A 1.2kg mass is oscillating without friction on a spring whose spring constant is 3400N/m. When the mass's displacement is 7.2cm. What is its acceleration?

Options

A. `-3.8m//s^(2)`
B. `-200m//s^(2)`
C. `-240m//s^(2)`
D. `-2.0 xx 10^(4) m//s^(2)`

Answer

To find the acceleration of a mass oscillating on a spring in simple harmonic motion (SHM), we can use the following steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass (m) = 1.2 kg
- Spring constant (k) = 3400 N/m
- Displacement (x) = 7.2 cm = 7.2 × 10^(-2) m

2. **Understand the Formula for Acceleration in SHM:**
- The acceleration (a) of an object in simple harmonic motion can be calculated using the formula:
\[ a = -\frac{k}{m} \cdot x \]
- Here, \( k \) is the spring constant, \( m \) is the mass, and \( x \) is the displacement from the equilibrium position.

3. **Substitute the Values into the Formula:**
- Plugging in the values we have:
\[ a = -\frac{3400 \, \text{N/m}}{1.2 \, \text{kg}} \cdot (7.2 \times 10^{-2} \, \text{m}) \]

4. **Calculate the Value of Acceleration:**
- First, calculate \(-\frac{3400}{1.2}\):
\[ -\frac{3400}{1.2} = -2833.33 \, \text{m/s}^2 \]
- Now multiply this by the displacement:
\[ a = -2833.33 \cdot (7.2 \times 10^{-2}) = -204 \, \text{m/s}^2 \]

5. **Final Result:**
- The acceleration of the mass when it is displaced by 7.2 cm is:
\[ a = -204 \, \text{m/s}^2 \]
- The negative sign indicates that the acceleration is directed towards the equilibrium position.