📚 Phasor Diagrams: Intuitive Analysis of Superposition of Vibrations | 相量图：直观分析振动叠加
When two or more simple harmonic vibrations act together, the result can be surprisingly complex. Adding trigonometric functions directly is possible, but it is often algebraically messy and hides the physical meaning. Phasor diagrams convert each vibration into a rotating vector, making superposition visual and intuitive.
当两个或多个简谐振动同时作用时，结果可能出人意料地复杂。直接对三角函数进行加法运算虽然可行，但代数过程繁琐，且掩盖了物理本质。相量图将每个振动转化为旋转矢量，使叠加过程变得直观可视。
1. What Is a Phasor? | 什么是相量？
A phasor is a rotating vector whose length represents the amplitude of a simple harmonic oscillation, and whose angle with a fixed reference axis represents the phase. The instantaneous displacement is the projection of this phasor onto the reference axis.
相量是一个旋转矢量，其长度代表简谐振动的振幅，与固定参考轴的夹角代表相位。振动的瞬时位移就是该相量在参考轴上的投影。
x = A cos(ωt + φ₀)
Here, A is the amplitude, ω is the angular frequency, t is time, and φ₀ is the initial phase. The phasor rotates counterclockwise with angular speed ω.
其中A是振幅，ω是角频率，t是时间，φ₀是初相位。相量以角速度ω逆时针旋转。
2. Why Use Phasor Diagrams for Superposition? | 为什么用相量图分析叠加？
When two oscillations with the same frequency are added, the sum is another oscillation of the same frequency, but the amplitude and phase must be determined. Using trigonometric identities to find them is possible but tedious, especially for many components.
当两个同频率的振动相加时，合振动仍是同频率的振动，但振幅和相位需要确定。用三角恒等式求解虽然可行，但十分繁琐，尤其当分量较多时。
Phasor diagrams allow us to add oscillations by vector addition. The resultant phasor is simply the vector sum of the individual phasors, and its length and angle immediately give the amplitude and phase of the resultant vibration.
相量图允许我们通过矢量加法来叠加振动。合相量就是各分相量的矢量和，其长度和角度直接给出合振动的振幅和相位。
3. Representing a Vibration as a Phasor | 将振动表示为相量
For a vibration x = A cos(ωt + φ₀), draw an arrow of length A, starting from the origin, making an angle φ₀ with the positive x-axis at t = 0. At a later time, the angle becomes ωt + φ₀.
对于振动 x = A cos(ωt + φ₀)，从原点画一长度为A的箭头，在t = 0时与x轴正方向夹角为φ₀。在稍后时刻，角度变为ωt + φ₀。
- Length of phasor = amplitude A
- Angle with reference axis = phase (ωt + φ₀)
- Projection onto reference axis = displacement x
- 相量的长度 = 振幅 A
- 与参考轴的夹角 = 相位 (ωt + φ₀)
- 在参考轴上的投影 = 位移 x
4. Adding Two Same-Frequency Vibrations | 叠加两个同频率振动
Suppose two vibrations have the same angular frequency ω:
设有两个角频率均为ω的振动：
x₁ = A₁ cos(ωt + φ₁), x₂ = A₂ cos(ωt + φ₂)
Draw their phasors from the same origin. The resultant displacement is the sum of their projections, which equals the projection of the vector sum of the two phasors. Therefore, the resultant phasor is obtained by vector addition.
从同一点画出它们的相量。合位移等于两个投影之和，也等于两个相量矢量和的投影。因此，合相量通过矢量加法得到。
A = √(A₁² + A₂² + 2A₁A₂cos(φ₂ − φ₁))
This is the amplitude of the resultant vibration. The phase can be found from the tangent of the angle of the resultant phasor.
这就是合振动的振幅。合振动的相位可通过合相量与参考轴夹角的正切求出。
5. Special Cases: In-Phase, Anti-Phase, and Quadrature | 特殊情况：同相、反相与正交
The phase difference Δφ = φ₂ − φ₁ determines the nature of the superposition.
相位差 Δφ = φ₂ − φ₁ 决定了叠加的性质。
- If Δφ = 0 (in phase): A = A₁ + A₂, constructive reinforcement.
- If Δφ = π (anti-phase): A = |A₁ − A₂|, destructive cancellation.
- If Δφ = π/2 (quadrature): A = √(A₁² + A₂²), intermediate.
- 若 Δφ = 0（同相）：A = A₁ + A₂，相长增强。
- 若 Δφ = π（反相）：A = |A₁ − A₂|，相消减弱。
- 若 Δφ = π/2（正交）：A = √(A₁² + A₂²)，介于两者之间。
On a phasor diagram, in-phase phasors point in the same direction, anti-phase phasors point in opposite directions, and quadrature phasors are perpendicular.
在相量图中，同相相量指向同一方向，反相相量指向相反方向，正交相量相互垂直。
6. Phase Difference and Resultant Amplitude | 相位差与合振幅
The general formula for the resultant amplitude of two same-frequency vibrations is:
两个同频率振动合振幅的通用公式为：
A = √(A₁² + A₂² + 2A₁A₂cosΔφ)
where Δφ is the constant phase difference. This equation is central to interference phenomena.
其中Δφ为恒定相位差。该方程是干涉现象的核心。
Notice that the cross term 2A₁A₂cosΔφ depends only on the phase difference, not on time. This is why phasor diagrams are useful: they isolate the phase relationship.
注意交叉项2A₁A₂cosΔφ只依赖于相位差，与时间无关。这正是相量图的价值所在：它将相位关系单独分离出来。
7. Application to Wave Interference | 在波的干涉中的应用
In double-slit interference, two waves arriving at a point have a phase difference Δφ = (2π/λ)·ΔL, where ΔL is the path difference. The resultant amplitude at that point is found by adding the two wave phasors.
在双缝干涉中，到达某一点的两列波具有相位差 Δφ = (2π/λ)·ΔL，其中ΔL为光程差。该点的合振幅通过将两个波相量相加得出。
If A₁ = A₂ = A₀, the resultant intensity is proportional to:
若A₁ = A₂ = A₀，合强度正比于：
I ∝ 4A₀²cos²(Δφ/2)
This formula emerges naturally from the phasor diagram, where the two phasors bend into a symmetric resultant.
该公式可以很自然地从相量图中得出：两个等长相量合成一个对称的合相量。
8. Beats: Superposition of Different Frequencies | 拍：不同频率的叠加
When two vibrations have slightly different frequencies, their phase difference changes continuously with time. On a phasor diagram, the relative phasor rotates steadily, causing the resultant amplitude to grow and shrink periodically. This produces beats.
当两个振动频率略有不同时，它们的相位差随时间连续变化。在相量图中，相对相量稳定旋转，导致合振幅周期性地增大和减小，从而产生拍。
The beat frequency is the difference of the two frequencies:
拍频等于两频率之差：
f_beat = |f₂ − f₁|
The phasor diagram clarifies why the amplitude modulation occurs at this frequency: the relative phasor completes one full revolution relative to the other at a rate equal to the frequency difference.
相量图清楚地解释了为什么振幅调制以该频率发生：相对相量相对于另一个相量完成一次完整旋转的速率等于频率差。
9. Phasors and the Reference Circle | 相量与参考圆
A phasor diagram is closely related to the reference circle of simple harmonic motion. The tip of the phasor traces a circle of radius A, and the projection onto the x-axis gives x = A cos(ωt + φ₀).
相量图与简谐运动的参考圆密切相关。相量端点画出半径为A的圆，在x轴上的投影给出x = A cos(ωt + φ₀)。
This geometric interpretation allows visualisation of velocity and acceleration as well: the velocity phasor is perpendicular to the displacement phasor and leads it by π/2, while the acceleration phasor is opposite to the displacement.
这种几何解释也使速度和加速度可视化：速度相量垂直于位移相量并超前π/2，而加速度相量与位移相量方向相反。
10. Worked Example | 例题解析
Two vibrations act on a particle: x₁ = 3cos(ωt + π/3) and x₂ = 4cos(ωt − π/6). Find the amplitude of the resultant vibration.
两个振动作用于一个质点：x₁ = 3cos(ωt + π/3) 和 x₂ = 4cos(ωt − π/6)。求合振动的振幅。
Draw the two phasors: A₁ = 3 at 60°, A₂ = 4 at −30°. The phase difference is Δφ = (−30°) − (60°) = −90°.
画两个相量：A₁ = 3，方向60°；A₂ = 4，方向−30°。相位差为 Δφ = (−30°) − (60°) = −90°。
A = √(3² + 4² + 2·3·4·cos(−90°)) = √(9 + 16 + 0) = √25 = 5
The resultant amplitude is 5. The phasor diagram shows a right triangle with legs 3 and 4, confirming the result.
合振幅为5。相量图显示一个直角边为3和4的直角三角形，验证了结果。
11. Common Mistakes and Tips | 常见错误与提示
Students often treat two vibrations as adding like scalars. This is only correct when they are exactly in phase. In general, phasor addition is vector addition, and the phase difference must be included.
学生常将两个振动像标量一样直接相加。这仅在两振动完全同相时才正确。一般情况下，相量加法是矢量加法，必须计入相位差。
- Always draw the phasors with correct relative angles.
- Remember that phase is measured relative to a common reference.
- Convert degrees to radians when using the formula in SI units.
- Check whether the problem asks for amplitude, intensity, or phase.
- 画相量时务必保证相对角度正确。
- 记住相位是相对于共同参考轴测量的。
- 在SI单位制中使用公式时，要将角度由度转换为弧度。
- 注意题目问的是振幅、强度还是相位。
12. Conclusion | 总结
Phasor diagrams transform algebraic trigonometry into geometric vector addition. They provide an intuitive and powerful method for analysing the superposition of vibrations, from simple same-frequency addition to interference and beats. Mastering this tool is essential for IB Physics, especially in wave and oscillation topics.
相量图将代数三角运算转化为几何矢量加法。它为分析振动叠加提供了一种直观而强大的方法，从简单的同频叠加到干涉和拍现象均适用。掌握这一工具对于IB物理，特别是波动与振动部分，至关重要。
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