To solve Question 1, we analyze the given problem:

The differential equation modeling the motion of the ball dropped into viscous oil is: $m \frac{d^2 x}{dt^2} = -r \frac{dx}{dt} - mg,$ where:

• $m$: Mass of the ball
• $r$: Constant related to viscosity
• $g$: Gravitational acceleration
• $x(t)$: Displacement of the ball over time $t$.

Part (i): General Solution for Displacement $x(t)$

Rewriting the equation: $\frac{d^2 x}{dt^2} + \frac{r}{m} \frac{dx}{dt} + g = 0.$

1. This is a second-order linear ordinary differential equation with constant coefficients. The characteristic equation is: $\lambda^2 + \frac{r}{m} \lambda = 0.$

2. Solving for $\lambda$, we get: $\lambda_1 = 0, \quad \lambda_2 = -\frac{r}{m}.$

3. The general solution for $x(t)$ is: $x(t) = C_1 + C_2 e^{-\frac{r}{m} t} + \frac{mg}{r}.$

   Here, $\frac{mg}{r}$ accounts for the steady-state solution under constant force $mg$.

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Part (ii): Velocity Approaching Limiting Value $v \to \frac{mg}{r}$

The velocity is the derivative of $x(t)$: $v(t) = \frac{dx}{dt} = -C_2 \frac{r}{m} e^{-\frac{r}{m} t}.$

1. As $t \to \infty$, the exponential term $e^{-\frac{r}{m} t} \to 0$. Therefore: $v(t) \to 0 + \frac{mg}{r}.$

2. The limiting velocity is: $v_{\text{limiting}} = \frac{mg}{r}.$

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If further clarification or additional examples are needed, let me know! Here are some related questions to expand your understanding:

1. How does the solution change if $r = 0$ (no viscosity)?
2. What happens if the initial velocity of the ball is non-zero?
3. How can this equation be solved numerically for specific values of $m, r, g$?
4. What is the significance of the steady-state displacement $x(t)$?
5. How does changing $r/m$ affect the time taken to approach the limiting velocity?

Tip: Always check the behavior of the solution as $t \to \infty$ to verify physical accuracy!