Câu 23. [VNA] Một vật dao động điều hoà có gia tốc biên đổi theo phương trình a = 5cos(10t +3 )(m/s2). Ở thời điểm ban đầu, vật ở li độA. 5 cm. B. 2,5 cm.

Break Down the Problem

1. Identify the given variables:

   • The equation of acceleration: $a(t) = 5 \cos(10t + 3\pi)$ (in m/s²)
   • Initial displacement: either $A = 5$ cm or $B = 2.5$ cm.

2. Convert the displacements from cm to m:

   • $A = 0.05$ m
   • $B = 0.025$ m

3. Use the relationship between acceleration and displacement in simple harmonic motion.

Relevant Concepts

1. In simple harmonic motion, the acceleration $a$ is related to displacement $x$ by the formula: $a = -\omega^2 x$ where $\omega$ is the angular frequency.

2. From the given $a(t)$, the angular frequency $\omega$ can be identified as $\omega = 10$ rad/s.

Analysis and Detail

1. Using the relationship:

   • Rearranging gives us: $x = -\frac{a}{\omega^2}$

2. Substitute $a = 5 \cos(10t + 3\pi)$ into the equation while at $t = 0$: $a(0) = 5 \cos(3\pi) = 5 \cdot (-1) = -5 \, \text{m/s}^2$

3. Calculating displacement: $x = -\frac{-5}{10^2} = \frac{5}{100} = 0.05 \, \text{m} = 5 \, \text{cm}$

Verify and Summarize

• The calculation shows that when the acceleration is $-5 \, \text{m/s}^2$, the corresponding displacement is indeed $5 \, \text{cm}$.

Final Answer

The initial displacement is $5$ cm.