Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus C1.1 to C1.6考纲 C1.1 至 C1.6PHYSICS HL
An object has $a = -100\,x$ (SI). (a) state and confirm the two SHM features; (b) find $\omega$, $T$, $f$.某物体满足 $a = -100\,x$（SI）。(a) 写出并确认两个简谐特征；(b) 求 $\omega$、$T$、$f$。
For simple harmonic motion the acceleration must be proportional to the displacement (first power of $x$) and directed back toward equilibrium (opposite sign): the defining condition is $a = -\omega^{2}x$. (A1)
The given relation $a = -100\,x$ has acceleration proportional to $x$ and a minus sign, so it has both features and the motion is SHM. (A1)
Match coefficients with $a = -\omega^{2}x$: $\omega^{2} = 100\ \mathrm{s^{-2}}$, so $\omega = \sqrt{100} = 10\ \mathrm{rad\,s^{-1}}$ (positive root). (M1·A1)
$$ T = \frac{2\pi}{\omega} = \frac{2\pi}{10} \approx 0.63\ \mathrm{s}, \qquad f = \frac{1}{T} = \frac{\omega}{2\pi} \approx 1.6\ \mathrm{Hz}. $$(A1)
简谐运动要求加速度与位移成正比（$x$ 的一次方），且方向指回平衡位置（反号）：定义条件为 $a = -\omega^{2}x$。(A1)
所给关系 $a = -100\,x$ 中加速度与 $x$ 成正比且带负号，故两特征皆备，运动为简谐运动。(A1)
与 $a = -\omega^{2}x$ 比较系数：$\omega^{2} = 100\ \mathrm{s^{-2}}$，故 $\omega = \sqrt{100} = 10\ \mathrm{rad\,s^{-1}}$（取正根）。(M1·A1)
$$ T = \frac{2\pi}{\omega} = \frac{2\pi}{10} \approx 0.63\ \mathrm{s}, \qquad f = \frac{1}{T} = \frac{\omega}{2\pi} \approx 1.6\ \mathrm{Hz}. $$(A1)
Pendulum: $24$ swings in $30\ \mathrm{s}$. (a) find $T$, $f$, $\omega$; (b) effect on period of doubling the amplitude.单摆：$30\ \mathrm{s}$ 内 $24$ 次摆动。(a) 求 $T$、$f$、$\omega$；(b) 摆幅加倍对周期的影响。
The period is the time for one swing: $T = \dfrac{30}{24} = 1.25\ \mathrm{s}$. (A1)
Frequency: $f = \dfrac{1}{T} = \dfrac{1}{1.25} = 0.80\ \mathrm{Hz}$. (A1)
$$ \omega = 2\pi f = 2\pi (0.80) \approx 5.0\ \mathrm{rad\,s^{-1}}. $$(A1)
The period is unchanged. (A1)
For a simple pendulum $T = 2\pi\sqrt{L/g}$ contains no amplitude term, so (within the small-angle range where the motion stays simple harmonic) the period is independent of amplitude. Such oscillations are isochronous. (R1)
周期是一次摆动的时间：$T = \dfrac{30}{24} = 1.25\ \mathrm{s}$。(A1)
频率：$f = \dfrac{1}{T} = \dfrac{1}{1.25} = 0.80\ \mathrm{Hz}$。(A1)
$$ \omega = 2\pi f = 2\pi (0.80) \approx 5.0\ \mathrm{rad\,s^{-1}}. $$(A1)
周期不变。(A1)
单摆 $T = 2\pi\sqrt{L/g}$ 不含振幅项，故（在运动仍为简谐的小角度范围内）周期与振幅无关。此类振动称为等时振动。(R1)
$m = 0.30\ \mathrm{kg}$ on $k = 120\ \mathrm{N\,m^{-1}}$. (a) period; (b) new period with mass $\times 4$; (c) effect of a stiffer spring.$m = 0.30\ \mathrm{kg}$ 配 $k = 120\ \mathrm{N\,m^{-1}}$。(a) 周期；(b) 质量变四倍后的新周期；(c) 更硬弹簧的影响。
Use the data-booklet result $T = 2\pi\sqrt{m/k}$: (M1)
$$ T = 2\pi\sqrt{\frac{0.30}{120}} = 2\pi\sqrt{2.5\times 10^{-3}} \approx 0.31\ \mathrm{s}. $$(A1)
Since $T = 2\pi\sqrt{m/k} \propto \sqrt{m}$, multiplying $m$ by $4$ multiplies $T$ by $\sqrt{4} = 2$. (M1)
$$ T_{\text{new}} = 2T = 2(0.314) \approx 0.63\ \mathrm{s}. $$(A1)
The period would be shorter. (A1)
$T \propto 1/\sqrt{k}$, so a larger spring constant gives a smaller period: a stiffer spring exerts a larger restoring force for the same displacement, producing a larger acceleration and a faster oscillation. (R1)
用数据手册公式 $T = 2\pi\sqrt{m/k}$：(M1)
$$ T = 2\pi\sqrt{\frac{0.30}{120}} = 2\pi\sqrt{2.5\times 10^{-3}} \approx 0.31\ \mathrm{s}. $$(A1)
因 $T = 2\pi\sqrt{m/k} \propto \sqrt{m}$，$m$ 变为 $4$ 倍使 $T$ 乘以 $\sqrt{4} = 2$。(M1)
$$ T_{\text{new}} = 2T = 2(0.314) \approx 0.63\ \mathrm{s}. $$(A1)
周期会变短。(A1)
$T \propto 1/\sqrt{k}$，故弹簧常数越大周期越小：更硬的弹簧在相同位移下提供更大的恢复力，产生更大的加速度与更快的振动。(R1)
Seconds pendulum ($T = 2.0\ \mathrm{s}$), $g = 9.81\ \mathrm{m\,s^{-2}}$. (a) required length + one assumption; (b) factor period changes by on the Moon ($g \to g/6$).秒摆（$T = 2.0\ \mathrm{s}$），$g = 9.81\ \mathrm{m\,s^{-2}}$。(a) 所需摆长 + 一个假设；(b) 在月球（$g \to g/6$）周期变化的倍数。
Rearrange $T = 2\pi\sqrt{L/g}$ for $L$: (M1)
$$ L = \frac{gT^{2}}{4\pi^{2}} = \frac{(9.81)(2.0)^{2}}{4\pi^{2}} = \frac{39.24}{39.48} \approx 0.99\ \mathrm{m}. $$(A1)
Assumption: the amplitude is small (angle $\theta \lesssim 10^{\circ}$) so that $\sin\theta \approx \theta$ and the motion stays simple harmonic. (A1)
From $T = 2\pi\sqrt{L/g}$ with $L$ fixed, $T \propto 1/\sqrt{g}$. (M1)
Reducing $g$ by a factor of $6$ multiplies $T$ by $\sqrt{6}$, because $g$ appears under the square root in the denominator. (R1)
$$ \frac{T_{\text{Moon}}}{T_{\text{Earth}}} = \sqrt{\frac{g_{\text{Earth}}}{g_{\text{Moon}}}} = \sqrt{6} \approx 2.4. $$So the period lengthens by a factor of about $2.4$. (A1)
将 $T = 2\pi\sqrt{L/g}$ 对 $L$ 变形：(M1)
$$ L = \frac{gT^{2}}{4\pi^{2}} = \frac{(9.81)(2.0)^{2}}{4\pi^{2}} = \frac{39.24}{39.48} \approx 0.99\ \mathrm{m}. $$(A1)
假设：振幅很小（角度 $\theta \lesssim 10^{\circ}$），使 $\sin\theta \approx \theta$，运动保持简谐。(A1)
由 $T = 2\pi\sqrt{L/g}$ 且 $L$ 固定，$T \propto 1/\sqrt{g}$。(M1)
$g$ 减为 $1/6$ 使 $T$ 乘以 $\sqrt{6}$，因 $g$ 在分母的平方根之下。(R1)
$$ \frac{T_{\text{Moon}}}{T_{\text{Earth}}} = \sqrt{\frac{g_{\text{Earth}}}{g_{\text{Moon}}}} = \sqrt{6} \approx 2.4. $$故周期变长约 $2.4$ 倍。(A1)
SHM with $x_{0} = 0.050\ \mathrm{m}$, $\omega = 20\ \mathrm{rad\,s^{-1}}$. (a) $v_{\max}$ and where; (b) $a_{\max}$ and where; (c) speed at $x = 0.030\ \mathrm{m}$ and why not $60\%$ of $v_{\max}$.简谐运动，$x_{0} = 0.050\ \mathrm{m}$、$\omega = 20\ \mathrm{rad\,s^{-1}}$。(a) $v_{\max}$ 及位置；(b) $a_{\max}$ 及位置；(c) $x = 0.030\ \mathrm{m}$ 处的速率及为何不是 $v_{\max}$ 的 $60\%$。
The speed is greatest at the equilibrium centre, where $v_{\max} = \omega x_{0}$. (M1)
$$ v_{\max} = (20)(0.050) = 1.0\ \mathrm{m\,s^{-1}} \quad (\text{at } x = 0). $$(A1)
The acceleration is greatest in magnitude at the extremes, where $a_{\max} = \omega^{2} x_{0}$. (M1)
$$ a_{\max} = (20)^{2}(0.050) = 20\ \mathrm{m\,s^{-2}} \quad (\text{at } x = \pm x_{0}). $$(A1)
Use $v = \pm\,\omega\sqrt{x_{0}^{2} - x^{2}}$: (M1)
$$ v = 20\sqrt{0.050^{2} - 0.030^{2}} = 20\sqrt{1.6\times 10^{-3}} = 20(0.040). $$(M1 for substitution)
$$ v = 0.80\ \mathrm{m\,s^{-1}}. $$(A1)
This is $80\%$ of $v_{\max}$, not $60\%$, because the speed depends on $\sqrt{x_{0}^{2} - x^{2}}$, not linearly on $x$: at $x = 0.6\,x_{0}$ the factor is $\sqrt{1 - 0.6^{2}} = \sqrt{0.64} = 0.8$. (R1)
速率在平衡中心最大，此处 $v_{\max} = \omega x_{0}$。(M1)
$$ v_{\max} = (20)(0.050) = 1.0\ \mathrm{m\,s^{-1}} \quad (\text{在 } x = 0). $$(A1)
加速度在端点大小最大，此处 $a_{\max} = \omega^{2} x_{0}$。(M1)
$$ a_{\max} = (20)^{2}(0.050) = 20\ \mathrm{m\,s^{-2}} \quad (\text{在 } x = \pm x_{0}). $$(A1)
用 $v = \pm\,\omega\sqrt{x_{0}^{2} - x^{2}}$：(M1)
$$ v = 20\sqrt{0.050^{2} - 0.030^{2}} = 20\sqrt{1.6\times 10^{-3}} = 20(0.040). $$(代入得 M1)
$$ v = 0.80\ \mathrm{m\,s^{-1}}. $$(A1)
这是 $v_{\max}$ 的 $80\%$ 而非 $60\%$，因为速率取决于 $\sqrt{x_{0}^{2} - x^{2}}$，不是 $x$ 的线性函数：在 $x = 0.6\,x_{0}$ 处因子为 $\sqrt{1 - 0.6^{2}} = \sqrt{0.64} = 0.8$。(R1)
Pendulum $T^{2}$ vs $L$ data given. (a) show $T^{2}$ vs $L$ linear through origin and state gradient; (b) gradient and $g$; (c) % uncertainty in $T$ and in $T^{2}$ at $L = 1.000\ \mathrm{m}$ ($T = 2.01 \pm 0.02\ \mathrm{s}$); (d) systematic source and intercept direction if lengths too short.给出单摆 $T^{2}$ 对 $L$ 的数据。(a) 证明 $T^{2}$ 对 $L$ 为过原点直线并说明斜率；(b) 斜率与 $g$；(c) $L = 1.000\ \mathrm{m}$（$T = 2.01 \pm 0.02\ \mathrm{s}$）处 $T$ 与 $T^{2}$ 的百分比不确定度；(d) 系统误差来源及摆长偏短时截距方向。
Start from $T = 2\pi\sqrt{L/g}$ and square both sides: (M1)
$$ T^{2} = \frac{4\pi^{2}}{g}\,L. $$This has the form $T^{2} = (\text{gradient})\times L$ with no intercept, so a plot of $T^{2}$ against $L$ is a straight line through the origin. (A1)
Comparing with $y = mx$, the gradient is $\dfrac{4\pi^{2}}{g}$. (A1)
Read the gradient from two well-separated points, e.g. $(0.200,\,0.805)$ and $(1.000,\,4.024)$: (M1)
$$ \text{gradient} = \frac{4.024 - 0.805}{1.000 - 0.200} = \frac{3.219}{0.800} \approx 4.02\ \mathrm{s^{2}\,m^{-1}}. $$(A1)
Since the gradient is $\dfrac{4\pi^{2}}{g}$: $g = \dfrac{4\pi^{2}}{\text{gradient}} = \dfrac{4\pi^{2}}{4.02} \approx 9.8\ \mathrm{m\,s^{-2}}$. (A1)
Percentage uncertainty in $T$: (M1)
$$ \frac{0.02}{2.01}\times 100\% \approx 1.0\%. $$Since $T^{2}$ raises $T$ to the power $2$, its percentage uncertainty is double: $2\times 1.0\% \approx 2.0\%$. (A1)
A systematic error such as measuring the length to the top of the bob instead of to its centre (so every $L$ is recorded too short) shifts the whole line. (B1)
If the true lengths are larger than recorded, each plotted point sits at too small an $L$, so the line of best fit is displaced and, extrapolated back, crosses the $T^{2}$ axis below the origin (a negative intercept). (M1)
A non-zero intercept is the fingerprint of this systematic error, not random scatter. (A1)
由 $T = 2\pi\sqrt{L/g}$ 两边平方：(M1)
$$ T^{2} = \frac{4\pi^{2}}{g}\,L. $$此式形如 $T^{2} = (\text{斜率})\times L$，无截距，故 $T^{2}$ 对 $L$ 作图为过原点的直线。(A1)
与 $y = mx$ 比较，斜率为 $\dfrac{4\pi^{2}}{g}$。(A1)
用相距较远的两点读斜率，如 $(0.200,\,0.805)$ 与 $(1.000,\,4.024)$：(M1)
$$ \text{斜率} = \frac{4.024 - 0.805}{1.000 - 0.200} = \frac{3.219}{0.800} \approx 4.02\ \mathrm{s^{2}\,m^{-1}}. $$(A1)
因斜率为 $\dfrac{4\pi^{2}}{g}$：$g = \dfrac{4\pi^{2}}{\text{斜率}} = \dfrac{4\pi^{2}}{4.02} \approx 9.8\ \mathrm{m\,s^{-2}}$。(A1)
$T$ 的百分比不确定度：(M1)
$$ \frac{0.02}{2.01}\times 100\% \approx 1.0\%. $$因 $T^{2}$ 把 $T$ 取二次幂，其百分比不确定度加倍：$2\times 1.0\% \approx 2.0\%$。(A1)
系统误差，如将摆长测到摆球顶端而非其中心（使每个 $L$ 都记录偏短），会使整条直线平移。(B1)
若真实摆长大于记录值，每个点都落在偏小的 $L$ 处，故最佳拟合直线被移位，反向外推时在原点下方与 $T^{2}$ 轴相交（负截距）。(M1)
非零截距是该系统误差的指纹，而非随机散布。(A1)
Cosine $x$-$t$ graph starting at maximum; peak $0.060\ \mathrm{m}$, cycle $0.40\ \mathrm{s}$. (a) amplitude, period, $\omega$; (b) where speed is greatest and its value; (c) phase and shape of $v$-$t$ and $a$-$t$; (d) displacement of greatest acceleration.从最大值开始的余弦 $x$-$t$ 图；峰 $0.060\ \mathrm{m}$，周期 $0.40\ \mathrm{s}$。(a) 振幅、周期、$\omega$；(b) 速率最大处及其值；(c) $v$-$t$ 与 $a$-$t$ 的相位与形状；(d) 加速度最大处的位移。
Read straight off the graph: amplitude $x_{0} = 0.060\ \mathrm{m}$ and period $T = 0.40\ \mathrm{s}$. (A1)
Angular frequency: $\omega = \dfrac{2\pi}{T}$.
$$ \omega = \frac{2\pi}{0.40} \approx 15.7\ \mathrm{rad\,s^{-1}}. $$(A1)
The velocity is the slope of the $x$-$t$ graph, which is steepest as the curve crosses $x = 0$ (the equilibrium centre). So the speed is greatest at $x = 0$. (M1)
Its value is $v_{\max} = \omega x_{0}$: (M1)
$$ v_{\max} = (15.7)(0.060) \approx 0.94\ \mathrm{m\,s^{-1}}. $$(A1)
Velocity leads displacement by $90^{\circ}$ (a quarter cycle); since $x$ is a cosine, $v$ is a (negative) sine. (A1)
Acceleration is $180^{\circ}$ out of phase with displacement (anti-phase); since $x$ is a cosine, $a$ is an inverted cosine. (A1)
The acceleration has its greatest magnitude at the extremes, $x = \pm x_{0}$. (A1)
From $a = -\omega^{2}x$, the magnitude of $a$ scales with $|x|$, so it peaks where $|x|$ is largest, namely at the turning points. (R1)
直接从图读出：振幅 $x_{0} = 0.060\ \mathrm{m}$、周期 $T = 0.40\ \mathrm{s}$。(A1)
角频率：$\omega = \dfrac{2\pi}{T}$。
$$ \omega = \frac{2\pi}{0.40} \approx 15.7\ \mathrm{rad\,s^{-1}}. $$(A1)
速度是 $x$-$t$ 图的斜率，曲线穿过 $x = 0$（平衡中心）时最陡。故速率在 $x = 0$ 最大。(M1)
其值为 $v_{\max} = \omega x_{0}$：(M1)
$$ v_{\max} = (15.7)(0.060) \approx 0.94\ \mathrm{m\,s^{-1}}. $$(A1)
速度比位移超前 $90^{\circ}$（四分之一周期）；因 $x$ 为余弦，$v$ 为（负）正弦。(A1)
加速度与位移相位相反（$180^{\circ}$）；因 $x$ 为余弦，$a$ 为倒余弦。(A1)
加速度在端点 $x = \pm x_{0}$ 处大小最大。(A1)
由 $a = -\omega^{2}x$，$a$ 的大小与 $|x|$ 成正比，故在 $|x|$ 最大处（即折返点）达峰。(R1)
$m = 0.40\ \mathrm{kg}$, $k = 160\ \mathrm{N\,m^{-1}}$, $x_{0} = 0.060\ \mathrm{m}$, frictionless. (a) $\omega$; (b) total energy and $v_{\max}$; (c) $E_{K}$ and $E_{P}$ at $x = 0.030\ \mathrm{m}$, confirm sum; (d) effect of doubling amplitude on $E$; (e) sketch $E_{K}$ and $E_{P}$ vs time, and how many maxima each per oscillation.$m = 0.40\ \mathrm{kg}$、$k = 160\ \mathrm{N\,m^{-1}}$、$x_{0} = 0.060\ \mathrm{m}$，无摩擦。(a) $\omega$；(b) 总能量与 $v_{\max}$；(c) $x = 0.030\ \mathrm{m}$ 处 $E_{K}$、$E_{P}$ 并验证之和；(d) 振幅加倍对 $E$ 的影响；(e) 画 $E_{K}$、$E_{P}$ 随时间变化及每次振动各达最大几次。
For a mass-spring system $\omega^{2} = k/m$: (M1)
$$ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{160}{0.40}} = \sqrt{400} = 20\ \mathrm{rad\,s^{-1}}. $$(A1)
Total energy $E = \tfrac{1}{2} m \omega^{2} x_{0}^{2}$: (M1)
$$ E = \tfrac{1}{2}(0.40)(20)^{2}(0.060)^{2} = \tfrac{1}{2}(0.40)(400)(3.6\times 10^{-3}) = 0.288\ \mathrm{J}. $$(A1)
At the centre all energy is kinetic, so $\tfrac{1}{2} m v_{\max}^{2} = E$, giving $v_{\max} = \omega x_{0} = (20)(0.060) = 1.2\ \mathrm{m\,s^{-1}}$. (A1)
Kinetic energy $E_{K} = \tfrac{1}{2} m \omega^{2}(x_{0}^{2} - x^{2})$: (M1)
$$ E_{K} = \tfrac{1}{2}(0.40)(400)(0.060^{2} - 0.030^{2}) = \tfrac{1}{2}(0.40)(400)(2.7\times 10^{-3}) = 0.216\ \mathrm{J}. $$(A1)
Potential energy is the remainder: $E_{P} = E - E_{K} = 0.288 - 0.216 = 0.072\ \mathrm{J}$, and $E_{K} + E_{P} = 0.288\ \mathrm{J} = E$, as required. (A1)
The total energy becomes four times as large. (A1)
$E = \tfrac{1}{2} m \omega^{2} x_{0}^{2} \propto x_{0}^{2}$ with $m$ and $\omega$ fixed, so doubling $x_{0}$ multiplies $E$ by $2^{2} = 4$. (R1)
Both $E_{K}$ and $E_{P}$ oscillate between $0$ and $E$, in antiphase with each other (when one is maximum the other is zero), and their sum is the constant horizontal line $E$. (M1)
Each reaches its maximum twice per oscillation, so the energy curves have double the frequency of the displacement. (A1)
弹簧振子 $\omega^{2} = k/m$：(M1)
$$ \omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{160}{0.40}} = \sqrt{400} = 20\ \mathrm{rad\,s^{-1}}. $$(A1)
总能量 $E = \tfrac{1}{2} m \omega^{2} x_{0}^{2}$：(M1)
$$ E = \tfrac{1}{2}(0.40)(20)^{2}(0.060)^{2} = \tfrac{1}{2}(0.40)(400)(3.6\times 10^{-3}) = 0.288\ \mathrm{J}. $$(A1)
在中心能量全为动能，故 $\tfrac{1}{2} m v_{\max}^{2} = E$，得 $v_{\max} = \omega x_{0} = (20)(0.060) = 1.2\ \mathrm{m\,s^{-1}}$。(A1)
动能 $E_{K} = \tfrac{1}{2} m \omega^{2}(x_{0}^{2} - x^{2})$：(M1)
$$ E_{K} = \tfrac{1}{2}(0.40)(400)(0.060^{2} - 0.030^{2}) = \tfrac{1}{2}(0.40)(400)(2.7\times 10^{-3}) = 0.216\ \mathrm{J}. $$(A1)
势能为余量：$E_{P} = E - E_{K} = 0.288 - 0.216 = 0.072\ \mathrm{J}$，且 $E_{K} + E_{P} = 0.288\ \mathrm{J} = E$，符合要求。(A1)
总能量变为四倍。(A1)
$E = \tfrac{1}{2} m \omega^{2} x_{0}^{2} \propto x_{0}^{2}$，$m$、$\omega$ 不变，故 $x_{0}$ 加倍使 $E$ 乘以 $2^{2} = 4$。(R1)
$E_{K}$ 与 $E_{P}$ 都在 $0$ 与 $E$ 之间振荡，彼此反相（一者最大时另一者为零），二者之和为恒定的水平线 $E$。(M1)
每次振动各达最大两次，故能量曲线的频率为位移频率的两倍。(A1)
Trolley $m = 0.25\ \mathrm{kg}$ between springs, net restoring force $F = -kx$, $k = 100\ \mathrm{N\,m^{-1}}$. (a) show SHM and $\omega^{2} = k/m$; (b) period; (c) acceleration at release from $x = 0.040\ \mathrm{m}$; (d) light damping: amplitude, energy, period.小车 $m = 0.25\ \mathrm{kg}$ 夹于弹簧间，合恢复力 $F = -kx$，$k = 100\ \mathrm{N\,m^{-1}}$。(a) 证明简谐且 $\omega^{2} = k/m$；(b) 周期；(c) 在 $x = 0.040\ \mathrm{m}$ 释放时的加速度；(d) 轻阻尼：振幅、能量、周期。
Apply Newton's second law along the track with net force $F = -kx$: $ma = -kx$. (M1)
$$ a = -\frac{k}{m}\,x. $$This has the form $a = -(\text{positive constant})\times x$, the defining condition of SHM, so the motion is simple harmonic. (M1)
Comparing with $a = -\omega^{2}x$ identifies $\omega^{2} = \dfrac{k}{m}$. (A1)
$\omega = \sqrt{k/m} = \sqrt{100/0.25} = \sqrt{400} = 20\ \mathrm{rad\,s^{-1}}$, so $T = \dfrac{2\pi}{\omega}$. (M1)
$$ T = \frac{2\pi}{20} \approx 0.31\ \mathrm{s}. $$(A1)
Released from rest at $x = 0.040\ \mathrm{m}$ (the amplitude), the magnitude of the acceleration is $|a| = \omega^{2}x$: (M1)
$$ |a| = (20)^{2}(0.040) = 400(0.040) = 16\ \mathrm{m\,s^{-2}}. $$(A1)
With a small frictional force, the amplitude decreases gradually with time (typically an exponential decay envelope). (A1)
The total energy also falls with time, since $E \propto x_{0}^{2}$ and the work done against friction is dissipated as thermal energy. (A1)
For light damping the period is almost unchanged from the undamped value: damping lowers the oscillation frequency only very slightly, so $T \approx 2\pi\sqrt{m/k}$ still holds to a good approximation. (R1)
沿轨道对合力 $F = -kx$ 应用牛顿第二定律：$ma = -kx$。(M1)
$$ a = -\frac{k}{m}\,x. $$此式形如 $a = -(\text{正常数})\times x$，即简谐运动的定义条件，故运动为简谐。(M1)
与 $a = -\omega^{2}x$ 比较得 $\omega^{2} = \dfrac{k}{m}$。(A1)
$\omega = \sqrt{k/m} = \sqrt{100/0.25} = \sqrt{400} = 20\ \mathrm{rad\,s^{-1}}$，故 $T = \dfrac{2\pi}{\omega}$。(M1)
$$ T = \frac{2\pi}{20} \approx 0.31\ \mathrm{s}. $$(A1)
在 $x = 0.040\ \mathrm{m}$（振幅）处由静止释放，加速度大小为 $|a| = \omega^{2}x$：(M1)
$$ |a| = (20)^{2}(0.040) = 400(0.040) = 16\ \mathrm{m\,s^{-2}}. $$(A1)
有小摩擦力时，振幅随时间逐渐减小（通常为指数衰减包络）。(A1)
总能量也随时间下降，因 $E \propto x_{0}^{2}$，且克服摩擦所做的功以热能形式耗散。(A1)
对轻阻尼，周期与无阻尼值几乎不变：阻尼只极微小地降低振动频率，故 $T \approx 2\pi\sqrt{m/k}$ 仍很好地成立。(R1)
SHM, $x_{0} = 0.080\ \mathrm{m}$, $T = 0.50\ \mathrm{s}$, released from rest at maximum at $t = 0$. (a) write $x(t)$, justify cosine, find $\omega$; (b) displacement and velocity at $t = 0.10\ \mathrm{s}$; (c) acceleration at $t = 0.10\ \mathrm{s}$ and check $a = -\omega^{2}x$.简谐运动，$x_{0} = 0.080\ \mathrm{m}$、$T = 0.50\ \mathrm{s}$，$t = 0$ 在最大位移处静止释放。(a) 写 $x(t)$、论证用余弦、求 $\omega$；(b) $t = 0.10\ \mathrm{s}$ 时的位移与速度；(c) $t = 0.10\ \mathrm{s}$ 时的加速度并验证 $a = -\omega^{2}x$。
The mass is released from rest at maximum displacement, so at $t = 0$ we have $x = x_{0}$; the curve that starts at its peak is the cosine, hence $x = x_{0}\cos\omega t$ (a sine would start at $x = 0$). (A1)
The angular frequency is $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{0.50} = 4\pi \approx 12.6\ \mathrm{rad\,s^{-1}}$, so $x = 0.080\cos(4\pi t)$. (A1)
Displacement: $x = 0.080\cos(4\pi \times 0.10) = 0.080\cos(1.257\ \mathrm{rad})$. (M1)
$$ x = 0.080(0.309) \approx 0.025\ \mathrm{m}. $$(A1)
Velocity is $v = -x_{0}\omega\sin\omega t = -(0.080)(12.57)\sin(1.257)$:
$$ v = -(0.080)(12.57)(0.951) \approx -0.96\ \mathrm{m\,s^{-1}}. $$The minus sign shows the mass is moving back toward equilibrium. (A1)
Acceleration from the defining condition $a = -\omega^{2}x$: (M1)
$$ a = -(12.57)^{2}(0.0247) = -(158)(0.0247) \approx -3.9\ \mathrm{m\,s^{-2}}. $$(A1)
The same value follows from $a = -x_{0}\omega^{2}\cos\omega t = -(0.080)(158)(0.309) \approx -3.9\ \mathrm{m\,s^{-2}}$, confirming $a = -\omega^{2}x$ holds at every instant. (R1)
质量在最大位移处由静止释放，故 $t = 0$ 时 $x = x_{0}$；从峰值开始的曲线是余弦，故 $x = x_{0}\cos\omega t$（正弦会从 $x = 0$ 开始）。(A1)
角频率 $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{0.50} = 4\pi \approx 12.6\ \mathrm{rad\,s^{-1}}$，故 $x = 0.080\cos(4\pi t)$。(A1)
位移：$x = 0.080\cos(4\pi \times 0.10) = 0.080\cos(1.257\ \mathrm{rad})$。(M1)
$$ x = 0.080(0.309) \approx 0.025\ \mathrm{m}. $$(A1)
速度为 $v = -x_{0}\omega\sin\omega t = -(0.080)(12.57)\sin(1.257)$：
$$ v = -(0.080)(12.57)(0.951) \approx -0.96\ \mathrm{m\,s^{-1}}. $$负号表示质量正朝平衡位置回移。(A1)
由定义条件 $a = -\omega^{2}x$ 求加速度：(M1)
$$ a = -(12.57)^{2}(0.0247) = -(158)(0.0247) \approx -3.9\ \mathrm{m\,s^{-2}}. $$(A1)
同一值也可由 $a = -x_{0}\omega^{2}\cos\omega t = -(0.080)(158)(0.309) \approx -3.9\ \mathrm{m\,s^{-2}}$ 得到，验证 $a = -\omega^{2}x$ 在每一时刻都成立。(R1)