To solve the problem step by step, we will analyze the given information and calculate the required quantities.
### Given Data:
- Natural length of the spring, \( L_0 = 40 \, \text{cm} = 0.4 \, \text{m} \)
- Spring constant, \( k = 500 \, \text{N/m} \)
- Mass of the block, \( m = 1 \, \text{kg} \)
- Length of the spring when the block is released, \( L = 45 \, \text{cm} = 0.45 \, \text{m} \)
### Step 1: Find the equilibrium position of the spring.
At equilibrium, the force exerted by the spring equals the weight of the block:
\[
kx_0 = mg
\]
Where:
- \( x_0 \) is the extension of the spring from its natural length.
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity).
Rearranging gives:
\[
x_0 = \frac{mg}{k}
\]
Substituting the values:
\[
x_0 = \frac{1 \times 9.81}{500} = 0.01962 \, \text{m} = 1.962 \, \text{cm}
\]
### Step 2: Determine the equilibrium length of the spring.
The equilibrium length of the spring can be calculated as:
\[
L_e = L_0 + x_0 = 40 \, \text{cm} + 1.962 \, \text{cm} = 41.962 \, \text{cm}
\]
### Step 3: Calculate the amplitude of the motion.
The amplitude \( A \) is the maximum displacement from the equilibrium position. The block is released from \( 45 \, \text{cm} \):
\[
A = L - L_e = 45 \, \text{cm} - 41.962 \, \text{cm} = 3.038 \, \text{cm} \approx 3 \, \text{cm}
\]
### Step 4: Calculate the maximum velocity.
The maximum velocity \( V_{\text{max}} \) in simple harmonic motion is given by:
\[
V_{\text{max}} = \omega A
\]
Where:
\[
\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{500}{1}} = 22.36 \, \text{rad/s}
\]
Thus,
\[
V_{\text{max}} = 22.36 \times 0.03 \approx 0.6708 \, \text{m/s} = 67.08 \, \text{cm/s}
\]
### Step 5: Calculate the maximum acceleration.
The maximum acceleration \( a_{\text{max}} \) is given by:
\[
a_{\text{max}} = \omega^2 A
\]
Calculating:
\[
a_{\text{max}} = (22.36)^2 \times 0.03 \approx 15 \, \text{m/s}^2
\]
### Step 6: Determine the minimum elastic potential energy.
The minimum elastic potential energy occurs when the spring is at its natural length:
\[
U = \frac{1}{2} k x^2
\]
At the natural length, \( x = 0 \):
\[
U_{\text{min}} = \frac{1}{2} \times 500 \times 0^2 = 0
\]
### Summary of Results:
1. Amplitude \( A \approx 3 \, \text{cm} \)
2. Maximum velocity \( V_{\text{max}} \approx 67.08 \, \text{cm/s} \)
3. Maximum acceleration \( a_{\text{max}} \approx 15 \, \text{m/s}^2 \)
4. Minimum elastic potential energy \( U_{\text{min}} = 0 \)
### Conclusion:
- The correct statements are:
- The block will perform SHM of amplitude approximately \( 3 \, \text{cm} \).
- The block will have maximum velocity approximately \( 67.08 \, \text{cm/s} \).
- The block will have maximum acceleration approximately \( 15 \, \text{m/s}^2 \).
- The minimum elastic potential energy of the spring will be \( 0 \).
