Comparing Amplitude and Frequency

The general SHM solution $x(t) = A\sin(\omega t)$ has two independent parameters:

• Amplitude $A$: how far it swings (vertical stretch).
• Angular frequency $\omega$: how fast it oscillates (horizontal compression).

Adjusting one does not affect the other. They control completely independent aspects of the wave's appearance.

The Formulas

For $x(t) = A\sin(\omega t)$:

$\text{Amplitude} = |A| \qquad \text{Period} \;T = \dfrac{2\pi}{\omega} \qquad \text{Frequency}\;f = \dfrac{\omega}{2\pi}$

The maximum velocity is $A\omega$, the maximum acceleration is $A\omega^{2}$, and the total energy is $\tfrac{1}{2}m(A\omega)^{2}$.

Step-by-Step Solution

Compare three oscillators:

1. $x_1 = 1\sin(t)$ — amplitude 1, $\omega = 1$
2. $x_2 = 2\sin(t)$ — amplitude 2, $\omega = 1$ (taller)
3. $x_3 = 1\sin(2t)$ — amplitude 1, $\omega = 2$ (faster)

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Step 1 — Compute the period and amplitude of each.

• $x_1$: $A = 1$, $\omega = 1$, $T = 2\pi \approx 6.28$
• $x_2$: $A = 2$, $\omega = 1$, $T = 2\pi$ (same)
• $x_3$: $A = 1$, $\omega = 2$, $T = \pi \approx 3.14$ (half)

Notice that doubling A doubles the height but doesn't change the period, while doubling ω halves the period but doesn't change the height.

Step 2 — Compute max velocities.

$v_{\max} = A\omega$:

• $x_1$: $v_{\max} = 1 \cdot 1 = 1$
• $x_2$: $v_{\max} = 2 \cdot 1 = 2$ (doubled by amplitude)
• $x_3$: $v_{\max} = 1 \cdot 2 = 2$ (doubled by frequency)

Both 2 and 3 have the same max velocity, but for different reasons.

Step 3 — Compute max accelerations.

$a_{\max} = A\omega^{2}$:

• $x_1$: $a_{\max} = 1$
• $x_2$: $a_{\max} = 2$ (doubled, by $A$)
• $x_3$: $a_{\max} = 4$ (quadrupled — $\omega^{2}$ enters squared)

Acceleration is much more sensitive to frequency than amplitude. A high-frequency oscillator can have huge accelerations even with small amplitude — that's why high-frequency vibrations damage parts.

Step 4 — Compute total energies.

For a mass-spring system, $E = \tfrac{1}{2}m\omega^{2}A^{2}$. With $m = 1$:

• $x_1$: $E = 0.5 \cdot 1 \cdot 1 = 0.5$
• $x_2$: $E = 0.5 \cdot 1 \cdot 4 = 2.0$ (4× larger because $A^{2}$)
• $x_3$: $E = 0.5 \cdot 4 \cdot 1 = 2.0$ (also 4× larger because $\omega^{2}$)

Both 2 and 3 have 4× the energy of 1, but for different reasons.

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Answer:

For $x(t) = A\sin(\omega t)$:

• $|A|$ controls the vertical extent of the wave
• $\omega$ controls how fast it oscillates
• $T = 2\pi/\omega$, $f = \omega/(2\pi)$
• $v_{\max} = A\omega$, $a_{\max} = A\omega^{2}$
• $E \propto A^{2}\omega^{2}$

The two parameters act independently on shape, but both affect velocity, acceleration, and energy.

Try It

• Adjust the A and ω sliders separately.
• Watch the wave grow taller (A) or oscillate faster (ω).
• The HUD shows the live values of period, frequency, and max velocity/acceleration.