Representing and Analyzing SHM
Introduced| English | Meaning |
|---|---|
| amplitude/ˈæmplɪtjuːd/ | The maximum displacement from an equilibrium position. |
| phase/feɪz/ | phase |
Drawing the wobble
- Trace a bobbing mass over time and you get a smooth wave on the graph.
- The same S-curve describes springs, pendulums, and sound.
- Its height, speed, and steepness all follow from one equation.
- A single cosine captures the whole back-and-forth.
The position over time
- The displacement in SHM traces a cosine:
- $A$ is the amplitude 振幅, the maximum displacement.
- $\omega$ is the angular frequency and $\varphi$ the phase 相位, the starting point.
Displacement varies sinusoidally with time in simple harmonic motion
In $x = A\cos(\omega t + \varphi)$, the symbol A stands for the ____.
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$A$ is the amplitude -- the largest displacement from equilibrium.
Speed and acceleration follow
- Differentiate to get the velocity and acceleration:
- Maximum speed $v_{max} = A\omega$; maximum acceleration $a_{max} = A\omega^2$.
An oscillator has $A = 0.2\ \text{m}$ and $\omega = 4\ \text{rad/s}$. Its maximum speed (in m/s)?
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$v_{max} = A\omega = 0.2 \times 4 = 0.8\ \text{m/s}$.
For the same oscillator ($A = 0.2\ \text{m}$, $\omega = 4\ \text{rad/s}$), the maximum acceleration (in m/s²)?
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$a_{max} = A\omega^2 = 0.2 \times 16 = 3.2\ \text{m/s}^2$.
Opposite extremes
- Speed is greatest at equilibrium (where $x = 0$) and zero at the ends.
- Acceleration is greatest at the ends and zero at equilibrium.
- So $x$ and $a$ are perfectly out of step, while $v$ runs a quarter-cycle ahead.
SHM as a sine wave
The displacement of a simple harmonic oscillator traces a cosine curve in time.
Where in the motion is the speed greatest?
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Speed peaks where $x = 0$ (the middle) and is zero at the ends.
In SHM, displacement and acceleration are exactly out of step (opposite phase).
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$a = -\omega^2 x$, so $a$ is always opposite to $x$.
An oscillator has $A = 0.1\ \text{m}$ and $\omega = 5\ \text{rad/s}$.
- Max speed: $v_{max} = A\omega = 0.1 \times 5 = 0.5\ \text{m/s}$, reached at the middle.
- Max acceleration: $a_{max} = A\omega^2 = 0.1 \times 25 = 2.5\ \text{m/s}^2$, reached at the ends.
A mass released from rest at the far end is best described using...
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At $t = 0$ the displacement is maximum, which cosine gives with $\varphi = 0$.
Do not swap where speed and acceleration peak — they sit at opposite places: speed is largest in the middle, acceleration largest at the extremes. If the motion starts at the far end, use cosine ($\varphi = 0$); if it starts at equilibrium moving outward, use sine. And $\omega$ here is in rad/s, not the frequency $f$.
SHM is one cosine: $x = A\cos(\omega t + \varphi)$, with amplitude $A$ and phase $\varphi$. Its derivatives give $v_{max} = A\omega$ (at equilibrium) and $a_{max} = A\omega^2$ (at the extremes). Displacement and acceleration are exactly out of step, and velocity leads position by a quarter cycle.
Write from a maximum speed of 1 m/s
- Amplitude is 0.50 m and angular frequency is 2 rad/s.
- Find the maximum speed.
An oscillator has amplitude 0.50 m and angular frequency 2 rad/s. What is its maximum speed, and where in the motion does that speed occur?
0.50 times 2 is 1.
What is the maximum speed?
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Amplitude times angular frequency is 0.50 times 2.