| English | 意味 |
|---|---|
| period/ˈpɪərɪəd/ | 周期 |
| frequency/ˈfriːkwənsi/ | 周波数 |
| angular frequency/ˈæŋɡjʊlə ˈfriːkwənsi/ | 角周波数 |
Counting the beats
- A metronome ticks at a steady rate you can count.
- A heavy mass on a soft spring bobs slowly; a stiff spring snaps back quickly.
- Each oscillator has its own natural rhythm.
- Two simple formulas predict exactly how fast each one repeats.
Period, frequency, and omega
- The period 周期 $T$ is the time for one full cycle.
- The frequency 频率 $f = 1/T$ counts cycles per second.
- The angular frequency 角频率 is $\omega = 2\pi f = \dfrac{2\pi}{T}$.
An oscillator has a period of $0.25\ \text{s}$. Its frequency (in Hz)?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$f = 1/T = 1/0.25 = 4\ \text{Hz}$.
A mass on a spring
- A mass $m$ on a spring of stiffness $k$ has period:
- A heavier mass swings slower (longer $T$); a stiffer spring is faster.
- It follows from $\omega = \sqrt{k/m}$.
You replace a mass-spring oscillator's mass with a heavier one. The period...
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
$T = 2\pi\sqrt{m/k}$, so a bigger $m$ gives a longer $T$.
If you quadruple the mass on a spring, the period becomes...
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
$T \propto \sqrt{m}$, so $\sqrt{4} = 2$ times the period.
A swinging pendulum
- A pendulum of length $L$ has period:
- A longer pendulum swings slower; stronger gravity is faster.
- Notice the mass of the bob does not appear.
What changes the period?
Sort each change by whether it changes an oscillator's period.
Making a pendulum's bob heavier (same length) changes its period.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
$T = 2\pi\sqrt{L/g}$ has no mass term -- the bob's mass does not matter.
A pendulum has $L = 1.0\ \text{m}$ and $g = 9.8\ \text{m/s}^2$. Its period (in s, to 1 decimal)?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
$T = 2\pi\sqrt{L/g} = 2\pi\sqrt{1/9.8} \approx 2.0\ \text{s}$.
A $0.2\ \text{kg}$ mass on a $k = 50\ \text{N/m}$ spring.
- Period: $T = 2\pi\sqrt{m/k} = 2\pi\sqrt{0.2/50} = 2\pi\sqrt{0.004} \approx 0.40\ \text{s}$.
- Frequency: $f = 1/T \approx 2.5\ \text{Hz}$.
For ideal SHM, a larger amplitude gives a longer period.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The period is independent of amplitude -- big and small swings take the same time.
A remarkable fact -- for ideal SHM the period does not depend on amplitude, so a big swing and a small swing take the same time. The pendulum formula holds only for small angles; the spring period ignores gravity and the pendulum period ignores the bob's mass. And mind the square root: quadrupling the mass only doubles the period.
The period $T$ (one cycle), frequency $f = 1/T$, and angular frequency $\omega = 2\pi/T$ describe the rhythm. A spring gives $T = 2\pi\sqrt{m/k}$; a pendulum gives $T = 2\pi\sqrt{L/g}$. The period is independent of amplitude, and each formula ignores one thing -- the spring ignores gravity, the pendulum ignores the bob's mass.