While a particle executes simple harmonic motion, the rate of change of acceleration is maximum and minimum respectively at

Options

A. the mean position and extreme positions
B. the extreme positions and mean position
C. the mean position alternatively
D. the extreme positions alternatively.

Explanation

To solve the problem of determining where the rate of change of acceleration is maximum and minimum during simple harmonic motion (SHM), we can follow these steps:

### Step 1: Understand the Acceleration in SHM
In simple harmonic motion, the acceleration \( A \) of a particle is given by the formula:
\[ A = -\omega^2 x \]
where \( \omega \) is the angular frequency and \( x \) is the displacement from the mean position.

**Hint:** Recall that in SHM, the acceleration is always directed towards the mean position and is proportional to the displacement.

### Step 2: Find the Rate of Change of Acceleration
To find the rate of change of acceleration with respect to time, we differentiate the acceleration with respect to time:
\[ \frac{dA}{dt} = \frac{d}{dt}(-\omega^2 x) \]
Using the chain rule, this becomes:
\[ \frac{dA}{dt} = -\omega^2 \frac{dx}{dt} \]
Since \( \frac{dx}{dt} \) is the velocity \( v \), we can rewrite this as:
\[ \frac{dA}{dt} = -\omega^2 v \]

**Hint:** Remember that velocity is the derivative of displacement with respect to time.

### Step 3: Express Velocity in Terms of Amplitude and Displacement
The velocity \( v \) in SHM can be expressed as:
\[ v = \omega \sqrt{A^2 - x^2} \]
where \( A \) is the amplitude of the motion.

**Hint:** This formula shows how velocity depends on the position \( x \) in SHM.

### Step 4: Substitute Velocity into the Rate of Change of Acceleration
Substituting the expression for \( v \) into the rate of change of acceleration gives:
\[ \frac{dA}{dt} = -\omega^2 \left(\omega \sqrt{A^2 - x^2}\right) = -\omega^3 \sqrt{A^2 - x^2} \]

### Step 5: Analyze the Rate of Change of Acceleration
Now, we need to analyze \( \frac{dA}{dt} \) at two key positions:
- **At the Mean Position (x = 0):**
\[ \frac{dA}{dt} = -\omega^3 \sqrt{A^2 - 0^2} = -\omega^3 A \]
This value is maximum in magnitude.

- **At the Extreme Position (x = A):**
\[ \frac{dA}{dt} = -\omega^3 \sqrt{A^2 - A^2} = 0 \]
This value is minimum (zero).

### Conclusion
Thus, we conclude that:
- The rate of change of acceleration is **maximum** at the **mean position**.
- The rate of change of acceleration is **minimum** at the **extreme position**.

The correct answer is that the rate of change of acceleration is maximum at the mean position and minimum at the extreme position.

**Final Answer:** Maximum at the mean position, minimum at the extreme position.