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IB Diploma Programme · IB Diploma · Physics · HL נושא 11 4:16 קריאת קול באנגלית · תרגום אנגלי + סינית שרוף בתוך הסרטון
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A swing gains a large amplitude when pushes arrive with the right timing.
当推力时机恰当时，秋千会获得较大的振幅。
The forcing frequency and damping 阻尼 help determine the response.
驱动力频率和阻尼共同决定了系统的响应。
Simple harmonic motion has acceleration proportional to displacement and directed toward equilibrium.
简谐运动的加速度与位移成正比，并指向平衡位置。
Period is time per cycle; frequency is its reciprocal.
周期是每次全振动所用的时间；频率是周期的倒数。
Damping transfers energy away from the oscillating system.
阻尼把能量从振动系统中转移出去。
Resonance is a large response to periodic forcing near a natural frequency.
共振是在接近固有频率的周期性驱动下产生的大幅响应。
Damping is energy transfer out of an oscillating system.
阻尼是能量从振动系统中转移出去的过程。
Velocity is greatest near equilibrium for ideal simple harmonic motion, while acceleration magnitude is greatest at extreme displacement.
对于理想简谐运动，速度在平衡位置附近最大，而加速度大小在最大位移处最大。
Resonance can occur near the natural frequency under periodic driving, with amplitude limited by damping.
在周期性驱动力作用下，系统可能在固有频率附近发生共振，其振幅受阻尼限制。
In the lab: Measure time for several complete oscillations and divide.
在实验室中，测量多次完整振荡的时间并求平均。
Define a cycle consistently and use a small displacement when the model requires it.
始终如一地定义一个周期，并在模型要求时使用小位移。
Keep pendulum paths clear and record damping effects rather than assuming perfect motion.
保持单摆路径畅通无阻，记录阻尼效应，而非假设理想运动。
Known: 20 cycles take 32 seconds.
已知：20 个周期耗时 32 秒。
T equals total time divided by cycles, which is 32 divided by 20, equaling 1.6 seconds.
T 等于总时间除以周期数，即 32 除以 20，等于 1.6 秒。
Frequency equals 1 divided by T, which is 1 divided by 1.6, equaling 0.625 Hertz.
频率等于 1 除以 T，即 1 除以 1.6，等于 0.625 赫兹。
A shorter total timing interval would make reaction-time error a larger fraction of the measurement.
更短的总计时区间会使反应时间误差在测量中所占比例更大。
Your turn: 30 cycles take 45 s.
轮到你了：30 个周期耗时 45 秒。
Find period.
求周期。
Pause the video and work it out.
请暂停视频，自己算一算。
Check: 45 seconds divided by 30 cycles gives a period of 1.5 seconds.
核对：45 秒 ÷ 30 个周期，周期为 1.5 秒。
Not every repeated motion is simple harmonic motion.
并非所有重复运动都是简谐运动。
A force proportional to displacement but directed away from equilibrium does not produce simple harmonic motion.
与位移成正比但方向背离平衡位置的力不会产生简谐运动。
Not true: Every periodic motion satisfies the condition for simple harmonic motion.
这种说法不正确。 并非所有周期性运动都满足简谐运动的条件。
SHM needs a restoring force proportional to displacement and directed TOWARD equilibrium.
简谐运动需要一个与位移成正比且指向平衡位置的回复力。
Motion that merely repeats - a bouncing ball, a circular orbit - is periodic without satisfying that condition.
仅仅重复的运动，例如弹跳的球或圆周轨道，虽然是周期性的，但并不满足该条件。
Simple harmonic motion has acceleration proportional to displacement and directed toward equilibrium.
简谐运动的加速度与位移成正比，且方向指向平衡位置。
Velocity is greatest near equilibrium for ideal SHM, while acceleration magnitude is greatest at extreme displacement.
对于理想简谐运动，速度在平衡位置附近最大，而加速度大小在最大位移处最大。
Watch for: not every repeated motion is simple harmonic motion.
请注意：并非所有重复运动都是简谐运动。
Forty cycles take sixteen seconds, so one cycle takes sixteen divided by forty: the period is 0.40 seconds.
40个周期耗时16秒，所以一个周期为16除以40：周期是0.40秒。
The frequency is one over the period, 2.5 hertz, two and a half cycles every second.
频率是周期的倒数，为2.5赫兹，即每秒两个半周期。
Not every periodic motion is simple harmonic.
并非所有周期性运动都是简谐运动。
The displacement repeats every period.
位移每经过一个周期重复一次。
Count complete cycles, not crossings: there are two zero crossings in each cycle, and the frequency is one over the period.
数完整的周期，而不是过零点：每个周期有两次经过零点，频率等于周期的倒数。
Twenty cycles take 32.0 plus or minus 0.4 seconds.
20个周期耗时32.0±0.4秒。
Find the period and the maximum absolute uncertainty, treating the cycle count as exact.
求周期和最大绝对不确定度，将周期数视为精确值。
Why time several cycles?
为什么要测量多个周期的时间？
T equals 1.60 plus or minus 0.02 seconds, relative uncertainty 1.25 percent.
T等于1.60正负0.02秒，相对不确定度为1.25%。
A single cycle with the same plus or minus 0.4 second timing uncertainty has much larger relative uncertainty.
若单次周期的计时不确定度同为正负0.4秒，则其相对不确定度要大得多。
Define the cycle consistently.
请统一定义周期。
Pause and check yourself. Explain each point on screen in your own words, then change one number in the last example and say which assumption must still hold.
暂停一下，自我检查：用自己的话解释屏幕上的每个要点，再改动最后一个例子中的一个数字，说明哪个假设仍然必须成立。