Oscillations — 题型分析

Question Type 1: Definition of SHM

如何识别

题目包含 "State what is meant by simple harmonic motion"。

标准解题方法

解题要点

1. 加速度与位移成正比（proportional to displacement）
2. 加速度方向与位移相反（in opposite direction to displacement / directed towards equilibrium）

评分标准

评分模式

• B1: acceleration (directly) proportional to displacement
• B1: acceleration in opposite direction to displacement / directed towards equilibrium

完整原题

Example 1 — 9702/41/O/N/20 Q3(a) (2 marks):

State what is meant by simple harmonic motion.

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MS:

• B1: acceleration (directly) proportional to displacement
• B1: acceleration in opposite direction to displacement

Example 2 — 9702/41/M/J/21 Q3(a) (2 marks):

State what is meant by simple harmonic motion.

📝 MS 展开查看

MS:

• B1: acceleration proportional to displacement
• B1: acceleration (directed) towards equilibrium / opposite to displacement

Example 3 — 9702/41/O/N/22 Q3(a) (2 marks):

State the defining equation for simple harmonic motion. Identify the meaning of each of the symbols used to represent physical quantities.

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MS:

• B1: $a = -\omega^2 x$
• B1: $a$ = acceleration, $\omega$ = angular frequency, $x$ = displacement

常见陷阱

常见陷阱

• 必须同时提到"正比"和"相反方向"两个条件
• 只写 $a = -\omega^2 x$ 不够——还需要文字说明

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Question Type 2: SHM Calculations

如何识别

题目给出振动参数（振幅、频率、质量等），求速度、加速度、力、能量等。

标准解题方法

解题步骤

1. 确定振幅 $x_0$ 和角频率 $\omega$（或频率 $f$）
2. 最大速度：$v_0 = \omega x_0$
3. 最大加速度：$a_0 = \omega^2 x_0$
4. 最大力：$F_0 = m a_0 = m \omega^2 x_0$
5. 特定位置速度：$v = \pm \omega \sqrt{x_0^2 - x^2}$
6. 总能量：$E = \frac{1}{2} m \omega^2 x_0^2$

评分标准

评分模式

• C1: 写出正确公式（如 $v = \omega x_0$）
• C1: 正确代入数值（注意单位统一）
• A1: 最终答案（含单位）

完整原题

Example 1 — 9702/41/M/J/20 Q3 (9 marks):

The piston in the cylinder of a car engine moves with SHM. The distance moved between max and min height is 9.8 cm. Mass = 640 g. The piston completes 2700 oscillations in 1.0 minute. Determine: (a)(i) the amplitude (a)(ii) the frequency (a)(iii) the maximum speed (a)(iv) the speed when the top of the piston is 2.3 cm below its maximum height (b) the resultant force giving rise to its maximum acceleration

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MS:

• (a)(i) A1: amplitude $= 9.8 / 2 = 4.9$ cm
• (a)(ii) A1: $f = 2700 / 60 = 45$ Hz
• (a)(iii) C1: $v_0 = \omega x_0$ or $v_0 = 2\pi f x_0$
• (a)(iii) A1: $v_0 = 2\pi \times 45 \times 0.049 = 13.9$ m s$^{-1}$
• (a)(iv) C1: $v = \omega \sqrt{x_0^2 - x^2}$ or $v^2 = \omega^2 (x_0^2 - x^2)$
• (a)(iv) A1: $v = 2\pi \times 45 \times \sqrt{0.049^2 - (0.049 - 0.023)^2} = 9.9$ m s$^{-1}$
• (b) C1: $a_0 = \omega^2 x_0$ or $a_0 = (2\pi f)^2 x_0$
• (b) C1: $F = ma_0$
• (b) A1: $F = 0.640 \times (2\pi \times 45)^2 \times 0.049 = 2.5 \times 10^3$ N

Example 2 — 9702/41/O/N/22 Q3 (11 marks):

An object oscillates vertically with SHM. Fig. 3.2 shows $v$-$x$ graph. Fig. 3.3 shows $E_P$-$x$ graph. (i) Determine the amplitude $x_0$ (ii) Show that $\omega = 1.7$ rad s$^{-1}$ (iii) Determine the mass $M$ of the object

📝 MS 展开查看

MS:

• (i) A1: $x_0 = 0.10$ m (from graph)
• (ii) C1: $v_0 = 0.17$ m s$^{-1}$ (from graph), $v_0 = \omega x_0$
• (ii) C1: $\omega = 0.17 / 0.10 = 1.7$ rad s$^{-1}$
• (iii) C1: $E_{P,\max} = \frac{1}{2} m \omega^2 x_0^2$ or $E_{K,\max} = \frac{1}{2} m v_0^2$
• (iii) A1: $m = 2 \times 0.050 / (1.7^2 \times 0.10^2) = 3.5$ kg

常见陷阱

常见陷阱

• 振幅是总位移的一半（不是总位移）
• 角频率 $\omega$ 和频率 $f$ 不要混淆：$\omega = 2\pi f$
• 速度最大的位置在平衡位置（$x = 0$），加速度最大在极端位置（$x = \pm x_0$）
• $v$-$x$ 图是椭圆，$v$-$t$ 图是正弦/余弦

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Question Type 3: Damping and Resonance

如何识别

题目涉及 "damped oscillations"、"light/critical/heavy damping"、"resonance"、"forced oscillations" 等关键词。

标准解题方法

解题要点

• Light damping: 振幅逐渐减小，周期略增（仍可振荡多次）
• Critical damping: 最快回到平衡位置（不振荡）
• Heavy damping: 缓慢回到平衡位置（甚至不回），不振荡
• Resonance: 驱动力频率 = 系统自然频率时，振幅最大
• 如果给定频率下振幅恒定 → 存在阻尼（否则无阻尼时振幅会无限增大）

完整原题

Example — 9702/41/M/J/21 Q3(c) (3 marks):

The oscillator is switched on. The frequency is varied, keeping amplitude constant. The amplitude of oscillation of the trolley is a maximum at the frequency calculated in (b). (i) State the name of the effect giving rise to this maximum. (ii) At any given frequency, the amplitude of oscillation of the trolley is constant. Explain how this indicates that there are resistive forces opposing the motion.

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MS:

• (i) B1: resonance
• (ii) B1: without resistive forces the amplitude would continue to increase
• (ii) B1: constant amplitude shows energy input = energy dissipated

常见陷阱

常见陷阱

• 共振条件是驱动频率 = 自然频率（不是其他频率）
• 轻阻尼 ≠ 无阻尼——轻阻尼振幅仍会缓慢减小
• 临界阻尼是"最快回到平衡"（not "最快停止"）
• 画 $x$-$t$ 图时注意：轻阻尼是逐渐减小的正弦波；临界阻尼是快速回到零（不越过）；重阻尼是缓慢趋于零

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Question Type 4: Energy in SHM — Graph Sketching

如何识别

要求 sketch 能量（$E_K$, $E_P$, $E_{total}$）随位移或时间的变化图。

标准解题方法

图形要点

• $E_K$-$x$: 开口向下抛物线（$x=0$ 最大，$x=\pm x_0$ 为零）
• $E_P$-$x$: 开口向上抛物线（$x=0$ 为零，$x=\pm x_0$ 最大）
• $E_{total}$: 水平线（常数）
• $E_K$-$t$ 和 $E_P$-$t$: 正弦平方曲线（频率是振动频率的 2 倍）

常见陷阱

常见陷阱

• 总能量不变（守恒），不是动能或势能不变
• $E_K$ 和 $E_P$ 的相位差 $\pi/2$（时间上差 $T/4$）
• 能量转换频率是振动频率的 2 倍（因为每个周期动能两次最大）