Period and frequency of SHM oscillators
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Flip to reveal answersDefine the period T of an oscillation.
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Question
Define the period T of an oscillation.
Answer
The **time for one complete oscillation** (one full cycle), measured in seconds.
Question
Define the frequency f of an oscillation.
Answer
The **number of oscillations per second**, measured in hertz (Hz). f = 1 ÷ T.
Question
How are period and frequency related?
Answer
$f = \dfrac{1}{T}$ and $T = \dfrac{1}{f}$ — they are reciprocals. **Given** in the data booklet (T = 1/f).
Question
Period of a mass-spring oscillator?
Answer
$T = 2\pi\sqrt{\dfrac{m}{k}}$ — depends on the mass m and spring constant k. **Given**.
Question
Period of a simple pendulum?
Answer
$T = 2\pi\sqrt{\dfrac{l}{g}}$ — depends on the length l and gravity g. **Given**.
Question
Does the bob's mass affect a pendulum's period?
Answer
**No** — mass does not appear in T = 2π√(l/g), so the period is unchanged.
Question
Does gravity affect a mass-spring's period?
Answer
**No** — g does not appear in T = 2π√(m/k); only the mass and stiffness matter.
Question
A pendulum's length is made 4× longer. New period?
Answer
**×√4 = ×2** — the period doubles, because T ∝ √l.
Question
A spring's stiffness k is doubled. New period?
Answer
**×1/√2 ≈ 0.71** — a stiffer spring oscillates faster, so a shorter period (T ∝ 1/√k).
Question
What is angular frequency ω, and its link to f and T?
Answer
How fast the cycle turns (2π radians per cycle): **ω = 2πf = 2π ÷ T**. Unit: rad s⁻¹.
Question
Why does a factor inside the root only change the period by its square root?
Answer
Both period formulas have a √, so a quantity ×4 inside the root comes out as ×√4 = ×2.
Question
What does the spring constant k describe?
Answer
The spring's **stiffness** — a bigger k means a stiffer spring that pulls back harder and oscillates faster.
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Topic 3.1 hub
Simple harmonic motion
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