A 10 kg metal block is attached to a spring constant `1000 Nm^(-1)`. A block is displaced from equilibrium position by 10 cm and released. The maximum acceleration of the block is

Options

A. `10 ms^(-2)`
B. `100 ms^(-2)`
C. `200 ms^(-2)`
D. `0.1 ms^(-2)`

Explanation

To solve the problem step by step, we will follow the principles of simple harmonic motion (SHM) and use the given parameters.

### Step 1: Identify the parameters
We are given:
- Mass of the block, \( m = 10 \, \text{kg} \)
- Spring constant, \( k = 1000 \, \text{N/m} \)
- Displacement from equilibrium (amplitude), \( A = 10 \, \text{cm} = 0.1 \, \text{m} \)

### Step 2: Calculate the angular frequency (\( \omega \))
The angular frequency for a mass-spring system is given by the formula:
\[ \omega = \sqrt{\frac{k}{m}} \]
Substituting the values:
\[ \omega = \sqrt{\frac{1000 \, \text{N/m}}{10 \, \text{kg}}} = \sqrt{100} = 10 \, \text{rad/s} \]

### Step 3: Calculate the maximum acceleration (\( A_{\text{max}} \))
The maximum acceleration in simple harmonic motion is given by:
\[ A_{\text{max}} = \omega^2 A \]
Substituting the values we have:
\[ A_{\text{max}} = (10 \, \text{rad/s})^2 \times 0.1 \, \text{m} = 100 \times 0.1 = 10 \, \text{m/s}^2 \]

### Step 4: Conclusion
The maximum acceleration of the block is:
\[ \boxed{10 \, \text{m/s}^2} \]

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