📚 IB Mathematics: Resonance in Forced Vibrations | IB数学：受迫振动中的共振
In this article, we explore the mathematics behind forced vibrations, focusing on the phenomenon of resonance. We will derive the steady-state amplitude, identify the resonant frequency, and understand how damping affects the response of a system. This topic is essential for IB Mathematics Analysis and Approaches HL, especially in the context of modelling real-world oscillatory systems.
本文将探讨受迫振动背后的数学原理，重点分析共振现象。我们将推导稳态振幅、确定共振频率，并理解阻尼如何影响系统的响应。这一主题是 IB 数学分析与方法 HL 的重要考点，尤其适用于对现实世界中振荡系统进行建模。
1. The Forced Vibration Equation | 受迫振动方程
For a damped harmonic oscillator subjected to an external periodic force, the displacement x(t) satisfies the second-order linear differential equation:
对于受周期外力作用的阻尼谐振子，其位移 x(t) 满足如下二阶线性微分方程：
m x” + c x’ + k x = F₀ cos(ωt)
Here, m is the mass, c is the damping coefficient, k is the spring constant, F₀ is the amplitude of the driving force, and ω is the angular frequency of the external force.
其中，m 是质量，c 是阻尼系数，k 是劲度系数，F₀ 是驱动力的振幅，ω 是外力的角频率。
2. Homogeneous Solution and Transient Behaviour | 齐次解与暂态行为
The homogeneous equation m x” + c x’ + k x = 0 models the free damped motion. Its solution decays exponentially over time, representing the transient behaviour of the system.
齐次方程 m x” + c x’ + k x = 0 描述自由阻尼运动，其解随时间按指数衰减，代表系统的暂态行为。
For underdamped motion, the homogeneous solution is:
对于欠阻尼运动，齐次解为：
x_h(t) = e−γt (A cos(ω_d t) + B sin(ω_d t))
where γ = c/(2m) is the damping rate and ω_d = √(ω₀² − γ²) is the damped natural frequency, with ω₀ = √(k/m).
其中 γ = c/(2m) 为阻尼率，ω_d = √(ω₀² − γ²) 为阻尼固有频率，且 ω₀ = √(k/m)。
As t → ∞, the homogeneous solution vanishes, so the steady-state response is determined entirely by the particular solution.
当 t → ∞ 时，齐次解趋于零，因此稳态响应完全由特解决定。
3. Particular Solution and Steady State | 特解与稳态
To find the steady-state response, we assume a particular solution of the same frequency as the driving force:
为求稳态响应，我们假设一个与驱动力同频率的特解：
x_p(t) = A cos(ωt − δ)
Substituting this into the differential equation and matching coefficients gives the amplitude A and phase lag δ.
将此代入微分方程并比较系数，即可得到振幅 A 和相位滞后 δ。
The steady-state amplitude depends on the driving frequency ω, the natural frequency ω₀, the damping coefficient c, and the mass m.
稳态振幅取决于驱动频率 ω、固有频率 ω₀、阻尼系数 c 和质量 m。
4. Amplitude Formula | 振幅公式
After solving the algebraic equations, the steady-state amplitude is:
求解代数方程后，稳态振幅为：
A = F₀ / √( (k − mω²)² + (cω)² )
Equivalently, using ω₀ and γ, this can be written as:
等价地，利用 ω₀ 和 γ，可写为：
A = (F₀/m) / √( (ω₀² − ω²)² + (2γω)² )
This formula shows how the amplitude varies with the driving frequency. When ω is near ω₀, the denominator becomes small, leading to a large amplitude.
该公式展示了振幅随驱动频率的变化。当 ω 接近 ω₀ 时，分母变小，导致振幅显著增大。
5. Resonant Frequency | 共振频率
Resonance occurs when the driving frequency maximizes the amplitude. To find this frequency, we minimise the square of the denominator:
当驱动频率使振幅达到最大时，即发生共振。为求该频率，我们最小化分母的平方：
D(ω) = (ω₀² − ω²)² + (2γω)²
Differentiating with respect to ω and setting the derivative to zero gives the resonant frequency:
对 ω 求导并令导数为零，得到共振频率：
ω_res = √(ω₀² − 2γ²)
Note that resonance exists only if ω₀² > 2γ², i.e. the damping is not too large.
注意，只有满足 ω₀² > 2γ²（即阻尼不太大）时，共振才存在。
6. Maximum Amplitude | 最大振幅
Substituting ω_res back into the amplitude formula gives the maximum steady-state amplitude:
将 ω_res 代回振幅公式，可得最大稳态振幅：
A_max = F₀ / (2m γ √(ω₀² − γ²) )
In terms of the damping ratio ζ = γ/ω₀, this becomes:
用阻尼比 ζ = γ/ω₀ 表示，则：
A_max = (F₀/k) / (2ζ √(1 − ζ²) )
As ζ approaches zero, the maximum amplitude tends to infinity, representing undamped resonance. In practice, damping always limits the amplitude.
当 ζ 趋于零时，最大振幅趋向无穷，对应无阻尼共振。现实中，阻尼总会限制振幅的大小。
7. Phase Lag and Resonance | 相位滞后与共振
The phase lag δ describes how much the displacement lags behind the driving force. It satisfies:
相位滞后 δ 描述位移落后于驱动力的程度，满足：
tan δ = (cω) / (k − mω²) = (2γω) / (ω₀² − ω²)
At low frequencies (ω ≪ ω₀), the displacement is nearly in phase with the force (δ ≈ 0). At high frequencies, the displacement is nearly opposite to the force (δ ≈ π).
在低频时（ω ≪ ω₀），位移几乎与驱动力同相（δ ≈ 0）。在高频时，位移几乎与驱动力反相（δ ≈ π）。
Exactly at resonance, the phase lag is π/2, meaning the velocity is in phase with the driving force, which maximises energy transfer.
恰好在共振频率处，相位滞后为 π/2，即速度与驱动力同相，从而使能量传递达到最大。
8. Quality Factor and Bandwidth | 品质因数与带宽
The sharpness of the resonance peak is measured by the quality factor Q, defined as:
共振峰的尖锐程度用品质因数 Q 来度量，定义为：
Q = ω₀ / (2γ) = 1/(2ζ)
A high Q means low damping, a narrow resonant peak, and a large amplitude at resonance. The bandwidth Δω is the range of frequencies for which the amplitude is at least 1/√2 of its maximum value:
高 Q 意味着低阻尼、尖锐的共振峰以及共振时的大振幅。带宽 Δω 是振幅不低于最大值 1/√2 的频率范围：
Δω ≈ ω₀ / Q
This relation is widely used in electrical circuits, mechanical systems, and acoustics.
该关系广泛应用于电路、机械系统和声学中。
9. Worked Example: Forced Oscillator | 例题：受迫振子
Consider a mass m = 2 kg, spring constant k = 200 N/m, damping coefficient c = 4 N·s/m, and driving amplitude F₀ = 10 N. Find the resonant frequency and the maximum amplitude.
考虑质量 m = 2 kg，劲度系数 k = 200 N/m，阻尼系数 c = 4 N·s/m，驱动力幅值 F₀ = 10 N 的系统。求共振频率和最大振幅。
First compute ω₀ = √(k/m) = √(200/2) = 10 rad/s. The damping rate γ = c/(2m) = 4/(4) = 1 rad/s.
首先计算 ω₀ = √(k/m) = √(200/2) = 10 rad/s。阻尼率 γ = c/(2m) = 4/4 = 1 rad/s。
The resonant frequency is ω_res = √(ω₀² − 2γ²) = √(100 − 2) = √98 ≈ 9.90 rad/s.
共振频率为 ω_res = √(ω₀² − 2γ²) = √(100 − 2) = √98 ≈ 9.90 rad/s。
The maximum amplitude is A_max = F₀ / (2m γ √(ω₀² − γ²)) = 10 / (2·2·1·√(100−1)) ≈ 10 / (4·9.95) ≈ 0.251 m.
最大振幅为 A_max = F₀ / (2m γ √(ω₀² − γ²)) = 10 / (2·2·1·√(100−1)) ≈ 10 / (4·9.95) ≈ 0.251 m。
10. Damping Effects on Resonance | 阻尼对共振的影响
As damping increases, the resonant frequency shifts slightly lower, and the maximum amplitude decreases. When γ ≥ ω₀/√2, no resonance peak exists; the amplitude decreases monotonically with frequency.
随着阻尼增大，共振频率略向低频移动，最大振幅减小。当 γ ≥ ω₀/√2 时，不再存在共振峰；振幅随频率单调递减。
The table below summarises the key characteristics for light, moderate, and heavy damping.
下表概述了轻阻尼、中等阻尼和强阻尼下的关键特征。
| Damping | Resonant peak | Phase at low ω | Phase at high ω |
| Light (γ small) | Sharp, high | 0 | π |
| Moderate | Broad, lower | 0 | π |
| Heavy (γ large) | No peak | 0 | π |
IB exam questions often ask you to interpret graphs of amplitude versus driving frequency, so practising with different damping values is essential.
IB 考试常要求你解读振幅随驱动频率变化的图像，因此练习不同阻尼值下的曲线非常关键。
11. Real-World Applications | 现实应用
Resonance has both beneficial and destructive effects. Bridges, buildings, and aircraft wings are designed to avoid resonance with environmental vibrations. Conversely, musical instruments, medical ultrasound, and MRI machines rely on resonance to function effectively.
共振既有有益的一面，也有破坏性的影响。桥梁、建筑物和飞机机翼的设计需要避免与环境振动发生共振。相反，乐器、医学超声和核磁共振成像则依赖共振来有效工作。
Mathematically, understanding the differential equation and its amplitude formula allows engineers to tune systems to desired frequencies or add damping to suppress dangerous oscillations.
在数学上，理解微分方程及其振幅公式使工程师能够将系统调谐到所需频率，或通过增加阻尼来抑制危险的振荡。
12. Summary and Exam Tips | 总结与考试提示
To master forced vibrations for the IB exam, remember the key formulas: the amplitude A, resonant frequency ω_res, and quality factor Q. Always check whether damping allows a resonance peak to exist.
为掌握 IB 考试中的受迫振动，请牢记关键公式：振幅 A、共振频率 ω_res 和品质因数 Q。始终检查阻尼是否允许共振峰存在。
When solving problems, sketch the amplitude-frequency graph, identify the limit at ω = 0 where A = F₀/k, and note how the peak shifts with damping. Practice transforming the differential equation into algebraic form using the assumed solution.
解题时，先画出振幅-频率曲线草图，识别 ω = 0 时 A = F₀/k 的极限，并注意峰随阻尼的变化。通过假设解的形式，练习将微分方程转化为代数方程。
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