To solve this problem, we use the principle of conservation of energy for a spring-mass system undergoing simple harmonic motion.

Energy conservation equation:

$E = \frac{1}{2} k A^2 = \frac{1}{2} k x^2 + \frac{1}{2} m v^2$

Where:

• $E$ is the total mechanical energy,
• $k = 0.880 \, \text{N/m}$ is the spring constant,
• $A = 1.90 \, \text{m}$ is the amplitude of oscillation,
• $x = 1.85 \, \text{m}$ is the position from equilibrium,
• $m = 2.30 \, \text{kg}$ is the mass of the block,
• $v$ is the velocity we need to find.

Step 1: Total energy

$E = \frac{1}{2} k A^2$ Substitute the values: $E = \frac{1}{2} (0.880) (1.90)^2 = 1.5868 \, \text{J}$

Step 2: Potential energy at $x = 1.85 \, \text{m}$

$U = \frac{1}{2} k x^2$ Substitute the values: $U = \frac{1}{2} (0.880) (1.85)^2 = 1.5118 \, \text{J}$

Step 3: Kinetic energy

The kinetic energy is the remaining part of the total energy: $K = E - U$ $K = 1.5868 - 1.5118 = 0.075 \, \text{J}$

Step 4: Velocity

Kinetic energy is given by: $K = \frac{1}{2} m v^2$ Solve for $v$: $v = \sqrt{\frac{2K}{m}}$ Substitute the values: $v = \sqrt{\frac{2(0.075)}{2.30}} = \sqrt{0.0652} = 0.268 \, \text{m/s}$

Final Answer:

The block's speed is: $\boxed{2.68 \times 10^{-1} \, \text{m/s}}$

Let me know if you need further explanation or assistance!

Related Questions:

1. How does the spring constant affect the maximum speed of the block?
2. What happens to the velocity when the block is at the equilibrium position?
3. What is the total energy if the amplitude changes to 2.5 m?
4. How would the speed change if the mass of the block were doubled?
5. What is the significance of simple harmonic motion in this scenario?

Tip:

Always use energy conservation in problems involving oscillations; it simplifies velocity calculations without requiring time-dependent equations.