Câu 10. [VNA] Một vật dao động điều hoà với phương trình x = 4cos10t (cm). Tốc độ dao động cựcđại của chất điểm làA. 0,4 m/s. B. 0,04 m/s. C. 4 m/s. D. 40 m/s

1. Break Down the Problem

We need to find the maximum speed of the harmonic oscillator described by the equation $x = 4 \cos(10t)$ (in cm).

2. Relevant Concepts

The velocity $v$ of the harmonic oscillation can be derived from the position function by differentiating it with respect to time $t$: $v(t) = \frac{dx}{dt} = -A \omega \sin(\omega t)$ where:

• $A$ is the amplitude of the oscillation,
• $\omega$ is the angular frequency.

From the given equation:

• Amplitude $A = 4$ cm,
• Angular frequency $\omega = 10$ rad/s.

3. Analysis and Detail

To find the maximum speed, we need the maximum value of $v(t)$, which occurs when $\sin(\omega t)$ equals 1: $v_{\text{max}} = A \omega$

Substituting the values: $v_{\text{max}} = 4 \text{ cm} \times 10 \text{ rad/s} = 40 \text{ cm/s}$

4. Verify and Summarize

Convert $40 \text{ cm/s}$ to $\text{m/s}$: $40 \text{ cm/s} = 0.4 \text{ m/s}$

Final Answer

The maximum speed of the harmonic oscillator is $\text{A. } 0.4 \text{ m/s}$.