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GRE Subject Test · GRE Physics נושא 1 4:52 קריאת קול באנגלית · תרגום אנגלי + סינית שרוף בתוך הסרטון
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תרגום טקסט
A planet on a circular orbit accelerates even when its speed is constant, because velocity includes direction.
行星沿圆轨道运动时，即使速率恒定也在加速，因为速度包含方向。
The velocity keeps turning, so the acceleration points to the centre.
速度方向不断改变，所以加速度指向圆心。
This topic uses Newtonian, Lagrangian and Hamiltonian descriptions for rotation, central forces and oscillations, and uses each conservation law only where its conditions hold.
本主题用牛顿、拉格朗日和哈密顿描述来分析旋转、中心力和振动，并且只在条件成立时使用各条守恒定律。
Newton’s second law is a vector statement.
牛顿第二定律是一个矢量表述。
Choose coordinates and identify constraints before components.
在确定分量之前，先选择坐标系并识别约束条件。
Energy conservation applies when work from nonconservative forces is accounted for; momentum conservation needs zero net external impulse.
当计入非保守力所做的功时，能量守恒成立。 动量守恒需要合外力冲量为零。
For small oscillations, linearise around a stable equilibrium.
对于小振幅振动，需在稳定平衡点附近进行线性化。
A restoring term proportional to displacement gives harmonic motion.
与位移成正比的恢复力项将产生简谐运动。
Lagrange’s equations use L=T−V and generalised coordinates.
拉格朗日方程使用 L 等于 T 减 V，以及广义坐标。
This undergraduate formalism is distinct from school-level force substitution.
这种大学形式的力学与中学阶段的力替换法不同。
Choose an inertial frame and a coordinate x measured from spring equilibrium.
选择一个惯性参考系，并以弹簧平衡位置为原点建立坐标x。
For a conservative spring, the Lagrangian avoids solving constraint forces.
对于保守弹簧，使用拉格朗日量可以避免求解约束力。
The Lagrangian is kinetic energy minus potential energy: L equals one half m x-dot squared, minus one half k x squared.
拉格朗日量等于动能减去势能：L 等于二分之一 m 乘 x 点的平方，减去二分之一 k 乘 x 的平方。
Put L into Lagrange's equation: d by d t of the partial of L with respect to x-dot, minus the partial of L with respect to x, equals zero.
把 L 代入拉格朗日方程：L 对 x 点的偏导数再对时间求导，减去 L 对 x 的偏导数，等于零。
For the spring this gives m x-double-dot plus k x equals zero: simple harmonic motion.
对弹簧而言，得到 m 乘 x 的二阶导数加 k x 等于零，即简谐运动。
With m equal to 2.0 kilograms and k equal to 18 newtons per metre, omega is the square root of k over m: the square root of 9 per second squared, which is 3.0 radians per second.
取 m 等于 2.0 千克、k 等于 18 牛顿每米，角频率 ω 等于 k 除以 m 的平方根，即 9 每二次方秒的平方根，等于每秒 3.0 弧度。
The period is two pi over omega, two pi over 3.0 radians per second, which is about 2.09 seconds.
周期等于 2π 除以 ω，即 2π 除以每秒 3.0 弧度，约为 2.09 秒。
Positive stiffness gives stable motion.
正刚度给出稳定运动。
The units of k/m are inverse seconds squared.
k 除以 m 的单位是秒的负二次方。
Left: a pendulum's angular acceleration against angle.
左图：单摆的角加速度随角度的变化。
The exact curve, minus five sine theta, and the small-angle line, minus five theta, agree near zero and part only at larger angles, which is why the linear model works for small swings.
精确曲线负五倍正弦θ与小角度直线负五倍θ在零附近重合，只在较大角度处分开，所以线性模型适用于小摆幅。
Right: with angular momentum conserved, angular speed falls as one over r squared, from 4 radians per second at 1.5 metres to 1 at 3 metres.
右图：角动量守恒时，角速度与 r 的平方成反比，从 1.5 米处的每秒 4 弧度降到 3 米处的每秒 1 弧度。
A bead of mass m moves without friction along a straight rail.
质量为m的珠子沿无摩擦直导轨运动。
Its potential is V of x equals a x to the fourth, minus b x squared, with a and b positive.
其势能为V(x) = ax^4 − bx^2，其中a、b为正数。
Find every equilibrium, classify stability, and derive the small-oscillation frequency at each stable equilibrium.
求所有平衡位置，判断稳定性，并推导每个稳定平衡点处的小振幅振动频率。
State why the frequency formula cannot describe motion near x equals zero.
说明为何该频率公式无法描述x = 0附近的运动。
Force is F equals negative V prime, which equals negative 4 a x cubed plus 2 b x.
力为 F = -V' = -4ax^3 + 2bx。
Equilibrium requires x equals zero or x squared equals b divided by 2 a.
平衡条件要求 x = 0 或 x^2 = b/(2a)。
Curvature is V double prime equals 12 a x squared minus 2 b. It is negative 2 b at zero and positive 4 b at either outer point.
曲率为 V'' = 12ax^2 - 2b：在零点处为 -2b，在两侧外点处均为 4b。
Zero is unstable; both outer points are stable.
零点不稳定；两个外点均稳定。
Write x equals x sub 0 plus xi and retain terms linear in xi: m xi double dot equals negative 4 b xi, so omega equals the square root of 4 b over m. At zero the displacement grows rather than oscillates.
令 x = x_0 + ξ，并保留关于 ξ 的一次项：mξ̈ = -4bξ，因此 ω = √(4b/m)。
The approximation requires the absolute value of xi small compared with the square root of b divided by 2 a.
在零点处，位移会增长而非振荡。 该近似要求 |ξ| 远小于 √(b/(2a))。
Award equilibrium, curvature, classification, linearisation, frequency and condition: one mark each.
评分标准：平衡、曲率、分类、线性化、频率及条件，各一分。
Conservation of kinetic energy does not hold in every collision, even when momentum is conserved.
动能守恒并非在所有碰撞中都成立，即使动量守恒也是如此。
Lagrangian: Kinetic energy minus potential energy for the standard mechanical system.
拉格朗日量：标准力学系统的动能减去势能。
angular momentum: The moment of linear momentum about a reference point.
角动量：关于参考点的线动量的力矩。
Retrieve the meanings without reading the definitions, then change a parameter in the original transfer and test the limiting case.
在不阅读定义的情况下回忆含义，然后改变原始传递中的一个参数并检验极限情况。