📚 Representing s.h.m. graphically | 简谐运动的图像表示
In CIE A-Level Physics, graphical representation of simple harmonic motion is a core skill. You need to interpret and sketch displacement-time, velocity-time, acceleration-time and energy graphs, and you must understand the phase relationships between them.
在 CIE A-Level 物理中，简谐运动的图像表示是一项核心技能。你需要解读并绘制位移-时间、速度-时间、加速度-时间和能量图像，并且必须理解它们之间的相位关系。
These graphs are not just drawings; they show the defining features of s.h.m. such as sinusoidal variation, constant period and the link between acceleration and displacement.
这些图像不仅仅是示意图；它们展示了简谐运动的定义特征，例如正弦变化、恒定周期以及加速度与位移之间的联系。
1. What s.h.m. graphs show | 简谐运动图像展示什么
ed equilibrium position and always acts towards that position. This is written as F = -kx.
当质点所受合力与其离开固定平衡位置的位移成正比，并且始终指向该平衡位置时，就会发生简谐运动。这可以写作 F = -kx。
y. The negative sign tells us that acceleration and displacement are always opposite in direction.
等价地，加速度为 a = -ω²x，其中 ω 是角频率。负号告诉我们，加速度与位移的方向始终相反。
When plotted against time, displacement, velocity and acceleration are all sinusoidal waves with the same period T. The graphs therefore repeat regularly and show clear maximum and zero values.
当以时间为横轴绘图时，位移、速度和加速度都是具有相同周期 T 的正弦波。因此这些图像有规律地重复，并显示出明确的最大值和零值。
2. Displacement-time graph | 位移-时间图像
If timing starts when the particle passes through the equilibrium position moving in the positive direction, the displacement-time graph is a sine curve:
如果从质点经过平衡位置并向正方向运动时开始计时，位移-时间图像是正弦曲线：
x = x₀ sin(ωt)
Here x₀ is the amplitude, the maximum displacement from equilibrium. The graph starts at zero, rises to +x₀, returns to zero, falls to -x₀ and repeats every period T.
其中 x₀ 是振幅，即离开平衡位置的最大位移。图像从零开始，上升到 +x₀，返回零，下降到 -x₀，然后每隔周期 T 重复一次。
). This is not a different motion; it is only a phase shift of π/2 rad in the time origin.
如果从最大正位移处开始计时，同一运动可以用 x = x₀ cos(ωt) 描述。这不是不同的运动，只是时间原点发生了 π/2 rad 的相位移动。
t gives the velocity. The steepest parts occur at x = 0, and the slope is zero at x = ±x₀.
在位移-时间图像上，任意一点的斜率表示速度。最陡的部分出现在 x = 0 处，而在 x = ±x₀ 处斜率为零。
3. Velocity-time graph | 速度-时间图像
is the gradient of the displacement-time graph. For x = x₀ sin(ωt), differentiating gives:
速度是位移的变化率，因此它是位移-时间图像的斜率。对于 x = x₀ sin(ωt)，求导得到：
v = dx/dt = ωx₀ cos(ωt)
s when the particle passes through the equilibrium position. This is where the displacement graph is steepest.
最大速率为 v₀ = ωx₀，出现在质点经过平衡位置时。这正是位移图像最陡的位置。
The velocity-time graph is also sinusoidal but it leads the displacement graph by π/2 rad, or 90°. When displacement is zero, velocity is maximum; when displacement is maximum, velocity is zero.
速度-时间图像也是正弦曲线，但它比位移图像超前 π/2 rad，即 90°。当位移为零时，速度最大；当位移最大时，速度为零。
article changes direction, so the instantaneous velocity must be zero. This is shown by the graph touching the time axis at those instants.
在端点 x = ±x₀ 处，质点改变运动方向，因此瞬时速度必须为零。图像在这些时刻与时间轴相切。
4. Acceleration-time graph | 加速度-时间图像
, so it is the gradient of the velocity-time graph. Differentiating v = ωx₀ cos(ωt) gives:
加速度是速度的变化率，因此它是速度-时间图像的斜率。对 v = ωx₀ cos(ωt) 求导得到：
a = dv/dt = -ω²x₀ sin(ωt) = -ω²x
This is the defining equation of s.h.m. The acceleration-time graph has the same shape as the displacement-time graph but is inverted, so acceleration is π rad, or 180°, out of phase with displacement.
这是简谐运动的定义方程。加速度-时间图像与位移-时间图像形状相同但方向相反，因此加速度与位移相位差为 π rad，即 180°。
occurs at the extreme positions. Acceleration is zero at the equilibrium position, where the restoring force is zero.
最大加速度为 a₀ = ω²x₀，出现在端点。加速度在平衡位置为零，因为该处的回复力为零。
On the acceleration-time graph,
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