The average acceleration of a particle performing SHM over one complete oscillation is

Options

A. `(omega^(2)A)/(2)`
B. `(omega^(2)A)/(sqrt(2))`
C. zero
D. `A omega^(2)`

Explanation

To find the average acceleration of a particle performing Simple Harmonic Motion (SHM) over one complete oscillation, we can follow these steps:

### Step 1: Understand the Equation of Motion for SHM
The position of a particle in SHM can be described by the equation:
\[ x(t) = A \sin(\omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

### Step 2: Derive the Acceleration
The acceleration \( a(t) \) of the particle can be derived from the position function:
\[ a(t) = -\omega^2 x(t) \]
Substituting the position equation into the acceleration equation gives:
\[ a(t) = -\omega^2 A \sin(\omega t + \phi) \]

### Step 3: Calculate the Average Acceleration Over One Complete Cycle
The average acceleration \( \bar{a} \) over one complete oscillation (from \( t = 0 \) to \( t = T \), where \( T \) is the time period) is given by:
\[ \bar{a} = \frac{1}{T} \int_0^T a(t) \, dt \]
Substituting the expression for acceleration:
\[ \bar{a} = \frac{1}{T} \int_0^T -\omega^2 A \sin(\omega t + \phi) \, dt \]

### Step 4: Evaluate the Integral
The integral of \( \sin(\omega t + \phi) \) over one complete cycle is zero because the positive area above the time axis cancels out the negative area below it. Therefore:
\[ \int_0^T \sin(\omega t + \phi) \, dt = 0 \]
This leads to:
\[ \bar{a} = -\frac{\omega^2 A}{T} \cdot 0 = 0 \]

### Step 5: Conclusion
Thus, the average acceleration of a particle performing SHM over one complete oscillation is:
\[ \bar{a} = 0 \]

### Summary
The average acceleration of a particle in SHM over one complete oscillation is zero because the positive and negative accelerations cancel each other out over one complete cycle.

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