Solution: 2D Oscillation with Constant Force
1. Equation of Motion
The particle is acted upon by an elastic force \( \vec{F}_{el} = -k\vec{r} \) (since relaxed length is negligible) and a constant external force \( \vec{F}_{ext} = F_0 \hat{i} \).
$$ m\vec{a} = -k\vec{r} + F_0 \hat{i} $$Separating into components:
$$ m\ddot{x} = -kx + F_0 \quad , \quad m\ddot{y} = -ky $$
2. Equilibrium Position (Center of Oscillation)
The equilibrium position is where the net force is zero.
- \( -kx_{eq} + F_0 = 0 \implies x_{eq} = F_0/k \)
- \( -ky_{eq} = 0 \implies y_{eq} = 0 \)
Center of Oscillation: \( C \left( \frac{F_0}{k}, 0 \right) \).
3. Motion Analysis
The motion is Simple Harmonic Motion (SHM) about the center \( C \). Since the particle is released from rest at \( P(x_0, y_0) \), the motion occurs along the straight line connecting \( P \) and \( C \).
The other extreme position \( P’ \) is symmetric to \( P \) with respect to \( C \).
$$ \frac{P + P’}{2} = C \implies P’ = 2C – P $$ $$ x’ = 2(F_0/k) – x_0 \quad , \quad y’ = 2(0) – y_0 = -y_0 $$
4. Period
The restoring force coefficient is \( k \). The time period depends only on mass and spring constant.
Oscillates between \( (x_0, y_0) \) and \( \left( \frac{2F_0}{k} – x_0, -y_0 \right) \).
Period \( T = 2\pi\sqrt{\frac{m}{k}} \).
Period \( T = 2\pi\sqrt{\frac{m}{k}} \).