Blog

📚 A-Level Mathematics MA03 Exam Report June 2022: Key Topics Explained | A-Level 数学 MA03 2022年6月考试报告知识点精讲

The June 2022 examiner report for A-Level Mathematics Paper 3 (MA03) reveals recurring pitfalls that prevented many candidates from securing top marks. A closer look shows that errors were often rooted in incomplete understanding of partial fractions, mishandling of modulus inequalities, loose notation in differentiation, and misinterpretation of complex loci. This article breaks down the ten most significant topics highlighted in the report, offering clear explanations and exam-focused advice to help you avoid the same mistakes.

2022年6月A-Level数学纯数3（MA03）的考官报告揭示了许多考生反复出现的失分点。分析表明，错误往往源于部分分式理解不完整、模不等式处理不当、微分符号使用不严谨以及对复数轨迹的误读。本文将基于报告提炼出最重要的十个专题，提供清晰的解析和贴近考试的指导，帮助你避开这些雷区。

1. Partial Fractions and Improper Rational Expressions | 部分分式与假分式

The examiners noted that when a rational expression is improper (i.e. the degree of the numerator is equal to or greater than that of the denominator), many candidates forgot to carry out polynomial division first. Skipping this step led to an incorrect partial fraction form and blocked the resolution of subsequent integration or series expansion tasks.

考官指出，当有理式为假分式（即分子次数大于或等于分母次数）时，许多考生忘记先进行多项式除法。跳过这一步会导致部分分式形式错误，并阻碍后续积分或级数展开的完成。

• Always check the degrees: if deg(num) ≥ deg(den), perform long division to write the expression as Q(x) + remainder/denominator.
• 务必先检查次数：若分子次数 ≥ 分母次数，应先用长除法化为 Q(x) + 余式/分母 的形式。
• For a repeated linear factor (ax+b)², the decomposition must contain A/(ax+b) + B/(ax+b)².
• 对于重线性因式 (ax+b)²，分解结果必须包含 A/(ax+b) + B/(ax+b)² 两项。
• If the denominator includes an irreducible quadratic such as x²+1, the corresponding numerator is linear, e.g. Cx+D.
• 若分母包含不可约二次式如 x²+1，对应分子应为一次式 Cx+D。

2. Modulus Equations and Inequalities | 模方程与不等式

A frequent error in the exam was discarding solutions when solving modulus equations. Candidates typically considered the positive branch but forgot that the expression inside the modulus could also equal the negative of the right-hand side.

考试中的一个常见错误是解模方程时丢失解。考生通常只考虑正分支，却忘了模内部表达式也可能等于右边的相反数。

|f(x)| = a ⇒ f(x) = a or f(x) = −a (a ≥ 0)

|f(x)| = a ⇒ f(x) = a 或 f(x) = −a (a ≥ 0)

For inequalities such as |2x − 3| < 5, the correct approach is to rewrite as a compound inequality −5 < 2x − 3 < 5, rather than squaring both sides unnecessarily. Squaring can sometimes introduce extraneous conditions and should be used with caution when the inequality sign involves 'greater than'.

对于 |2x − 3| < 5 这类不等式，正确思路是改写为复合不等式 −5 < 2x − 3 < 5，而不必两边平方。平方有时会引入额外限制，且当不等号为“大于”时需格外谨慎。

3. Exponential and Logarithmic Equations | 指数与对数方程

The report highlighted a tendency to forget the domain restrictions on logarithmic functions. Equations like log₂(x) + log₂(x − 2) = 3 require x > 0 and x − 2 > 0, meaning solutions must be x > 2. Many candidates solved the algebra correctly but accepted all algebraic solutions without checking against the domain.

报告强调考生容易遗忘对数函数的定义域限制。例如 log₂(x) + log₂(x − 2) = 3 要求 x > 0 且 x − 2 > 0，即解须满足 x > 2。很多考生代数运算正确，却直接接受所有代数解而未作定义域检验。

Another common slip was seen in equations involving eˣ. When solving e²ˣ − 4eˣ + 3 = 0, using the substitution y = eˣ transforms it into a quadratic. Candidates then often solved for y correctly but forgot to discard the negative y value, since eˣ > 0 for all real x.

另一个常见失误出现在含 eˣ 的方程中。解 e²ˣ − 4eˣ + 3 = 0 时，令 y = eˣ 可化为二次方程。考生往往能正确解出 y，却忘记舍去负值，因为对所有实数 x，恒有 eˣ > 0。

4. Trigonometric Equations and the R-formula | 三角方程与R公式

When solving equations like 3 sinθ + 4 cosθ = 2, the expected method is to use the R-formula to combine the left-hand side into R sin(θ + α) or R cos(θ − α). The June 2022 report noted that many candidates lost accuracy by rounding α too early, leading to final answers falling outside the required tolerance.

解方程 3 sinθ + 4 cosθ = 2 时，预期的方法是用R公式将左边合并为 R sin(θ + α) 或 R cos(θ − α)。2022年6月报告指出，许多考生因过早对 α 取整而导致精度损失，最终答案超出允许误差范围。

Furthermore, examiners stressed the importance of finding all solutions within the given interval. For example, after obtaining a principal value from sin(θ + 36.9°) = 0.4, you must use the symmetry of the sine curve: θ + 36.9° = 180° − principal value, and then add/subtract 360° periods before solving for θ. Missing the secondary value or period additions was a common cause of incomplete solution sets.

此外，考官强调在给定区间内求出所有解的重要性。例如，由 sin(θ + 36.9°) = 0.4 得到主值后，还需利用正弦曲线的对称性：θ + 36.9° = 180° − 主值，再通过加减 360° 周期求出 θ。漏掉第二解或周期增量是解集不完整的常见原因。

5. Differentiation of Inverse Trig and Implicit Functions | 反三角函数与隐函数微分

The derivatives of the inverse trigonometric functions, though provided in the formula booklet, were frequently misapplied. The derivative of arctan(x) is 1/(1 + x²), but candidates often omitted the chain rule when the argument was a function of x. For instance, differentiating arctan(3x) requires multiplying by 3, giving 3/(1 + 9x²).

反三角函数的导数虽然在公式表中给出，但常被误用。arctan(x) 的导数是 1/(1 + x²)，但当自变量是 x 的函数时，考生经常忽略链式法则。例如求 arctan(3x) 的导数需乘以 3，结果为 3/(1 + 9x²)。

Implicit differentiation caused difficulties when terms involved products of x and y. For the equation x² + xy + y² = 7, differentiating xy with respect to x gives y + x(dy/dx), not just y. The report warned that missing the dy/dx term on the y-factor leads to a completely wrong gradient and was heavily penalised.

隐函数微分中，涉及 x 与 y 的乘积项时常出现问题。对于方程 x² + xy + y² = 7，xy 对 x 求导得 y + x(dy/dx)，而不仅是 y。报告提醒，漏掉 y 因子上的 dy/dx 项会导致梯度完全错误且被严重扣分。

6. Integration by Substitution and Reverse Chain Rule | 代换积分与逆链式法则

When evaluating a definite integral by substitution, the given substitution often makes the algebra cleaner, but many candidates neglected to change the limits. For ∫₀² x√(x²+1) dx with u = x²+1, the new limits become u = 1 and u = 5. Writing the final answer with the original limits—or worse, mixing variables—was a persistent error.

用代换法计算定积分时，给定的代换通常会让代数更简洁，但很多考生忘记变换上下限。对 ∫₀² x√(x²+1) dx 令 u = x²+1，新的积分限应为 u = 1 和 u = 5。使用原积分限作答——甚至变量混杂——是持续出现的错误。

In reverse chain rule situations, such as ∫ f'(x)/f(x) dx = ln|f(x)| + C, candidates often forgot the absolute value. This can cause issues when f(x) may be negative over the interval. The same care is needed for ∫ f'(x) e^(f(x)) dx = e^(f(x)) + C.

在逆链式法则情形下，如 ∫ f'(x)/f(x) dx = ln|f(x)| + C，考生常常遗漏绝对值。当 f(x) 在积分区间内可能为负时，就会出问题。对于 ∫ f'(x) e^(f(x)) dx = e^(f(x)) + C 也需同样留意。

7. Numerical Integration – The Trapezium Rule | 数值积分——梯形法则

Examiners pointed out that many candidates did not use the correct number of strips or ordinates specified in the question. If a question states “use 4 strips”, you need 5 ordinates (x-values) and the width h = (b − a)/4. Misreading strips as ordinates caused the entire answer to be scaled incorrectly.

考官指出，许多考生没有使用题目指定的正确条数或纵坐标数量。若题目说明“用4条带”，你需要5个纵坐标（x值），且带宽 h = (b − a)/4。将条数误读为纵坐标数会导致整个结果比例错误。

Another common mistake was forgetting the factor ½ in the trapezium rule formula: Area ≈ h/2 [y₀ + yₙ + 2(y₁ + y₂ + … + yₙ₋₁)]. Some candidates wrote h[sum] instead of h/2[sum]. Others used an inconsistent number of decimal places in working, leading to final answers not matching the mark scheme’s accuracy requirements.

另一个常见错误是忘记梯形法则公式中的因子 ½：面积 ≈ h/2 [y₀ + yₙ + 2(y₁ + y₂ + … + yₙ₋₁)]。部分考生写成 h[和] 而非 h/2[和]。还有些人在运算中使用不一致的小数位数，导致最终答案不符合评分标准中的精度要求。

8. Solving Differential Equations | 解微分方程

Many candidates lost marks by not separating variables correctly. For dy/dx = (x+1)/y, the correct separation is y dy = (x+1) dx. Writing dx on the wrong side or attempting to integrate without separation was a fundamental error. After integration, don’t forget the constant of integration; the general solution must contain an arbitrary constant unless additional conditions are given.

很多考生因未正确分离变量而失分。对于 dy/dx = (x+1)/y，正确的分离形式是 y dy = (x+1) dx。将 dx 写在错误的一侧或未经分离直接积分都是根本性错误。积分后别忘了积分常数；除非给出额外条件，通解必须包含任意常数。

When an initial condition is provided, substitute it early to find the constant, but be careful with the form of the solution. For a logarithmic solution like ln|y| = … , candidates often left the answer as y = e^(…) without considering the absolute value. A better approach is to combine constants and write y = A e^(…) where A = ± e^C, and then determine the sign using the initial condition.

当给出初始条件时，尽早代入解出常数，但需注意解的形式。对于对数形式的解如 ln|y| = … ，考生常不加绝对值直接写成 y = e^(…)。更好的做法是合并常数写成 y = A e^(…)，其中 A = ± e^C，再利用初始条件确定符号。

9. Complex Numbers – Loci and Polynomial Roots | 复数——轨迹与多项式根

Loci problems in the Argand diagram were handled poorly when candidates misinterpreted |z − a| = |z − b| as a circle. In fact, it represents the perpendicular bisector of the line segment joining points a and b. Recognising the shape is essential for sketching and for finding intersections with other loci.

阿干特图上的轨迹问题中，考生常将 |z − a| = |z − b| 误解为圆。事实上，它表示连接点 a 和点 b 线段的垂直平分线。正确识别形状对于作图和求与其他轨迹的交点至关重要。

When solving polynomial equations with complex roots, the fundamental theorem requires that non-real roots occur in conjugate pairs. If the polynomial has real coefficients and one root is 2 + i, then 2 − i is also a root. The report revealed that candidates often failed to use this fact to construct a quadratic factor with real coefficients, which would simplify the division process.

解带复数根的多项式方程时，基本定理要求非实根成共轭对出现。若多项式系数为实数且有一根为 2 + i，则 2 − i 也是根。报告显示，考生常未利用这一事实构造具有实系数的二次因式，从而导致除法过程复杂化。

10. Vectors: Equations of Lines and Points of Intersection | 向量：直线方程与交点

The report underlined that when finding the intersection of two lines given in parametric form r = a + λb and r = c + μd, the parameters λ and μ are generally different. Setting the equations equal gives a system, and candidates must solve for both parameters. A common mistake was to assume λ = μ, which only happens if the lines are the same or meet at a very specific coincident point.

报告强调，求两条参数式直线 r = a + λb 和 r = c + μd 的交点时，参数 λ 和 μ 通常不同。令两方程相等得到方程组后，考生必须解出两个参数。一个常见错误是假设 λ = μ，只有当两线重合或交于一个非常特殊的重合点时才成立。

For dot product and angle problems, the formula cosθ = (a·b)/(|a||b|) was generally recalled, but the calculation of the modulus (magnitude) of a vector in 3D was sometimes inaccurate. Ensure you compute √(x² + y² + z²) correctly, especially when vectors involve parameters. Also, to prove two lines are perpendicular, show a·b = 0; the converse is true only for non-zero vectors.

关于点积与夹角问题，公式 cosθ = (a·b)/(|a||b|) 通常能被记住，但三维向量模（大小）的计算有时不准确。务必正确计算 √(x² + y² + z²)，尤其是当向量含参时。此外，要证明两线垂直，需证明 a·b = 0；其逆命题仅对非零向量成立。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

📚 OxfordAQA International A-Level Further Mathematics (9665) Mechanics Topic Test: A Detailed Breakdown of Question Types | OxfordAQA 国际 A-Level 进阶数学(9665) 力学专题测试题型深度解析

The Mechanics module in OxfordAQA International A-Level Further Mathematics (9665) challenges students with a wide variety of problem types, ranging from projectile motion to rigid body statics and differential equations of motion. Understanding the structure of these question types and mastering the underlying physics is essential for success in the topic test. This article provides an in-depth analysis of the most common mechanics questions, offering strategies, formulas, and tips to help you tackle each type with confidence.

OxfordAQA 国际 A-Level 进阶数学 (9665) 的力学模块涵盖了多种多样的题型，从抛体运动到刚体静力学和运动微分方程。理解这些题型结构并掌握背后的物理原理，是攻克专题测试的关键。本文将深入解析最常见的力学题型，提供策略、公式和技巧，帮助你自信应对每一类问题。

1. Kinematics in One and Two Dimensions | 一维与二维运动学

One-dimensional constant acceleration problems are typically solved using the SUVAT equations. These link initial velocity u, final velocity v, acceleration a, time t and displacement s. The most common task is to identify three known quantities and select the appropriate equation to find a fourth. For two-dimensional kinematics, position vectors are given as functions of time, and differentiation yields velocity and acceleration vectors. Integration of acceleration with initial conditions returns velocity and position.

一维匀加速问题通常使用 SUVAT 方程求解。这些方程关联初速度 u、末速度 v、加速度 a、时间 t 和位移 s。最常见的方法是找出三个已知量，选择合适的方程求出第四个量。对于二维运动学，位置向量常以时间函数形式给出，通过求导得到速度向量和加速度向量；而对加速度积分并代入初始条件则可还原速度和位置。

v = u + at    s = ut + ½at²    v² = u² + 2as    s = ½(u + v)t

Students often confuse the direction of vectors when dealing with vertical motion under gravity. It is crucial to define a positive direction consistently – for example, upwards as positive, making gravitational acceleration g = -9.8 m s⁻². In two‑dimensional vector problems, the velocity vector is the first derivative of the position vector r, and acceleration is the second derivative. When integrating, remember to include the constant of integration determined by initial velocity or initial position.

在处理垂直重力运动时，学生常混淆向量的方向。必须统一规定正方向——例如取向上为正，则重力加速度 g = -9.8 m s⁻²。在二维向量问题中，速度向量是位置向量 r 的一阶导数，加速度是二阶导数。积分时务必加上由初速度或初位置决定的积分常数。

2. Projectile Motion | 抛体运动

Projectile motion questions split the motion into horizontal and vertical components. The horizontal component has constant velocity (zero acceleration), while the vertical component has constant downward acceleration g. Key results include time of flight, maximum height, horizontal range, and the equation of the trajectory. You will often be asked to find the angle of projection θ for a given range or to locate the position of a projectile at a specific time.

抛体运动问题将运动分解为水平与竖直分量。水平方向为匀速运动（加速度为零），竖直方向具有恒定的向下加速度 g。核心结果包括飞行时间、最大高度、水平射程以及轨迹方程。你常会被要求求给定射程的抛射角 θ，或计算特定时刻抛体的位置。

x = u cosθ t    y = u sinθ t – ½gt²    Range = (u² sin 2θ)/g

To derive the trajectory equation, eliminate t from the parametric equations for x and y. This gives y = x tanθ – (g x²)/(2u² cos²θ). When solving for maximum height, set vertical velocity to zero: v_y = u sinθ – gt = 0. Common mistakes include forgetting that the vertical velocity at the highest point is zero but the horizontal velocity remains u cosθ, and using inconsistent signs for g when the launch point and target are at different heights.

推导轨迹方程时，从 x 和 y 的参数方程中消去 t，得到 y = x tanθ – (g x²)/(2u² cos²θ)。求最大高度时，令竖直速度为零：v_y = u sinθ – gt = 0。常见错误包括：忘记最高点竖直速度为零而水平速度仍为 u cosθ，以及当起抛点与目标点高度不同时 g 的符号使用不一致。

3. Newton’s Laws and Connected Particles | 牛顿定律与连接体

Connected particle questions involve two or more masses linked by a light inextensible string, often passing over a smooth pulley. Applying Newton’s second law to each particle individually (and using the constraint that the string length is constant) allows you to find the acceleration of the system and the tension in the string. Pulley problems at the edge of a table, inclined plane setups, and lift problems all fall into this category.

连接体问题涉及两个或多个由轻质且不可伸长的绳子连接的物体，绳子常跨过光滑滑轮。对每个物体单独应用牛顿第二定律（并利用绳长不变的约束条件），即可求出系统的加速度和绳中张力。桌边滑轮、斜面组合以及电梯问题都属于这一类。

ΣF = ma    T – mg = ma (for a hanging particle)

Always draw clear free‑body force diagrams showing weight, tension, normal reaction, and friction where applicable. For a particle on a rough inclined plane, resolve weight into components parallel and perpendicular to the plane. Remember that the tension is the same throughout a light string passing over a smooth pulley, and the accelerations of connected particles have the same magnitude. Once the acceleration is found, SUVAT equations can then be applied to find velocities and displacements.

始终绘制清晰的受力分析图，标明重力、张力、法向反作用力以及摩擦力（如适用）。对于粗糙斜面上的物体，将重力沿斜面方向与垂直于斜面方向分解。记住，轻绳跨过光滑滑轮时张力处处相等，且连接体的加速度大小相同。一旦求得加速度，便可进一步用 SUVAT 方程求速度和位移。

4. Work, Energy, and Power | 功、能与功率

Work‑energy problems involve calculating the work done by a force (force × displacement in the direction of the force) and linking it to the change in kinetic energy and gravitational potential energy. The work‑energy principle states that the total work done by all forces equals the change in kinetic energy. Power is the rate of doing work: P = Fv for a constant force acting on a moving object.

功能问题涉及计算力所做的功（力 × 沿力方向的位移），并将其与动能和重力势能的变化联系起来。功能原理指出，所有力所做的总功等于动能的变化量。功率是做功的快慢：对于作用在运动物体上的恒力，有 P = Fv。

KE = ½mv²    GPE = mgh    Work = Fs cosθ    P = Fv

Typical questions ask for the speed reached by a car travelling up an incline against resistance, or the height to which a particle rises when projected upwards with a known initial kinetic energy. The work done against friction converts mechanical energy into heat, so it must be subtracted from the total mechanical energy balance. In power problems, be mindful that maximum speed occurs when the driving force equals the total resistive force.

典型问题包括：汽车沿斜坡行驶克服阻力时的速度，或以已知初动能竖直上抛的物体所能达到的高度。克服摩擦力所做的功将机械能转化为内能，因此必须从机械能平衡中扣除。在功率问题中，最大速度发生在驱动力等于总阻力之时。

5. Momentum and Direct Collisions | 动量与直接碰撞

In direct collisions, particles move along the same straight line. The principle of conservation of linear momentum states that total momentum before impact equals total momentum after impact, provided no external forces act. The coefficient of restitution e defines the ratio of relative speed after collision to relative speed before collision: e = (v₂ – v₁)/(u₁ – u₂). For a perfectly elastic collision e = 1, for a perfectly inelastic collision e = 0.

在直接碰撞中，质点沿同一直线运动。动量守恒原理表明，若没有外力作用，碰撞前的总动量等于碰撞后的总动量。恢复系数 e 定义为碰撞后相对速度与碰撞前相对速度的比值：e = (v₂ – v₁)/(u₁ – u₂)。完全弹性碰撞 e = 1，完全非弹性碰撞 e = 0。

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂    e = (separation speed) / (approach speed)

When solving, write two equations: conservation of momentum and the restitution equation. Solve simultaneously for the unknown velocities. For a collision with a fixed wall, the speed of the wall is zero, and the speed of the particle after collision is e × (speed before collision), with reversed direction. The loss of kinetic energy is also a common follow‑up question, calculated as initial KE minus final KE.

解题时列出两个方程：动量守恒方程和恢复系数方程，联立求解未知速度。对于与固定墙壁的碰撞，墙壁速度为零，碰撞后质点的速率为 e ×（碰撞前速率），方向相反。动能损失也是常见的后续问题，通过初动能减末动能计算。

6. Oblique Collisions and the Coefficient of Restitution | 斜碰撞与恢复系数

Oblique collisions occur when particles approach each other at an angle. The impulse acts along the line joining the centres at the moment of impact. The component of velocity perpendicular to the line of centres remains unchanged, while the component parallel to the line of centres obeys the restitution law. This splits the problem into two independent directions, making vector resolution essential.

斜碰撞发生在质点以一定角度相互接近时。冲量沿着碰撞瞬间的中心连线方向作用。垂直于中心连线的速度分量保持不变，而平行于该连线的速度分量遵循恢复系数定律。这样就将问题分解为两个独立的方向，向量分解至关重要。

Parallel: e = (v₂ₚ – v₁ₚ)/(u₁ₚ – u₂ₚ)    Perpendicular: v₁ₚ = u₁ₚ, v₂ₚ = u₂ₚ (note: use proper components)

Questions frequently ask for the angle of deflection, the final speed, or the impulse magnitude. Begin by drawing a clear diagram with the line of centres identified. Resolve velocities into components parallel and perpendicular to this line. Apply momentum conservation along the line of centres (or simply the restitution equation for the parallel components if masses are equal and no external impulse). After finding the parallel components after collision, recombine with the unchanged perpendicular components using Pythagoras and trigonometry to find the final speed and direction.

常见问题要求计算偏转角、末速率或冲量大小。首先画出清晰的示意图，标出中心连线。将速度沿平行和垂直于该线的方向分解。沿中心连线方向应用动量守恒（如果质量相等且无外部冲量，也可直接对平行分量使用恢复系数方程）。求出碰撞后的平行分量后，与不变的垂直分量合成，利用勾股定理和三角函数求出末速率和方向。

7. Circular Motion | 圆周运动

Circular motion questions involve a particle moving with constant speed in a circle (horizontal circular motion) or varying speed in a vertical circle. The acceleration is directed towards the centre: a = v²/r = rω². Newton’s second law applied towards the centre gives the centripetal force equation: F = mv²/r = mrω². For a conical pendulum or a car on a banked track, resolve forces and equate the horizontal component to the required centripetal force.

圆周运动问题包括质点以恒定速率做水平圆周运动，或在竖直面内以变速率做圆周运动。加速度指向圆心：a = v²/r = rω²。将牛顿第二定律沿半径方向应用于圆心，得到向心力方程：F = mv²/r = mrω²。对于锥摆或倾斜弯道上的汽车，需分解力并将水平分量与所需的向心力等量列出。

Centripetal acceleration = v²/r    Centripetal force = mv²/r = mrω²

Vertical circle problems are more demanding because the speed changes. Use conservation of energy between the highest and lowest points, then apply radial force equations at critical points. At the top of a circular loop, the minimum speed for a particle on a string is when tension T = 0, giving v_min = √(gr). For a particle attached to a rod, the speed at the top can be zero because the rod can provide a supporting force. Be careful to distinguish between the two cases.

竖直圆周运动更具挑战性，因为速率会变化。利用最高点和最低点之间的能量守恒，然后在临界点应用径向力方程。在圆周轨道顶端，对于系于绳上的质点，最小速度对应张力 T = 0，即 v_min = √(gr)。而对于连在杆上的质点，在顶端的速率可以为零，因为杆能够提供支持力。务必区分这两种情形。

8. Simple Harmonic Motion | 简谐运动

Simple harmonic motion (SHM) is defined by a restoring force proportional to the displacement from equilibrium: acceleration a = –ω²x, where ω is the angular frequency. The displacement as a function of time can be written as x = A sin(ωt + φ) or x = A cos(ωt + φ). SHM appears in spring‑mass systems and simple pendulums, but also in many more abstract mechanics contexts where this differential equation arises.

简谐运动的定义是回复力与偏离平衡位置的位移成正比：加速度 a = –ω²x，其中 ω 为角频率。位移关于时间的函数可写为 x = A sin(ωt + φ) 或 x = A cos(ωt + φ)。简谐运动不仅出现在弹簧–质量系统和单摆中，也出现在许多导出该微分方程的抽象力学情境中。

a = –ω²x    v² = ω²(A² – x²)    T = 2π/ω

When solving SHM problems, identify the equilibrium position first, then find the effective spring constant or relevant parameter to determine ω. The velocity v² = ω²(A² – x²) is extremely useful for linking speed and displacement. For a spring, ω = √(k/m); for a simple pendulum, ω = √(g/l) for small angles. Remember that the maximum acceleration occurs at maximum displacement, and the maximum speed is ωA at the equilibrium position. The energy in SHM is proportional to A².

解简谐运动问题时，先确定平衡位置，再求出等效弹性系数或相关参数以确定 ω。关系式 v² = ω²(A² – x²) 在联系速度与位移时极为有用。对于弹簧，ω = √(k/m)；对于小角度单摆，ω = √(g/l)。注意，最大加速度出现在最大位移处，而最大速率为平衡位置处的 ωA。简谐运动的能量与振幅 A² 成正比。

9. Statics of Rigid Bodies and Moments | 刚体静力学与力矩

Statics problems require that the resultant force in any direction is zero and that the total moment about any point is zero. For a rigid body in equilibrium, you can take moments about any convenient point to eliminate unknown forces. Typical question elements include uniform rods, ladders against walls, hinged beams, and objects on the point of tilting. Friction is often introduced with the inequality F ≤ μR for static equilibrium.

静力学问题要求任意方向上的合力为零，且对任意点的合力矩为零。对于处于平衡状态的刚体，可以对任何方便的点取矩，以消去未知力。典型题型包括均匀杆、倚墙梯子、铰接梁以及即将翻倒的物体。摩擦力常用不等式 F ≤ μR 来描述静力平衡。

ΣF = 0    ΣM = 0    F ≤ μR

A ladder problem usually involves resolving horizontally and vertically, and taking moments about the foot of the ladder (or the wall) to find the normal reaction at the wall and the friction at the ground. Learn to identify the weight acting through the centre of the rod, and remember that the reaction at a rough surface is not necessarily normal – it has both a normal component and a friction component. When finding the maximum overhang or the position at which a rod begins to slip, the friction is limiting: F = μR.

梯子问题通常涉及水平与竖直方向分解，并对梯脚（或墙）取矩，以求出墙上的法向反作用力和地面的摩擦力。要学会识别通过杆中心的重量作用点，并记住粗糙表面的反作用力不一定是法向的——它既有法向分量也有摩擦力分量。在计算最大伸出量或杆开始滑动的临界位置时，摩擦力为极限值：F = μR。

10. Centres of Mass and Their Applications | 质心及其应用

Centres of mass questions involve finding the average position of mass distribution for a system of particles, uniform laminas, or composite solids. For discrete particles, the centre of mass coordinates are given by weighted averages: x̄ = Σ(mᵢ xᵢ)/Σmᵢ, ȳ = Σ(mᵢ yᵢ)/Σmᵢ. For uniform plane figures, the centre of mass can be found using integration or by standard results for common shapes (rectangle, triangle, sector of a circle). Questions then test the stability of a suspended or inclined object by checking whether the vertical through the centre of mass passes through the base.

质心问题涉及求质点系、均匀薄片或组合体的质量分布平均位置。对于离散质点，质心坐标由加权平均给出：x̄ = Σ(mᵢ xᵢ)/Σmᵢ，ȳ = Σ(mᵢ yᵢ)/Σmᵢ。对于均匀平面图形，质心可通过积分或常见形状（矩形、三角形、扇形）的标准结果求得。题目常通过判断过质心的竖直线是否穿过底面，来考察悬挂或倾斜物体的稳定性。

x̄ = Σ(mᵢxᵢ)/Σmᵢ    ȳ = Σ(mᵢyᵢ)/Σmᵢ    Triangle: distance from base = h/3

Composite bodies are treated by splitting the shape into known parts, finding the area and centre of each part, and using the weighted average. A common pitfall is subtracting a cut‑out: treat the missing area as a negative mass. When an object is suspended from a point, the equilibrium position is found by locating the centre of mass directly below the point of suspension. When placed on an inclined plane, toppling occurs when the line of action of the weight falls outside the base.

对于组合体，将其分解为已知部分，求出每部分的面积和形心，然后使用加权平均。常见的易错点是处理挖空部分：应将缺失面积视为负质量。当物体从一点悬挂时，平衡位置对应于质心位于悬挂点正下方。当放在斜面上时，若重力作用线超出底面范围，物体便会倾倒。

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

更多咨询请联系16621398022（同微信）