The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. Equation (1.1) gives 144, 14.4, and 1.44 kPa. It is dominant or comparable at 1 micrometer, a noticeable correction at 10 micrometers, and a small correction at 100 micrometers. “Small bubble” is therefore not a purely geometric description; it changes the pressure balance.

解答。由式(1.1)，结果为144、14.4和1.44 kPa。在1微米处，它占主导或至少与环境压力相当；在10微米处，它是明显修正；在100微米处，它较小。因此，“小气泡”不只是几何描述，还意味着压力平衡发生变化。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. Since $u^2r^2\propto r^{-2}$, the energy outside a lower radius $r_b$ is proportional to $1/r_b$. Therefore $T_{r>aR}/T=1/a$. Half the energy lies outside $2R$. A computational boundary only a few bubble radii away can alter the inertia substantially unless an appropriate exterior representation is used.

解答。由于$u^2r^2\propto r^{-2}$，从下限$r_b$向外积分的动能与$1/r_b$成正比，因此$T_{r>aR}/T=1/a$。一半动能位于$2R$之外。因此，如果计算边界只距气泡几个半径，且未采用适当外域处理，就可能显著改变惯性。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. Along a particle trajectory, $dr/dt=R^2\dot R/r^2$. Multiply by $3r^2$ to obtain $d(r^3)/dt=d(R^3)/dt$. Thus $r(t)^3-R(t)^3=r(t_0)^3-R(t_0)^3$. The relation expresses conservation of the liquid volume between the bubble and that material spherical surface.

解答。沿质点轨迹有$dr/dt=R^2\dot R/r^2$，乘以$3r^2$得$d(r^3)/dt=d(R^3)/dt$，因此$r(t)^3-R(t)^3=r(t_0)^3-R(t_0)^3$。该关系表达了气泡与该物质球面之间液体体积守恒。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. The chain rule gives $d\phi(R(t),t)/dt=\partial_t\phi+\dot R\partial_r\phi$. Since $\partial_r\phi|_R=\dot R$, subtract $\dot R^2$ from the total derivative. The correct partial derivative is $-2\dot R^2-R\ddot R$, which produces the $3/2$ inertia coefficient after adding $u^2/2$.

解答。链式法则给出$d\phi(R(t),t)/dt=\partial_t\phi+\dot R\partial_r\phi$。由于$\partial_r\phi|_R=\dot R$，应从全导数中再减去$\dot R^2$。正确偏导为$-2\dot R^2-R\ddot R$，加上$u^2/2$后即可得到$3/2$惯性系数。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. First calculate $p_{g0}=p_0-p_v+2\sigma/R_0=127786\ \mathrm{Pa}$. Equation (2.5) gives $R_B=18.242\ \mu\mathrm m$ and $p_{\mathrm{Blake}}=-2923.5\ \mathrm{Pa}$ absolute. This negative absolute pressure is a tensile metastable-liquid value within the model, not a negative gas pressure and not a universal measured inception threshold.

解答。先算得$p_{g0}=p_0-p_v+2\sigma/R_0=127786\ \mathrm{Pa}$。由式(2.5)，得到$R_B=18.242\ \mu\mathrm m$和绝对压力$p_{\mathrm{Blake}}=-2923.5\ \mathrm{Pa}$。这个负绝对压力表示模型中亚稳液体的拉伸状态，不是负气体压力，也不是普适实测初生阈值。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. $P=1-e^{-JV\tau}=0.5$ implies $JV\tau=\ln2$. Doubling volume gives $P_2=1-e^{-2\ln2}=0.75$. A threshold defined by a fixed event probability therefore shifts when observation volume changes, even if the microscopic rate law does not.

解答。由$P=1-e^{-JV\tau}=0.5$得$JV\tau=\ln2$。体积加倍后，$P_2=1-e^{-2\ln2}=0.75$。因此，即使微观速率定律不变，以固定事件概率定义的阈值也会随观测体积变化。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. Set $\ddot R\to0$ in $R\ddot R+3\dot R^2/2=\Delta p/\rho$. Then $U_\infty=\sqrt{2\Delta p/(3\rho)}$. For $\Delta p=98986\ \mathrm{Pa}$ and $\rho=998\ \mathrm{kg\,m^{-3}}$, this is $8.132\ \mathrm{m\,s^{-1}}$. The result assumes the driving pressure remains available; it does not apply to a fixed gas mass whose pressure falls as the bubble grows.

解答。在$R\ddot R+3\dot R^2/2=\Delta p/\rho$中令$\ddot R\to0$，得到$U_\infty=\sqrt{2\Delta p/(3\rho)}$。取$\Delta p=98986\ \mathrm{Pa}$和$\rho=998\ \mathrm{kg\,m^{-3}}$，可得$8.132\ \mathrm{m\,s^{-1}}$。该结果假定驱动压差始终存在，不适用于压力随膨胀下降的固定质量气体。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. Since $t_c\propto R_{\max}\Delta p^{-1/2}$, the first change gives $18.369\ \mu\mathrm s$, and the second gives $4.592\ \mu\mathrm s$. Under independent small uncertainties, $s_{t_c}^2/t_c^2\approx(s_R/R)^2+\tfrac14(s_\rho/\rho)^2+\tfrac14(s_{\Delta p}/\Delta p)^2$. Correlated uncertainties require covariance terms.

解答。由$t_c\propto R_{\max}\Delta p^{-1/2}$，第一种变化得到$18.369\ \mu\mathrm s$，第二种得到$4.592\ \mu\mathrm s$。对于独立小不确定度，有$s_{t_c}^2/t_c^2\approx(s_R/R)^2+\tfrac14(s_\rho/\rho)^2+\tfrac14(s_{\Delta p}/\Delta p)^2$。若误差相关，还需加入协方差项。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. Equation (5.2) gives $U=-\sqrt{(2\Delta p/3\rho)(8-1)}=-21.514\ \mathrm{m\,s^{-1}}$. The value is independent of $R_{\max}$ in this scale-free idealization at a fixed radius ratio. Collapse time, however, remains proportional to $R_{\max}$.

解答。式(5.2)给出$U=-\sqrt{(2\Delta p/3\rho)(8-1)}=-21.514\ \mathrm{m\,s^{-1}}$。在这一无额外长度尺度的理想化中，固定半径比处的速度与$R_{\max}$无关，但塌缩时间仍与$R_{\max}$成正比。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. Rearranging (5.2) gives $R/R_{\max}=[1+3\rho c^2M^2/(2\Delta p)]^{-1/3}=0.1443$. Therefore a numerical collapse continued to $0.05R_{\max}$ is already outside this conservative physical warning level. It can still be used as a mathematical benchmark of the incompressible equation, but not as a credible physical late-collapse trajectory.

解答。整理式(5.2)得$R/R_{\max}=[1+3\rho c^2M^2/(2\Delta p)]^{-1/3}=0.1443$。因此，将塌缩继续积分至$0.05R_{\max}$，已经超出这一保守物理警戒范围。它仍可作为不可压缩方程的数学基准，但不能作为可信的真实塌缩末期轨迹。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. Set time derivatives to zero in (6.1): $x=-P_a/[3\kappa p_{g0}-2\sigma/R_0]=-0.0021653$. The radius decreases by approximately $21.65\ \mathrm{nm}$. A truly slow physical experiment may be closer to isothermal than the code's $\kappa=1.4$; the calculation is conditional on the stated closure.

解答。在式(6.1)中令时间导数为零，得$x=-P_a/[3\kappa p_{g0}-2\sigma/R_0]=-0.0021653$，半径约减少$21.65\ \mathrm{nm}$。真实缓慢实验可能比代码中的$\kappa=1.4$更接近等温；此计算以所指定闭合为条件。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. $\beta_\mu$ doubles, while the undamped $\omega_0$ in (6.2) is unchanged. The damped free frequency $\sqrt{\omega_0^2-\beta^2}$ and forced-response peak can still change because they depend on damping. Thus “viscosity does not change frequency” is only correct when referring specifically to the undamped stiffness-to-mass frequency.

解答。$\beta_\mu$加倍，而式(6.2)中的无阻尼$\omega_0$不变。但有阻尼自由频率$\sqrt{\omega_0^2-\beta^2}$和受迫响应峰值位置仍可能改变，因为它们依赖阻尼。因此，“黏度不改变频率”仅在专指由刚度与质量决定的无阻尼频率时成立。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. A reversible potential satisfies $dW_g=-p_g\,dV$. Integrating gives $W_g=-p_{g0}V_0\ln(V/V_0)+\mathrm{constant}$. For an isothermal ideal gas, internal energy is constant; this logarithmic expression is therefore not the gas internal energy. It is the mechanical potential appropriate to the prescribed pressure law and thermal reservoir.

解答。可逆势满足$dW_g=-p_g\,dV$，积分得$W_g=-p_{g0}V_0\ln(V/V_0)+\mathrm{constant}$。等温理想气体的内能恒定，因此这个对数表达式不是气体内能，而是与所指定压力定律及热库相对应的力学势。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. The liquid-side pressure contains $-4\mu U/R$. Differentiation contributes $-4\mu\dot U/R+4\mu U^2/R^2$. Multiplication by $R/(\rho c)$ in (7.1) gives $-4\mu\dot U/(\rho c)$ on the right. Moving it to the left adds $4\mu/(\rho c)$ to the coefficient of $\dot U$. The effect is algebraically required, not an empirical correction.

解答。液侧压力包含$-4\mu U/R$，求导产生$-4\mu\dot U/R+4\mu U^2/R^2$。在式(7.1)中乘以$R/(\rho c)$后，右端出现$-4\mu\dot U/(\rho c)$。将其移到左端，就使$\dot U$的系数增加$4\mu/(\rho c)$。这一项由代数必然产生，并非经验修正。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. Equation (8.1) gives $f_2=\sqrt{12\sigma/(\rho R^3)}/(2\pi)=148.09\ \mathrm{kHz}$, compared with $342.37\ \mathrm{kHz}$ for the radial mode under the teaching gas closure. Different deformation families have different restoring mechanisms. A single measured oscillation frequency should not be assigned to radial motion without checking the shape.

解答。由式(8.1)，$f_2=\sqrt{12\sigma/(\rho R^3)}/(2\pi)=148.09\ \mathrm{kHz}$，而教学气体闭合下径向频率为$342.37\ \mathrm{kHz}$。不同变形类型具有不同回复机制。若未检查形状，就不能把某个实测振荡频率直接认定为径向振动。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. Not through (9.1), because $\nabla\Pi=0$. Boundaries, interfaces, or nonuniform loading are needed to generate the relevant pressure gradients. This does not say that uniform pressure on one side of a plate produces no force; that is a different boundary-value problem with a pressure difference across the structure.

解答。按式(9.1)不会，因为$\nabla\Pi=0$。需要边界、界面或非均匀加载形成相关压力梯度。这并不是说板一侧的均匀压力不产生力；后者是结构两侧存在压差的另一类边值问题。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. Equation (12.3) gives zero post-pulse free amplitude, because $\sin\pi=0$. The mode moved during the pulse; it did not remain stationary. The force first supplied and later recovered energy over one full oscillation. This is a counterexample to predicting structural response from total impulse alone when pulse duration is not negligible.

解答。式(12.3)给出脉冲后自由振幅为零，因为$\sin\pi=0$。模态在脉冲期间发生过运动，并不是始终静止。外力在完整振动周期内先输入能量，再回收能量。这说明当脉冲时长不可忽略时，不能只凭总冲量预测结构响应。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. $v=\sqrt{2G_c/m_A}=3.162\ \mathrm{m\,s^{-1}}$, and $I_p=m_Av=0.06325\ \mathrm{Pa\,s}$. These are not sufficient detachment thresholds. The estimate assumes all relevant kinetic energy becomes separation work and ignores bending, added liquid inertia, flaws, mode mix, and incomplete load coverage.

解答。$v=\sqrt{2G_c/m_A}=3.162\ \mathrm{m\,s^{-1}}$，且$I_p=m_Av=0.06325\ \mathrm{Pa\,s}$。这些数值不是充分脱离阈值。该估计假定相关动能全部转为分离功，并忽略弯曲、液体附加惯性、缺陷、混合模态和载荷覆盖不完整等因素。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. Equation (11.1) gives $R_b/a_0=5.105$, corresponding to a volume ratio of about 133.0. This calculation provides no evidence that sufficient heat exists to complete vaporization. The parameters deliberately do not identify a measured PFC formulation.

解答。式(11.1)给出$R_b/a_0=5.105$，对应体积比约133.0。这个计算并不能证明体系具有足够热量完成汽化。所列参数有意不对应任何已测量的特定全氟碳配方。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. Pressure amplitude halves because $p'\propto1/r$. The sphere area increases by four, while $p'^2$ decreases by four, so (7.8) remains unchanged in the absence of absorption. This reasoning fails if near-field terms, reflections, nonlinear dissipation, or sensor averaging dominate.

解答。由于$p'\propto1/r$，压力振幅减半。球面积增至四倍，而$p'^2$减至四分之一，因此无吸收时式(7.8)保持不变。如果近场项、反射、非线性耗散或传感器平均效应占主导，这个推理就会失效。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. The expression presumes a fixed-mass ideal gas and a specified polytropic process. At severe collapse, gas nonideality, phase change, thermal gradients, compressibility, chemistry, and nonsphericity may invalidate those premises. Even a valid model output is not an experimental measurement. The correct report labels it a conditional estimate and gives the model's validity limits.

解答。该表达式预设固定质量理想气体与指定多方过程。剧烈塌缩时，气体非理想性、相变、温度梯度、可压缩性、化学过程和非球形性都可能使这些前提失效。即使模型输出有效，它也不是实验测量。正确报告应将其标为条件性估计，并说明模型适用范围。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. $2^p=(1.00-0.92)/(0.92-0.90)=4$, so $p=2$. Richardson extrapolation gives $Q_*=0.90+(0.90-0.92)/(4-1)=0.89333$. The inference is conditional on the asymptotic model. Interface topology changes, stochastic nucleation, and unresolved shock peaks can make three-grid extrapolation misleading.

解答。$2^p=(1.00-0.92)/(0.92-0.90)=4$，因此$p=2$。Richardson外推得$Q_*=0.90+(0.90-0.92)/(4-1)=0.89333$。这一推断依赖渐近误差模型；界面拓扑变化、随机成核和未解析激波峰值，都可能使三网格外推产生误导。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. No. Infinitely many pairs yield the same $C$ and hence the same pressure law. An additional equilibrium relation or independent measurement is needed. An optimizer returning two precise-looking numbers does not remove this structural nonidentifiability.

解答。不能。无穷多组参数都能给出相同$C$，从而产生相同压力定律。需要附加平衡关系或独立测量。优化器即使返回两个看似精确的数值，也不能消除这种结构性不可辨识。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. It grows by a factor of four because the denominator is proportional to $\Delta t^2$. Higher temporal resolution is valuable, but acceleration estimation needs an appropriate observation model, smoothing with quantified bias, or direct fitting of a dynamical model. Do not confuse temporal resolution with derivative precision.

解答。由于分母与$\Delta t^2$成正比，噪声尺度增大至四倍。更高时间分辨率仍然有价值，但加速度估计需要适当观测模型、偏差可量化的平滑方法，或直接拟合动力学模型。不要将时间分辨率与导数精度混为一谈。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. [S390]: optoacoustic transmitter, focused ultrasound, nucleation, and liquid jetting. [S72]: photothermal phase change inside a hydrogel-composite stamp, surface bulging, and adhesion modification. [S343]: blister motion coupled to cavitation in a liquid-transfer setting. [S340]: stress-wave-mediated thin-film delamination. The common presence of a laser does not make the four source models interchangeable.

解答。[S390]：光声发射器、聚焦超声、成核与液体喷射。[S72]：水凝胶复合印章内部光热相变、表面鼓起与黏附改变。[S343]：液体转移环境中鼓包运动与空化耦合。[S340]：应力波介导的薄膜分层。共同使用激光，并不使四种源模型可以相互替代。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

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Solution. Its nucleation process forms liquid perfluorocarbon droplets through solvent exchange. The new phase and free-energy problem differ from vapor embryos forming in a liquid. A relevant title word is not enough; phase identity and process must match the claim.

解答。该文的成核过程通过溶剂交换形成液态全氟碳液滴。其新生相与自由能问题，不同于液体中形成蒸气胚核的过程。标题中存在相同关键词并不足够，物相与过程必须和论点匹配。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. The motion and traction are linked by the fluid equations. Prescribing both independently usually overconstrains the boundary or injects energy inconsistent with either physical source. Prescribe kinematics and calculate reaction pressure, or prescribe a thermodynamic pressure law and solve for motion, unless a carefully formulated coupling supplies a consistent relation.

解答。运动与牵引通过流体方程相联系。独立指定二者通常会使边界过度约束，或注入与任何真实源都不一致的能量。除非具有严格相容的耦合关系，否则应指定运动学并计算反作用压力，或指定热力学压力定律并求解运动。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. Determine whether the added mass represents liquid already included in the solved fluid domain. If so, it is double-counted. Added-mass corrections belong to reduced models or omitted fluid regions, with a clearly defined partition. A coupled model should recover a known wet-versus-dry structural benchmark without arbitrary duplicate inertia.

解答。应判断附加质量是否表示已包含在求解流体域中的液体。如果是，就发生了重复计算。附加质量修正应当用于降阶模型或被省略的流体区域，并明确划分。耦合模型应能在没有任意重复惯性的情况下复现已知干态与浸液结构基准。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. Compare full growth and collapse histories, rebound radius and period, and their variation with independently controlled ambient temperature or pulse duration. A single extra peak is less informative than a controlled family of trajectories. Fit shared parameters on one subset and test on held-out conditions; do not refit every trace independently and call the result prediction.

解答。可比较完整增长与塌缩历程、回弹半径和周期，以及它们随独立控制的环境温度或脉冲时长的变化。相比只增加一个峰值，受控的一组轨迹更有信息。应在部分数据上拟合共享参数，再在留出条件上检验；不要对每条曲线分别重新拟合后仍称之为预测。

The diagram locates the physical system; its key lists the symbols in the formula. Read the definition in the surrounding derivation when a letter has different roles.

图示标出物理体系，图例列出公式中的符号。若某字母有多种作用，应以邻近推导中的定义为准。

Solution. Fix geometry, support, payload mass, and a declared source normalization such as fixed volume excursion or fixed work. Sweep loading duration; measure or calculate mode-projected force, payload velocity, bending energy, and losses. Verify the uncoupled oscillator and fluid limit first. Compare at least one held-out duration and one altered support stiffness. A peak in useful motion may support the hypothesis, but only after showing that it is not caused by a hidden change in source energy or by numerical resonance.

解答。固定几何、支承、负载质量，并声明源归一化方式，例如固定体积变化或固定功。扫描加载时长，测量或计算模态投影力、负载速度、弯曲能和损失。首先验证未耦合振子与流体极限，然后至少检验一个留出时长和一个改变支承刚度的条件。有效运动出现峰值可能支持假设，但必须先证明它不是由隐含源能量变化或数值共振引起的。