IB Mathematics: Solving Harmonic Forcing with Complex Numbers | IB数学：用复数解处理谐波强迫项

📚 IB Mathematics: Solving Harmonic Forcing with Complex Numbers | IB数学：用复数解处理谐波强迫项

Many physical systems, from suspended bridges to electric circuits, experience a periodic external push. In the simplest model, a mass m is subject to a restoring force -kx and a driving force F cos(ωt). The motion satisfies a second-order linear differential equation. Standard trial-and-error with sines and cosines works, but the algebra grows rapidly when damping is included.

许多物理系统（从悬索桥到电路）都会受到周期性外力的作用。在最简单的模型中，质量 m 同时受到回复力 -kx 和驱动力 F cos(ωt)。运动满足一个二阶线性微分方程。用正弦和余弦逐项试解固然可行，但一旦加入阻尼，代数运算会迅速变得复杂。

1. The Problem: Real Oscillations | 问题：实数振荡

Consider a damped forced oscillator with equation m x″ + c x′ + k x = F cos(ωt). Bringing the system to resonance, computing phase shifts, or analysing amplitude responses all require solving this kind of equation. The homogeneous part gives the transient motion, but the particular solution determines the steady-state response that persists after transients die out.

考虑一个阻尼受迫振子，其方程为 m x″ + c x′ + k x = F cos(ωt)。无论是求共振、计算相位差，还是分析振幅响应，都需要求解这类方程。齐次部分给出瞬态运动，而特解决定瞬态衰减后仍然持续的稳态响应。

Writing x as a combination of cos(ωt) and sin(ωt) is possible, but the coefficients become complicated expressions, especially when the damping term c x′ couples the sine and cosine terms. The complex-number method avoids this by treating the whole forced response as a single complex exponential.

将 x 写成 cos(ωt) 和 sin(ωt) 的组合是可行的，但系数会变成复杂表达式，尤其当阻尼项 c x′ 把正弦项和余弦项耦合起来时。复数方法通过把整个受迫响应视为单个复指数函数，避免了这种复杂性。

2. Why Complex Numbers? | 为什么用复数？

The key advantage of complex numbers is linearity. A linear differential equation has the property that Re(solution to complex equation) is a solution to the original real equation. This lets us replace the awkward trigonometric forcing with a simple exponential, whose derivative is itself multiplied by a constant. Differentiation becomes multiplication, and integration becomes division.

复数的主要优势在于线性性。线性微分方程具有这样的性质：复数方程解出的解的实部，恰好是原实数方程的解。这使我们可以用简单的指数函数替代麻烦的三角函数，而指数函数求导后只相当于乘以一个常数。于是微分变成乘法，积分变成除法。

This approach is standard in IB Physics and Electrical Engineering, but it is also a powerful algebraic tool in IB Mathematics AA HL. Once the complex amplitude is found, everything else — amplitude, phase, resonance — follows automatically.

这一方法在 IB 物理和电气工程中很常见，但它也是 IB 数学 AA HL 中一个强大的代数工具。一旦求出复振幅，其余一切——振幅、相位、共振——都会自动得出。

3. Euler’s Formula: The Bridge | 欧拉公式：桥梁

Euler’s formula e^(iθ) = cos θ + i sin θ is the fundamental bridge. It shows that cos(ωt) = Re(e^(iωt)), and sin(ωt) = Im(e^(iωt)). More generally, a sine wave with phase shift can be written as Re(A_c e^(iωt)), where A_c is a complex number encoding amplitude and phase.

欧拉公式 e^(iθ) = cos θ + i sin θ 是关键的桥梁。它表明 cos(ωt) = Re(e^(iωt))，sin(ωt) = Im(e^(iωt))。更一般地，带有相位差的正弦波可以写成 Re(A_c e^(iωt))，其中 A_c 是编码振幅和相位的复数。

e^(iωt) = cos(ωt) + i sin(ωt)

Notice that differentiating e^(iωt) gives iω e^(iωt), and differentiating twice gives -ω² e^(iωt). Trigonometric addition formulas are replaced by simple powers of iω, which is precisely why the method is fast.

注意 e^(iωt) 对 t 求导得到 iω e^(iωt)，求导两次得到 -ω² e^(iωt)。三角函数的和角公式被 iω 的简单幂次取代，这正是该方法的快捷之处。

4. Complex Representation of Harmonic Forcing | 谐波强迫项的复数表示

For a forcing term F cos(ωt), define the complex forcing as F e^(iωt). Since the equation is linear, if x_c(t) satisfies the complex equation L[x_c] = F e^(iωt), then x = Re(x_c) satisfies L[x] = F cos(ωt). The same logic works for F sin(ωt) by taking the imaginary part.

对于强迫项 F cos(ωt)，定义复强迫项为 F e^(iωt)。由于方程是线性的，若 x_c(t) 满足复方程 L[x_c] = F e^(iωt)，则 x = Re(x_c) 满足 L[x] = F cos(ωt)。同理，对 F sin(ωt) 只需取虚部。

F cos(ωt) = Re(F e^(iωt))

The symbol L represents any linear differential operator. The advantage is that L operates on an exponential function, so the result is still the same exponential multiplied by some polynomial in iω.

符号 L 表示任意线性微分算子。好处在于 L 作用于指数函数时，结果仍然是同一个指数函数乘以 iω 的某个多项式。

5. Solving the Complex Equation | 求解复方程

Since the forcing has the form e^(iωt), we look for a particular solution of the same form: x_c(t) = A_c e^(iωt), with A_c a complex constant. Differentiating gives x_c′ = iω A_c e^(iωt) and x_c″ = -ω² A_c e^(iωt). Substitute these into the differential equation; the factor e^(iωt) cancels from every term, leaving an algebraic equation for A_c.

由于强迫项形式为 e^(iωt)，我们设相同形式的特解：x_c(t) = A_c e^(iωt)，其中 A_c 是复常数。求导得 x_c′

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）