the even ones are missing.

Open pipe: $f = \dfrac{v}{2L}$. Closed pipe: $f = \dfrac{v}{4L}$, which is exactly half. So $\dfrac{300}{2} = 150\text{ Hz}$ – the closed pipe sounds an octave lower.

Pressure is highest (or lowest) where molecules crowd together (or spread apart) – and that happens where the displacement graph crosses zero with a steep slope. Where the displacement is largest, the molecules are momentarily all shifted the same way, so the pressure there is normal. Displacement antinodes are pressure nodes, and vice versa.

The beat frequency is $|f_\text{string} - 440| = 3$, so the string is $3\text{ Hz}$ either side: $437\text{ Hz}$ or $443\text{ Hz}$. From the beats alone you cannot tell which – you would need to change the string's tension to see if the beats speed up or slow down.

A closed pipe resonates only at odd harmonics, so after the 1st harmonic ($170\text{ Hz}$) the next is the 3rd: $3 \times 170 = 510\text{ Hz}$. (The $340\text{ Hz}$ trap assumes a 2nd harmonic, which a closed pipe cannot make.)

Since $I \propto \dfrac{1}{r^2}$, dropping the intensity to $\tfrac{1}{4}$ needs $r^2$ to grow by 4, so $r$ must double: $2 \times 3.0 = 6.0\text{ m}$. (Multiplying the distance by 4 to match the $\tfrac14$ gives the $12\text{ m}$ trap.)

The shortest resonating column is a quarter wavelength: $L = \tfrac{\lambda}{4}$, so $\lambda = 4L = 0.80\text{ m}$. Then $v = f\lambda = (425)(0.80) = 340\text{ m/s}$. (Taking $L = \lambda$ gives the $85\text{ m/s}$ trap.)

The frequencies are in the ratio $250 : 750 : 1250 = 1 : 3 : 5$ – only odd multiples of $250\text{ Hz}$. That odd-harmonics-only pattern is the signature of a pipe closed at one end, and the lowest frequency, $250\text{ Hz}$, is the fundamental. (An open pipe would give $1 : 2 : 3$.)

Four beats means the string was $508$ or $516\text{ Hz}$. Raising the frequency made the beats slow (from 4 to 2), so the string moved closer to $512$ – it must have started below it, at $508\text{ Hz}$. (Had it been $516$, raising it would have made the beats faster.)

Successive resonances of a closed tube are half a wavelength apart, so $\tfrac{\lambda}{2} = 0.54 - 0.18 = 0.36\text{ m}$, giving $\lambda = 0.72\text{ m}$. (Taking the difference to be a whole $\lambda$ gives the $0.36\text{ m}$ trap.)

Fundamental: $f_1 = \dfrac{v}{4L} = \dfrac{340}{4(0.25)} = 340\text{ Hz}$. A closed pipe sounds only odd harmonics, so the next two are $3f_1 = 1020\text{ Hz}$ and $5f_1 = 1700\text{ Hz}$. (Option B lists all harmonics – that would be an open pipe.)