PH04 Formula Derivation: Simple Harmonic Motion | PH04 公式推导：简谐运动

📚 PH04 Formula Derivation: Simple Harmonic Motion | PH04 公式推导：简谐运动

This article unpacks the key formula derivations required in typical PH04 International Physics A exam papers, focusing on simple harmonic motion. We will derive displacement, velocity, acceleration, and energy relationships step by step, making the logic behind the mathematics clear.

本文深入解析PH04国际物理A考试中重要的公式推导，聚焦简谐运动。我们将逐步推导位移、速度、加速度以及能量关系式，使公式背后的数学逻辑清晰呈现。

1. Defining Simple Harmonic Motion | 定义简谐运动

Simple harmonic motion (SHM) is defined as oscillatory motion where the acceleration a is directly proportional to the displacement x from the equilibrium position and always directed towards that position.

简谐运动定义为加速度 a 与相对平衡位置的位移 x 成正比，且方向始终指向该平衡位置的振动。

a = -ω²x

Here ω is the angular frequency of the motion. The negative sign indicates that acceleration and displacement act in opposite directions.

其中 ω 是运动的角频率。负号表示加速度与位移方向相反。

2. Reference Circle as a Geometric Model | 参考圆作为几何模型

To derive the equations describing position as a function of time, we use a reference circle. A point moving with uniform circular motion of radius A and constant angular speed ω has a projection on a diameter that executes SHM.

为了推导位置随时间变化的方程，我们使用参考圆。一个半径为 A、角速度 ω 做匀速圆周运动的点，其在直径上的投影做简谐运动。

Suppose the point starts at an angle φ (phase constant) from the positive x-axis. At time t, the angular position is θ = ωt + φ.

假设该点从与正 x 轴夹角 φ（初相位）处开始运动。在时刻 t，角位置为 θ = ωt + φ。

The projection onto the horizontal axis gives the displacement x:

在水平轴上的投影给出位移 x：

x = A cos(ωt + φ)

If the initial displacement is zero at t = 0, we can set φ = -π/2 to obtain the more common sinusoidal form:

若初始位移在 t = 0 时为零，可令 φ = -π/2，得到更常见的正弦形式：

x = A sin(ωt)

3. Deriving the Velocity Equation | 推导速度方程

Velocity is the rate of change of displacement. Differentiating x = A sin(ωt) with respect to time t gives:

速度是位移的变化率。对 x = A sin(ωt) 关于时间 t 求导得：

v = dx/dt = ωA cos(ωt)

Using the trigonometric identity cos²θ + sin²θ = 1, we can also express v in terms of displacement x:

利用三角恒等式 cos²θ + sin²θ = 1，我们也可用位移 x 表示 v：

Since x = A sin(ωt), we have sin(ωt) = x/A. Then cos(ωt) = √(1 – sin²(ωt)) = √(1 – x²/A²) for cos(ωt) positive. Thus:

因为 x = A sin(ωt)，有 sin(ωt) = x/A。于是 cos(ωt) = √(1 – sin²(ωt)) = √(1 – x²/A²)（取正）。因此：

v = ωA √(1 – x²/A²) = ω √(A² – x²)

The maximum speed occurs when x = 0, giving v_max = ωA.

最大速率出现在 x = 0 处，即 v_max = ωA。

4. Deriving the Acceleration Equation | 推导加速度方程

Acceleration is the rate of change of velocity. Differentiating v = ωA cos(ωt) gives:

加速度是速度的变化率。对 v = ωA cos(ωt) 求导得：

a = dv/dt = -ω²A sin(ωt)

Since x = A sin(ωt), we directly obtain the fundamental SHM relationship:

由于 x = A sin(ωt)，我们直接得到简谐运动的基本关系：

a = -ω²x

This shows that acceleration is always opposite to displacement and proportional to it.

这表明加速度始终与位移反向且成正比。

5. Deriving the Velocity-Displacement Relation | 推导速度-位移关系式

A commonly tested derivation links velocity and displacement without explicit time. Starting from the definition a = dv/dt and using the chain rule:

一个常考的推导式是关联速度与位移、不含显式时间。从定义 a = dv/dt 出发，利用链式法则：

a = dv/dt = (dv/dx)(dx/dt) = v dv/dx

Substituting a = -ω²x:

代入 a = -ω²x：

v dv/dx = -ω²x

Rearranging and integrating both sides:

移项并对两边积分：

∫ v dv = ∫ -ω²x dx

½ v² = -½ ω² x² + C

To find the constant C, we use the condition that when displacement is maximum x = A, the velocity v = 0. Then C = ½ ω² A². Substituting back:

为确定常数 C，利用当位移最大 x = A 时速度 v = 0 的条件。得 C = ½ ω² A²。代回：

½ v² = -½ ω² x² + ½ ω² A²

Multiplying by 2 and rearranging gives the standard form:

乘以 2 并整理得标准形式：

v² = ω² (A² – x²)

Taking the square root yields v = ± ω √(A² – x²), with the sign indicating direction.

开平方得 v = ± ω √(A² – x²)，正负号表示方向。

6. Energy in Simple Harmonic Motion | 简谐运动中的能量

The kinetic energy E_k of the oscillating mass m is:

振动质量 m 的动能为：

E_k = ½ m v² = ½ m ω² (A² – x²)

The potential energy E_p is obtained by noting that the restoring force is F = –kx with k = mω². Since dE_p/dx = -F = kx, integrating gives:

势能 E_p 可通过回复力 F = –kx 且 k = mω² 求得。因 dE_p/dx = -F = kx，积分得：

E_p = ∫₀ˣ kx dx = ½ k x² = ½ m ω² x²

The total mechanical energy E_total is therefore:

因此总机械能为：

E_total = E_k + E_p = ½ m ω² (A² – x²) + ½ m ω² x² = ½ m ω² A²

This is a constant, confirming energy conservation in an ideal SHM system.

这是一个常量，证实了理想简谐运动系统中的能量守恒。

7. Deriving the Period of a Simple Pendulum | 推导单摆的周期

For a simple pendulum of length L and small angular displacement θ, the restoring force along the arc is F = –mg sinθ. For small angles (θ < 10°), sinθ ≈ θ (in radians). The tangential displacement s relates to angle by s = Lθ.

对于一个长度为 L、小角位移 θ 的单摆，沿圆弧的回复力为 F = –mg sinθ。对于小角度（θ < 10°），sinθ ≈ θ（弧度）。切向位移 s 与角度关系为 s = Lθ。

Thus the acceleration along the arc is a = d²s/dt² = -g sinθ ≈ -g θ = -(g/L) s. This matches the SHM form a = -ω² x provided we identify ω² = g/L. Hence:

因此切向加速度为 a = d²s/dt² = -g sinθ ≈ -g θ = -(g/L) s。这与简谐运动形式 a = -ω² x 吻合，只需令 ω² = g/L。因此：

ω = √(g/L)

Since the period T = 2π/ω, we obtain the well-known formula:

由于周期 T = 2π/ω，我们得到著名的公式：

T = 2π √(L/g)

8. Deriving the Period of a Mass-Spring System | 推导质量-弹簧系统的周期

For a mass m attached to a spring of force constant k, Hooke’s law gives the restoring force F = –kx. Newton’s second law states F = ma, so:

对于连接在劲度系数为 k 的弹簧上的质量 m，胡克定律给出回复力 F = –kx。牛顿第二定律 F = ma，因此：

ma = -kx => a = -(k/m) x

Comparing with a = -ω² x, we identify ω² = k/m. Hence the period:

与 a = -ω² x 比较，得 ω² = k/m。因此周期：

T = 2π √(m/k)

This derivation is independent of the amplitude, reflecting the isochronous nature of SHM.

这一推导与振幅无关，反映了简谐运动的等时性。

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