To solve the problem step by step, we will follow these steps:
### Step 1: Calculate the Spring Constant (k)
Given that the spring stretches 20 cm (0.2 m) when a mass \( m_1 = 40 \, \text{g} = 0.04 \, \text{kg} \) is hung from it, we can use Hooke's Law:
\[
F = kx
\]
Where:
- \( F = mg = 0.04 \, \text{kg} \times 10 \, \text{m/s}^2 = 0.4 \, \text{N} \)
- \( x = 0.2 \, \text{m} \)
Rearranging gives:
\[
k = \frac{F}{x} = \frac{0.4 \, \text{N}}{0.2 \, \text{m}} = 2 \, \text{N/m}
\]
### Step 2: Calculate the Angular Frequency (\( \omega \))
Now we need to find the angular frequency for the new mass \( m_2 = \frac{205}{32} \, \text{g} = \frac{205}{32000} \, \text{kg} \).
Convert \( m_2 \) to kg:
\[
m_2 = \frac{205}{32} \, \text{g} = \frac{205}{32 \times 1000} \, \text{kg} = \frac{205}{32000} \, \text{kg}
\]
Now, we can find \( \omega \):
\[
\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{2}{\frac{205}{32000}}}
\]
Calculating \( \omega \):
\[
\omega = \sqrt{\frac{2 \times 32000}{205}} = \sqrt{\frac{64000}{205}} \approx 16 \, \text{rad/s}
\]
### Step 3: Write the Equation of Motion
The equation of motion for simple harmonic motion (SHM) is given by:
\[
x(t) = x_0 \cos(\omega t)
\]
Where \( x_0 = 0.2 \, \text{m} \) and \( \omega \approx 16 \, \text{rad/s} \):
\[
x(t) = 0.2 \cos(16t)
\]
### Step 4: Calculate the Acceleration
The acceleration \( a(t) \) can be found using:
\[
a(t) = -\omega^2 x_0 \cos(\omega t)
\]
Substituting the values:
\[
a(t) = -16^2 \times 0.2 \cos(16t) = -256 \times 0.2 \cos(16t) = -51.2 \cos(16t) \, \text{m/s}^2
\]
### Step 5: Find Specific Values
To find the acceleration when \( x = 2.5 \, \text{cm} = 0.025 \, \text{m} \):
Set \( x(t) = 0.025 \):
\[
0.025 = 0.2 \cos(16t)
\]
Solving for \( \cos(16t) \):
\[
\cos(16t) = \frac{0.025}{0.2} = 0.125
\]
Now, we can find \( 16t \):
\[
16t = \cos^{-1}(0.125)
\]
### Step 6: Final Calculation of Acceleration
Substituting back into the acceleration equation:
\[
a(t) = -51.2 \cos(16t)
\]
Using \( \cos(16t) = 0.125 \):
\[
a(t) = -51.2 \times 0.125 = -6.4 \, \text{m/s}^2
\]
### Summary of Results
- Spring constant \( k = 2 \, \text{N/m} \)
- Angular frequency \( \omega \approx 16 \, \text{rad/s} \)
- Acceleration when \( x = 2.5 \, \text{cm} \) is approximately \( -6.4 \, \text{m/s}^2 \)
