Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus C4.1 to C4.6考纲 C4.1 至 C4.6PHYSICS HL
Standing wave on a string from an incident wave and its reflection. (a) two conditions for formation; (b) two ways it differs from a travelling wave.弦上由入射波与其反射波形成驻波。(a) 形成的两个条件；(b) 与行波的两点不同。
The two waves must have the same frequency (hence the same wavelength and speed on the same medium) and ideally equal amplitude. (B1)
They must travel in opposite directions along the same line, so that they continuously superpose. (B1)
A standing wave transfers no net energy along the string, whereas a travelling wave carries energy forward. (B1)
In a standing wave the amplitude varies with position (zero at nodes, maximum at antinodes) and all points between adjacent nodes are in phase; in a travelling wave every point has the same amplitude and the phase varies steadily with position. (B1)
两列波必须频率相同（在同一介质上故波长与波速相同），且理想情况下振幅相等。(B1)
它们必须沿同一直线反向传播，从而持续叠加。(B1)
驻波沿弦不传递净能量，而行波将能量向前输运。(B1)
驻波中振幅随位置变化（波节为零、波腹最大），且相邻波节间所有点同相；行波中每点振幅相同，相位随位置稳定变化。(B1)
Standing wave: adjacent nodes $0.18\ \mathrm{m}$ apart, wave speed $144\ \mathrm{m\,s^{-1}}$. (a) wavelength; (b) frequency; (c) node-to-antinode distance and phase across a node.驻波：相邻波节相距 $0.18\ \mathrm{m}$，波速 $144\ \mathrm{m\,s^{-1}}$。(a) 波长；(b) 频率；(c) 波节到波腹距离与跨波节的相位。
Adjacent nodes are half a wavelength apart: $\tfrac{\lambda}{2} = 0.18\ \mathrm{m}$. (M1)
$$ \lambda = 2 \times 0.18 = 0.36\ \mathrm{m}. $$(A1)
Use $v = f\lambda$: $f = \dfrac{v}{\lambda} = \dfrac{144}{0.36}$. (M1)
$$ f = 400\ \mathrm{Hz}. $$(A1)
A node and the nearest antinode are a quarter wavelength apart: $\tfrac{\lambda}{4} = 0.090\ \mathrm{m}$. (A1)
Two points on opposite sides of a node (in neighbouring loops) oscillate exactly antiphase, that is $\pi$ out of phase. (A1)
相邻波节相距半个波长：$\tfrac{\lambda}{2} = 0.18\ \mathrm{m}$。(M1)
$$ \lambda = 2 \times 0.18 = 0.36\ \mathrm{m}. $$(A1)
用 $v = f\lambda$：$f = \dfrac{v}{\lambda} = \dfrac{144}{0.36}$。(M1)
$$ f = 400\ \mathrm{Hz}. $$(A1)
波节与最近波腹相距四分之一波长：$\tfrac{\lambda}{4} = 0.090\ \mathrm{m}$。(A1)
波节两侧（相邻环内）的两点恰好反相，即相差 $\pi$。(A1)
Wire $L = 0.80\ \mathrm{m}$ fixed both ends, $v = 320\ \mathrm{m\,s^{-1}}$. (a) boundary condition and allowed wavelengths; (b) fundamental and third harmonic; (c) nodes and antinodes in the third harmonic.弦 $L = 0.80\ \mathrm{m}$ 两端固定，$v = 320\ \mathrm{m\,s^{-1}}$。(a) 边界条件与允许波长；(b) 基频与第三谐波；(c) 第三谐波的波节与波腹数。
A string fixed at both ends has a node at each end. (B1)
A whole number of half-wavelengths must fit the length: $L = n\tfrac{\lambda_n}{2}$, so $\lambda_n = \dfrac{2L}{n}$, $n = 1,2,3,\dots$ (A1)
With $f_n = \dfrac{nv}{2L}$: $f_1 = \dfrac{320}{2(0.80)} = \dfrac{320}{1.60} = 200\ \mathrm{Hz}$. (M1)
$$ f_3 = 3f_1 = 3 \times 200 = 600\ \mathrm{Hz}. $$(A1)
The third harmonic has $n = 3$ loops, hence $n + 1 = 4$ nodes counting both fixed ends. (A1)
There is one antinode at the centre of each loop, so there are $3$ antinodes. (A1)
两端固定的弦在两端各有一个波节。(B1)
整数个半波长须装入弦长：$L = n\tfrac{\lambda_n}{2}$，故 $\lambda_n = \dfrac{2L}{n}$，$n = 1,2,3,\dots$ (A1)
由 $f_n = \dfrac{nv}{2L}$：$f_1 = \dfrac{320}{2(0.80)} = \dfrac{320}{1.60} = 200\ \mathrm{Hz}$。(M1)
$$ f_3 = 3f_1 = 3 \times 200 = 600\ \mathrm{Hz}. $$(A1)
第三谐波有 $n = 3$ 个环，故含两固定端共 $n + 1 = 4$ 个波节。(A1)
每个环中央有一个波腹，故有 $3$ 个波腹。(A1)
Closed pipe $L = 0.50\ \mathrm{m}$, $v = 340\ \mathrm{m\,s^{-1}}$. (a) boundary conditions and fundamental; (b) next harmonic above the fundamental; (c) open-pipe fundamental and comparison.闭管 $L = 0.50\ \mathrm{m}$，$v = 340\ \mathrm{m\,s^{-1}}$。(a) 边界条件与基频；(b) 基频之上的下一谐波；(c) 开管基频及比较。
The open end is a displacement antinode and the closed end is a displacement node, so the shortest fit is a quarter wavelength and $f_n = \dfrac{nv}{4L}$ with odd $n$. (B1)
$$ f_1 = \frac{(1)(340)}{4(0.50)} = \frac{340}{2.00} = 170\ \mathrm{Hz}. $$(A1)
A closed pipe skips even harmonics, so the next one above $f_1$ is the third harmonic. (M1)
$$ f_3 = 3f_1 = 3 \times 170 = 510\ \mathrm{Hz}. $$(A1)
An open–open pipe uses $f_1 = \dfrac{v}{2L} = \dfrac{340}{2(0.50)} = 340\ \mathrm{Hz}$. (M1)
This is exactly double the closed-pipe fundamental: closing one end lowers the pitch by an octave. (A1)
开口端为位移波腹、闭口端为位移波节，故最短能装下的是四分之一波长，且 $f_n = \dfrac{nv}{4L}$，$n$ 取奇数。(B1)
$$ f_1 = \frac{(1)(340)}{4(0.50)} = \frac{340}{2.00} = 170\ \mathrm{Hz}. $$(A1)
闭管跳过偶数谐波，故 $f_1$ 之上的下一个是第三谐波。(M1)
$$ f_3 = 3f_1 = 3 \times 170 = 510\ \mathrm{Hz}. $$(A1)
开–开管用 $f_1 = \dfrac{v}{2L} = \dfrac{340}{2(0.50)} = 340\ \mathrm{Hz}$。(M1)
这恰为闭管基频的两倍：封闭一端使音调降低一个八度。(A1)
Driven mass-spring, natural frequency $f_0$, driving frequency $f_d$ variable. (a) define resonance and its condition; (b) steady-state frequency when $f_d \neq f_0$; (c) effect of increasing damping on the resonance curve; (d) one useful and one hazardous example.受迫弹簧振子，固有频率 $f_0$，驱动频率 $f_d$ 可变。(a) 定义共振及条件；(b) $f_d \neq f_0$ 时稳态频率；(c) 增大阻尼对共振曲线的影响；(d) 一个有用、一个有害的例子。
Resonance is the large-amplitude response that occurs when energy is transferred most efficiently from the driver to the system. (A1)
It occurs when the driving frequency equals a natural frequency of the system: $f_d = f_0$. (A1)
In the steady state a forced oscillator oscillates at the driving frequency $f_d$, not at its own natural frequency. (A1)
The natural frequency only governs how large the response is (how close $f_d$ is to $f_0$ sets the amplitude), not the frequency of oscillation. (R1)
Increasing the damping lowers the peak amplitude and broadens it, because energy is dissipated faster. (A1)
The peak also shifts to a slightly lower frequency than $f_0$. (A1)
Useful: tuning a radio so its circuit resonates with one broadcast frequency (or pushing a swing at its natural frequency). (B1)
Hazardous: wind or footfall driving a bridge or building near a natural frequency, as in the Tacoma Narrows or Millennium Bridge, against which engineers add dampers. (B1)
共振是当能量从驱动者向系统传递效率最高时出现的大振幅响应。(A1)
它发生在驱动频率等于系统某固有频率时：$f_d = f_0$。(A1)
稳态下受迫振子以驱动频率 $f_d$ 振动，而非自身固有频率。(A1)
固有频率只决定响应有多大（$f_d$ 与 $f_0$ 的接近程度决定振幅），而不决定振动频率。(R1)
增大阻尼降低峰值振幅并使其展宽，因为能量耗散更快。(A1)
峰也移向略低于 $f_0$ 的频率。(A1)
有用：调收音机使电路与某一广播频率共振（或以固有频率推秋千）。(B1)
有害：风或脚步以接近固有频率驱动桥梁或建筑，如塔科马海峡大桥或千禧桥，工程师为此加装阻尼器。(B1)
Open–open pipe; fundamental $f$ measured for several lengths $L$; $f$ vs $1/L$ data given. (a) show $f$ vs $1/L$ is linear through the origin and state the gradient; (b) gradient and speed of sound; (c) percentage uncertainty in $f$ at $L = 0.20$; (d) factor change if one end is closed; (e) one systematic error.开–开管；对若干长度 $L$ 测量基频 $f$；给出 $f$ 对 $1/L$ 的数据。(a) 证明 $f$ 对 $1/L$ 过原点线性并说明斜率；(b) 斜率与声速；(c) $L = 0.20$ 处 $f$ 的百分比不确定度；(d) 一端封闭时改变的倍数；(e) 一个系统误差。
For an open–open pipe the fundamental is $f = \dfrac{v}{2L}$. (M1)
$$ f = \frac{v}{2}\cdot\frac{1}{L}. $$This has the form $f = (\text{gradient})\times\tfrac{1}{L}$ with no intercept, so $f$ against $1/L$ is a straight line through the origin. (A1)
Comparing with $y = mx$, the gradient is $\dfrac{v}{2}$. (A1)
Read the gradient from two well-separated points, e.g. $(1.00,\,170)$ and $(5.00,\,850)$: (M1)
$$ \text{gradient} = \frac{850 - 170}{5.00 - 1.00} = \frac{680}{4.00} = 170\ \mathrm{m\,s^{-1}}. $$(A1)
Since the gradient equals $\tfrac{v}{2}$: $v = 2 \times 170 = 340\ \mathrm{m\,s^{-1}}$. (A1)
At that point $f = 850\ \mathrm{Hz}$ with absolute uncertainty $\pm 10\ \mathrm{Hz}$: (M1)
$$ \frac{10}{850}\times 100\% \approx 1.18\% \approx 1\%. $$(A1)
A closed pipe has $f_1 = \dfrac{v}{4L}$, which is half of $\dfrac{v}{2L}$. (M1)
So at every length the fundamental falls to one half of its open-pipe value (a factor of $\tfrac{1}{2}$). (A1)
The end correction: the displacement antinode lies slightly beyond the open end, so the effective length exceeds the measured $L$. (B1)
This makes every measured frequency consistently a little lower than the ideal $\tfrac{v}{2L}$, a systematic rather than random shift. (R1)
开–开管的基频为 $f = \dfrac{v}{2L}$。(M1)
$$ f = \frac{v}{2}\cdot\frac{1}{L}. $$此式形如 $f = (\text{斜率})\times\tfrac{1}{L}$，无截距，故 $f$ 对 $1/L$ 为过原点的直线。(A1)
与 $y = mx$ 比较，斜率为 $\dfrac{v}{2}$。(A1)
用相距较远的两点读斜率，如 $(1.00,\,170)$ 与 $(5.00,\,850)$：(M1)
$$ \text{斜率} = \frac{850 - 170}{5.00 - 1.00} = \frac{680}{4.00} = 170\ \mathrm{m\,s^{-1}}. $$(A1)
因斜率等于 $\tfrac{v}{2}$：$v = 2 \times 170 = 340\ \mathrm{m\,s^{-1}}$。(A1)
该点 $f = 850\ \mathrm{Hz}$，绝对不确定度 $\pm 10\ \mathrm{Hz}$：(M1)
$$ \frac{10}{850}\times 100\% \approx 1.18\% \approx 1\%. $$(A1)
闭管基频为 $f_1 = \dfrac{v}{4L}$，是 $\dfrac{v}{2L}$ 的一半。(M1)
故每个长度下基频都降为开管值的一半（$\tfrac{1}{2}$ 倍）。(A1)
端修正：位移波腹略在开口端之外，故有效长度大于所测 $L$。(B1)
这使每个所测频率都系统性地略低于理想的 $\tfrac{v}{2L}$，是系统偏移而非随机。(R1)
String $L = 0.60\ \mathrm{m}$ fixed both ends; resonant $f$ vs harmonic number $n$ data given. (a) why $f$ vs $n$ is linear through the origin; (b) gradient and its meaning; (c) wave speed; (d) new fundamental and graph change when tension is quadrupled, given $v = \sqrt{T/\mu}$.弦 $L = 0.60\ \mathrm{m}$ 两端固定；给出共振 $f$ 对谐波次数 $n$ 的数据。(a) 为何 $f$ 对 $n$ 过原点线性；(b) 斜率及含义；(c) 波速；(d) 张力变四倍时（已知 $v = \sqrt{T/\mu}$）的新基频与图的变化。
For a fixed-fixed string $f_n = \dfrac{nv}{2L} = \left(\dfrac{v}{2L}\right)n$. (M1)
This is of the form $f = (\text{constant})\times n$ with no intercept, so the graph is a straight line through the origin. (A1)
Read the gradient from two points, e.g. $(1,\,200)$ and $(5,\,1000)$: (M1)
$$ \text{gradient} = \frac{1000 - 200}{5 - 1} = \frac{800}{4} = 200\ \mathrm{Hz}. $$(A1)
The gradient equals $\dfrac{v}{2L}$, which is the fundamental frequency $f_1 = 200\ \mathrm{Hz}$. (A1)
The gradient is $\dfrac{v}{2L}$, so $v = 2L \times \text{gradient}$. (M1)
$$ v = 2(0.60)(200) = 1.20 \times 200. $$(M1 for substitution)
$$ v = 240\ \mathrm{m\,s^{-1}}. $$(A1)
Wave speed: $v = \sqrt{T/\mu}$, so quadrupling $T$ multiplies $v$ by $\sqrt{4} = 2$, giving $v' = 480\ \mathrm{m\,s^{-1}}$. (M1)
The new fundamental is $f_1' = \dfrac{v'}{2L} = \dfrac{480}{1.20} = 400\ \mathrm{Hz}$, double the original. (A1·A1)
Since the gradient of the $f$–$n$ graph is $f_1$, the line stays straight through the origin but its gradient doubles (it becomes steeper). (R1)
两端固定弦 $f_n = \dfrac{nv}{2L} = \left(\dfrac{v}{2L}\right)n$。(M1)
此式形如 $f = (\text{常数})\times n$，无截距，故图为过原点的直线。(A1)
用两点读斜率，如 $(1,\,200)$ 与 $(5,\,1000)$：(M1)
$$ \text{斜率} = \frac{1000 - 200}{5 - 1} = \frac{800}{4} = 200\ \mathrm{Hz}. $$(A1)
斜率等于 $\dfrac{v}{2L}$，即基频 $f_1 = 200\ \mathrm{Hz}$。(A1)
斜率为 $\dfrac{v}{2L}$，故 $v = 2L \times \text{斜率}$。(M1)
$$ v = 2(0.60)(200) = 1.20 \times 200. $$(代入得 M1)
$$ v = 240\ \mathrm{m\,s^{-1}}. $$(A1)
波速 $v = \sqrt{T/\mu}$，张力变四倍使 $v$ 变为 $\sqrt{4} = 2$ 倍，得 $v' = 480\ \mathrm{m\,s^{-1}}$。(M1)
新基频 $f_1' = \dfrac{v'}{2L} = \dfrac{480}{1.20} = 400\ \mathrm{Hz}$，为原来的两倍。(A1·A1)
因 $f$–$n$ 图的斜率即 $f_1$，直线仍过原点保持笔直，但斜率加倍（变陡）。(R1)
Guitar string $L = 0.65\ \mathrm{m}$ fixed both ends, $v = 260\ \mathrm{m\,s^{-1}}$. (a) show superposition gives $y = 2A\sin(kx)\cos(\omega t)$ and identify the amplitude factor; (b) show nodes are $\tfrac{\lambda}{2}$ apart; (c) fundamental frequency and third-harmonic wavelength; (d) why no net energy is transferred; (e) new fundamental at $L = 0.49\ \mathrm{m}$.吉他弦 $L = 0.65\ \mathrm{m}$ 两端固定，$v = 260\ \mathrm{m\,s^{-1}}$。(a) 证明叠加给出 $y = 2A\sin(kx)\cos(\omega t)$ 并指出振幅因子；(b) 证明波节相距 $\tfrac{\lambda}{2}$；(c) 基频与第三谐波波长；(d) 为何不传递净能量；(e) $L = 0.49\ \mathrm{m}$ 时的新基频。
Add the two waves and apply $\sin P + \sin Q = 2\sin\tfrac{P+Q}{2}\cos\tfrac{P-Q}{2}$ with $P = kx - \omega t$, $Q = kx + \omega t$: (M1)
$$ y = A\sin(kx-\omega t) + A\sin(kx+\omega t) = 2A\sin(kx)\cos(\omega t). $$(A1)
The position factor $2A\sin(kx)$ fixes the amplitude at each point $x$; the factor $\cos(\omega t)$ only sets the common time oscillation. (A1)
A node is a permanently stationary point, so $2A\sin(kx) = 0$, giving $kx = n\pi$ and $x = \dfrac{n\pi}{k}$. (M1)
Using $k = \dfrac{2\pi}{\lambda}$, consecutive nodes are separated by (M1)
$$ \Delta x = \frac{(n+1)\pi}{k} - \frac{n\pi}{k} = \frac{\pi}{k} = \frac{\pi}{2\pi/\lambda} = \frac{\lambda}{2}. $$(A1)
Fundamental: $f_1 = \dfrac{v}{2L} = \dfrac{260}{2(0.65)} = \dfrac{260}{1.30} = 200\ \mathrm{Hz}$. (M1·A1)
Third-harmonic wavelength: $\lambda_3 = \dfrac{2L}{3} = \dfrac{1.30}{3} \approx 0.43\ \mathrm{m}$. (A1)
The standing wave is the sum of two equal travelling waves carrying equal energy in opposite directions, so their energy flows cancel and the net flow is zero. (R1)
Energy is not lost but stored: it oscillates between kinetic (maximum at the antinodes as the string passes through equilibrium) and elastic potential (maximum at full displacement). (R1)
Because a node never moves, no energy can pass across it, so energy stays trapped within each loop. (A1)
Same wave speed, shorter length: $f_1' = \dfrac{v}{2L'} = \dfrac{260}{2(0.49)} = \dfrac{260}{0.98}$. (M1)
$$ f_1' \approx 265\ \mathrm{Hz}. $$(A1)
把两列波相加并用 $\sin P + \sin Q = 2\sin\tfrac{P+Q}{2}\cos\tfrac{P-Q}{2}$，取 $P = kx - \omega t$、$Q = kx + \omega t$：(M1)
$$ y = A\sin(kx-\omega t) + A\sin(kx+\omega t) = 2A\sin(kx)\cos(\omega t). $$(A1)
位置因子 $2A\sin(kx)$ 固定每个点 $x$ 的振幅；$\cos(\omega t)$ 只设定共同的时间振荡。(A1)
波节是始终静止的点，故 $2A\sin(kx) = 0$，得 $kx = n\pi$、$x = \dfrac{n\pi}{k}$。(M1)
用 $k = \dfrac{2\pi}{\lambda}$，相邻波节相距 (M1)
$$ \Delta x = \frac{(n+1)\pi}{k} - \frac{n\pi}{k} = \frac{\pi}{k} = \frac{\pi}{2\pi/\lambda} = \frac{\lambda}{2}. $$(A1)
基频：$f_1 = \dfrac{v}{2L} = \dfrac{260}{2(0.65)} = \dfrac{260}{1.30} = 200\ \mathrm{Hz}$。(M1·A1)
第三谐波波长：$\lambda_3 = \dfrac{2L}{3} = \dfrac{1.30}{3} \approx 0.43\ \mathrm{m}$。(A1)
驻波是两列等幅行波之和，它们沿相反方向携带等量能量，故能量流相消、净流为零。(R1)
能量并未损失而是被储存：在动能（弦经过平衡位置时波腹处最大）与弹性势能（最大位移处最大）之间振荡。(R1)
由于波节从不移动，能量无法越过它，故能量被困在每个环内。(A1)
波速不变、长度变短：$f_1' = \dfrac{v}{2L'} = \dfrac{260}{2(0.49)} = \dfrac{260}{0.98}$。(M1)
$$ f_1' \approx 265\ \mathrm{Hz}. $$(A1)
Resonance tube closed at the water surface, open at the top; tuning fork $480\ \mathrm{Hz}$, $v = 340\ \mathrm{m\,s^{-1}}$. (a) why loud sound only at certain lengths; (b) shortest resonant length; (c) next two lengths and the spacing; (d) role of damping and whether light or heavy damping makes a resonance easier to locate.共振管在水面处闭口、顶部开口；音叉 $480\ \mathrm{Hz}$，$v = 340\ \mathrm{m\,s^{-1}}$。(a) 为何只在某些长度处声音响亮；(b) 最短共振长度；(c) 接下来两个长度与间距；(d) 阻尼的作用以及轻阻尼还是重阻尼更易定位共振。
The air column has its own natural frequencies, set by its length and the open/closed boundary conditions. (M1)
A loud sound (resonance) is heard only when a natural frequency of the column matches the fork frequency, $f_0 = f_d = 480\ \mathrm{Hz}$, so only certain lengths resonate. (A1)
The column is closed at the water and open at the top, so the fundamental fit is $L = \dfrac{\lambda}{4}$. First find the wavelength: $\lambda = \dfrac{v}{f} = \dfrac{340}{480} \approx 0.7083\ \mathrm{m}$. (M1)
Then $L_1 = \dfrac{\lambda}{4}$: (M1)
$$ L_1 = \frac{0.7083}{4} \approx 0.177\ \mathrm{m}. $$(A1)
A closed column resonates at odd quarter-wavelengths, $L_n = \dfrac{(2m-1)\lambda}{4}$, so the next two are the $\tfrac{3\lambda}{4}$ and $\tfrac{5\lambda}{4}$ fits. (M1)
$$ L_2 = \frac{3\lambda}{4} = 3(0.177) \approx 0.531\ \mathrm{m}, \qquad L_3 = \frac{5\lambda}{4} = 5(0.177) \approx 0.885\ \mathrm{m}. $$(A1·A1)
Successive resonant lengths differ by $\dfrac{\lambda}{2} = \dfrac{0.7083}{2} \approx 0.354\ \mathrm{m}$. (A1)
Damping (here mainly sound energy radiated away and lost to the tube walls and air) removes energy from the resonating column, limiting the peak loudness. (M1)
Light damping gives a tall, narrow resonance peak; heavy damping gives a low, broad one. (A1)
A narrow peak means the loudness rises and falls sharply with length, so light damping makes the resonant length easier to pinpoint; heavy damping spreads the response over a range of lengths and blurs it. (R1)
空气柱有自己的固有频率，由其长度与开/闭边界条件决定。(M1)
只有当空气柱的某固有频率与音叉频率匹配时，$f_0 = f_d = 480\ \mathrm{Hz}$，才听到响亮的声音（共振），故只有某些长度共振。(A1)
空气柱在水面闭口、顶部开口，故基频对应 $L = \dfrac{\lambda}{4}$。先求波长：$\lambda = \dfrac{v}{f} = \dfrac{340}{480} \approx 0.7083\ \mathrm{m}$。(M1)
再求 $L_1 = \dfrac{\lambda}{4}$：(M1)
$$ L_1 = \frac{0.7083}{4} \approx 0.177\ \mathrm{m}. $$(A1)
闭口空气柱在奇数个四分之一波长处共振，$L_n = \dfrac{(2m-1)\lambda}{4}$，故接下来两个是 $\tfrac{3\lambda}{4}$ 与 $\tfrac{5\lambda}{4}$。(M1)
$$ L_2 = \frac{3\lambda}{4} = 3(0.177) \approx 0.531\ \mathrm{m}, \qquad L_3 = \frac{5\lambda}{4} = 5(0.177) \approx 0.885\ \mathrm{m}. $$(A1·A1)
相邻共振长度相差 $\dfrac{\lambda}{2} = \dfrac{0.7083}{2} \approx 0.354\ \mathrm{m}$。(A1)
阻尼（这里主要是辐射出去并被管壁与空气损耗的声能）从共振空气柱移走能量，限制峰值响度。(M1)
轻阻尼给出高而窄的共振峰；重阻尼给出低而宽的峰。(A1)
窄峰意味响度随长度急升急降，故轻阻尼使共振长度更易精确定位；重阻尼把响应铺展到一段长度范围而使其模糊。(R1)