To solve the problem, we need to analyze the motion of the particle in the conservative force field defined by the potential energy function \( U(y) = 2y^2 - 30y + 450 \).
### Step 1: Calculate the Potential Energy at \( y = 8 \, \text{m} \)
We start by substituting \( y = 8 \) into the potential energy equation:
\[
U(8) = 2(8^2) - 30(8) + 450
\]
Calculating each term:
\[
U(8) = 2(64) - 240 + 450
\]
\[
U(8) = 128 - 240 + 450
\]
\[
U(8) = 128 + 210 = 338 \, \text{J}
\]
### Step 2: Determine the Force Acting on the Particle
The force \( F \) associated with the potential energy is given by the negative gradient of the potential energy:
\[
F(y) = -\frac{dU}{dy}
\]
Calculating the derivative:
\[
\frac{dU}{dy} = \frac{d}{dy}(2y^2 - 30y + 450) = 4y - 30
\]
Thus, the force is:
\[
F(y) = - (4y - 30) = -4y + 30
\]
### Step 3: Calculate the Force at \( y = 8 \, \text{m} \)
Substituting \( y = 8 \) into the force equation:
\[
F(8) = -4(8) + 30
\]
\[
F(8) = -32 + 30 = -2 \, \text{N}
\]
### Step 4: Analyze the Motion of the Particle
The particle has a mass \( m = 3 \, \text{kg} \) and is initially moving with a velocity of \( 1 \, \text{m/s} \) towards the positive \( y \)-axis. We can find the acceleration using Newton's second law:
\[
F = ma \implies a = \frac{F}{m} = \frac{-2}{3} \approx -0.67 \, \text{m/s}^2
\]
### Step 5: Determine the Equilibrium Position
The equilibrium position occurs when the force is zero:
\[
-4y + 30 = 0 \implies 4y = 30 \implies y = 7.5 \, \text{m}
\]
### Step 6: Calculate the Amplitude of Oscillation
The particle is oscillating about the equilibrium position \( y = 7.5 \, \text{m} \). The amplitude \( A \) can be determined from the initial position \( y = 8 \):
\[
A = |y_{\text{initial}} - y_{\text{equilibrium}}| = |8 - 7.5| = 0.5 \, \text{m}
\]
### Step 7: Calculate the Angular Frequency \( \omega \)
Using the formula \( \omega^2 = \frac{k}{m} \), where \( k \) is the effective spring constant, we can find \( k \) from the force equation:
\[
k = 4 \, \text{N/m}
\]
Thus,
\[
\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{4}{3}} \approx 1.155 \, \text{rad/s}
\]
### Step 8: Calculate the Time Period \( T \)
The time period \( T \) is given by:
\[
T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\frac{4}{3}}} = 2\pi \sqrt{\frac{3}{4}} = \frac{3\pi}{2} \, \text{s}
\]
### Step 9: Maximum Speed of the Particle
The maximum speed \( v_{\text{max}} \) of the particle can be calculated as:
\[
v_{\text{max}} = \omega A = \sqrt{\frac{4}{3}} \cdot 0.5 \approx 0.577 \, \text{m/s}
\]
### Summary of Results
- Potential Energy at \( y = 8 \, \text{m} \): \( 338 \, \text{J} \)
- Force at \( y = 8 \, \text{m} \): \( -2 \, \text{N} \)
- Equilibrium Position: \( 7.5 \, \text{m} \)
- Amplitude of Oscillation: \( 0.5 \, \text{m} \)
- Angular Frequency: \( \sqrt{\frac{4}{3}} \, \text{rad/s} \)
- Time Period: \( \frac{3\pi}{2} \, \text{s} \)
- Maximum Speed: \( 0.577 \, \text{m/s} \)
