Passage XI) The differential equation of a particle undergoing SHM is given by `a(d^(2)x)/(dt^(2))`+bx = 0. The particle starts from the extreme position.
The ratio of the maximum acceleration to the maximum velocity of the particle is

Options

A. b/a
B. a/b
C. `sqrt(a/b)`
D. `sqrt(b/a)`

Answer

To solve the problem, we need to analyze the given differential equation of a particle undergoing Simple Harmonic Motion (SHM) and derive the required ratio of maximum acceleration to maximum velocity.

### Step-by-Step Solution:

1. **Understand the Differential Equation**:
The differential equation given is:
\[ a \frac{d^2x}{dt^2} + bx = 0 \]
This represents the motion of a particle in SHM.

2. **Rearranging the Equation**:
We can rearrange the equation to express the acceleration:
\[ a \frac{d^2x}{dt^2} = -bx \]
Dividing both sides by \(a\):
\[ \frac{d^2x}{dt^2} = -\frac{b}{a} x \]

3. **Identifying the Angular Frequency**:
The standard form of SHM is:
\[ \frac{d^2x}{dt^2} = -\omega^2 x \]
By comparing both equations, we can identify:
\[ \omega^2 = \frac{b}{a} \]
Therefore, the angular frequency \(\omega\) is:
\[ \omega = \sqrt{\frac{b}{a}} \]

4. **Maximum Acceleration and Maximum Velocity**:
- The maximum acceleration \(A_{max}\) in SHM is given by:
\[ A_{max} = \omega^2 A \]
- The maximum velocity \(V_{max}\) in SHM is given by:
\[ V_{max} = \omega A \]

5. **Finding the Ratio**:
We need to find the ratio of maximum acceleration to maximum velocity:
\[ \frac{A_{max}}{V_{max}} = \frac{\omega^2 A}{\omega A} \]
Simplifying this gives:
\[ \frac{A_{max}}{V_{max}} = \frac{\omega^2}{\omega} = \omega \]

6. **Substituting the Value of \(\omega\)**:
From our earlier calculation, we have:
\[ \omega = \sqrt{\frac{b}{a}} \]
Thus, the ratio of maximum acceleration to maximum velocity is:
\[ \frac{A_{max}}{V_{max}} = \sqrt{\frac{b}{a}} \]

### Final Answer:
The ratio of the maximum acceleration to the maximum velocity of the particle is:
\[ \sqrt{\frac{b}{a}} \]