Waves and Sound波与声

The energy carried by a wave is proportional to the square of the amplitude: $E \propto A^2$. Doubling the amplitude quadruples the energy transported. Wave speed, however, depends on the properties of the medium (tension and linear density for a string; bulk modulus and density for sound in a fluid), not on amplitude or frequency. This is why turning up the volume of a speaker (larger amplitude) does not change how fast the sound travels across a room — the speed is set by the air, not the source.波携带的能量与振幅的平方成正比：$E \propto A^2$。振幅加倍则能量变为四倍。然而波速取决于介质的性质（绳的张力与线密度；流体中声音的体积模量与密度），而非振幅或频率。这就是为什么调大扬声器音量（更大振幅）不会改变声音穿越房间的速度 — 速度由空气决定，而非声源。

For a transverse wave on a string, the speed is $v = \sqrt{T/\mu}$, where $T$ is the tension (N) and $\mu$ is the linear mass density (kg/m). Neither the frequency nor the amplitude of the wave appears — the medium's mechanical properties alone determine how fast the disturbance propagates. For sound in a gas, $v = \sqrt{\gamma P/\rho}$ (where $\gamma$ is the adiabatic index, $P$ is pressure, $\rho$ is density), or approximately $v \approx 331 + 0.6 T_C$ m/s, where $T_C$ is the Celsius temperature. At room temperature ($20^\circ$C) this gives $v \approx 343$ m/s. The temperature dependence is why a tuning fork played in a cold room produces the same frequency but shorter wavelengths than in a warm room.对绳上的横波，波速为 $v = \sqrt{T/\mu}$，其中 $T$ 是张力（N），$\mu$ 是线密度（kg/m）。频率与振幅均未出现 — 仅由介质的力学性质决定扰动传播速率。对气体中的声波，$v = \sqrt{\gamma P/\rho}$（$\gamma$ 为绝热指数，$P$ 为压强，$\rho$ 为密度），近似为 $v \approx 331 + 0.6 T_C$ m/s，其中 $T_C$ 为摄氏温度。室温（$20^\circ$C）下得 $v \approx 343$ m/s。温度依赖性解释了为何音叉在冷室中产生的频率相同但波长比暖室中更短。

A standing wave arises when a wave and its reflection superpose. On a string fixed at both ends, the reflection at each fixed end must produce a node (the string cannot move there). The constraint is that an integer number of half-wavelengths must fit in the length $L$: $n \cdot (\lambda_n / 2) = L$, so $\lambda_n = 2L/n$. Substituting into $v = f\lambda$ gives $f_n = v/\lambda_n = nv/(2L)$. The closed-open pipe forces a node at the closed end and antinode at the open end; fitting an odd number of quarter-wavelengths in $L$ gives $\lambda_n = 4L/n$ (odd $n$ only), so $f_n = nv/(4L)$. This is why a clarinet (approximated as a closed-open cylinder) sounds an octave lower than a flute (open-open) of the same length.驻波在波与其反射叠加时产生。在两端固定的弦上，每个固定端的反射必须产生波节（弦在该处不能运动）。约束条件是长度 $L$ 内必须容纳整数个半波长：$n \cdot (\lambda_n / 2) = L$，故 $\lambda_n = 2L/n$。代入 $v = f\lambda$ 得 $f_n = v/\lambda_n = nv/(2L)$。封闭–开口管在封闭端强制有波节、开口端有波腹；在 $L$ 内容纳奇数个四分之一波长给出 $\lambda_n = 4L/n$（仅奇数 $n$），故 $f_n = nv/(4L)$。这就是为什么单簧管（近似为封闭–开口圆柱体）比同等长度的长笛（开口–开口）低八度。

When a source moves at exactly the speed of sound ($v_s = v$), the denominator in the Doppler formula becomes zero — all the wavefronts pile up ahead of the source into a single pressure front. As the source exceeds the speed of sound ($v_s > v$), these piled-up wavefronts form a cone-shaped shock wave (Mach cone) that trails behind the source. An observer on the ground hears nothing until the cone reaches them, then experiences a sudden pressure pulse — the sonic boom. The half-angle of the cone satisfies $\sin\theta = v/v_s = 1/M$, where $M = v_s/v$ is the Mach number. BC Physics 11 explicitly lists "sonic boom" as an elaboration under characteristics of sound; Alberta 20–D2.9k covers the Doppler effect quantitatively and the sonic-boom case follows from it.当声源恰好以声速（$v_s = v$）运动时，多普勒公式分母为零 — 所有波阵面在声源前方堆积成一个压力前锋。当声源超过声速（$v_s > v$）时，这些堆积的波阵面形成拖在声源后方的锥形激波（马赫锥）。地面的观察者在锥面到达之前什么也听不到，然后经历一个突然的压力脉冲 — 即音爆。锥的半角满足 $\sin\theta = v/v_s = 1/M$，其中 $M = v_s/v$ 是马赫数。BC Physics 11 明确把"音爆"列为声的特性细化说明；阿尔伯塔 20–D2.9k 定量涵盖多普勒效应，音爆情形由此推导得出。

Superpose two cosine waves of equal amplitude $A$ but slightly different frequencies $f_1$ and $f_2$ at a fixed point in space. Their sum is:在空间固定点叠加两列等振幅 $A$、频率略有不同（$f_1$ 和 $f_2$）的余弦波。其和为：

Applying the sum-to-product identity $\cos\alpha + \cos\beta = 2\cos\!\left(\tfrac{\alpha+\beta}{2}\right)\cos\!\left(\tfrac{\alpha-\beta}{2}\right)$:应用和差化积公式 $\cos\alpha + \cos\beta = 2\cos\!\left(\tfrac{\alpha+\beta}{2}\right)\cos\!\left(\tfrac{\alpha-\beta}{2}\right)$：

The result is a rapid oscillation at the average frequency $\bar f = (f_1 + f_2)/2$ (the pitch you hear) modulated by a slow envelope at frequency $(f_1 - f_2)/2$. Because the envelope completes a full cycle of "loud-soft-loud" twice per period of the slow term, the beat rate you hear is $|f_1 - f_2|$, not $|f_1 - f_2|/2$. This is why the formula $f_{\text{beat}} = |f_1 - f_2|$ has no factor of $1/2$ — each full period of the envelope contains two amplitude maxima.结果是以平均频率 $\bar f = (f_1 + f_2)/2$（你听到的音调）快速振动、被以频率 $(f_1 - f_2)/2$ 缓慢变化的包络调制。由于包络在慢项的每个周期内完成两次"响-弱-响"完整循环，你听到的拍率为 $|f_1 - f_2|$，而非 $|f_1 - f_2|/2$。这就是为什么公式 $f_{\text{beat}} = |f_1 - f_2|$ 没有 $1/2$ 因子 — 包络的每个完整周期包含两个振幅极大值。