Maximum speed of a particle in simple harmonic motion is `v_(max)`. Then average speed of this particle in one time period is equal to

Options

A. `v_(max)/2`
B. `v_(max)/pi`
C. `(piv_(max))/2`
D. `(2v_(max))/(pi)`

Answer

To find the average speed of a particle in simple harmonic motion (SHM) over one complete time period, we can follow these steps:

### Step 1: Understand the motion of the particle
In SHM, the particle moves back and forth between two extreme positions, which are at a distance equal to the amplitude (A) from the mean position (equilibrium position). The total distance covered by the particle in one complete cycle (one time period) is four times the amplitude.

### Step 2: Calculate the total distance covered in one time period
The particle moves from:
- Mean position to maximum amplitude (A)
- Maximum amplitude back to mean position (A)
- Mean position to maximum compression (-A)
- Maximum compression back to mean position (-A)

Thus, the total distance covered in one complete cycle is:
\[ \text{Total Distance} = A + A + A + A = 4A \]

### Step 3: Determine the time period of the motion
The time period (T) of the SHM is the time taken to complete one full cycle. It is related to the angular frequency (ω) by the formula:
\[ T = \frac{2\pi}{\omega} \]

### Step 4: Calculate the average speed
The average speed (v_avg) can be calculated using the formula:
\[ v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}} \]

Substituting the values we found:
\[ v_{\text{avg}} = \frac{4A}{T} \]

Now, substituting the expression for the time period:
\[ v_{\text{avg}} = \frac{4A}{\frac{2\pi}{\omega}} = \frac{4A \cdot \omega}{2\pi} = \frac{2A \cdot \omega}{\pi} \]

### Step 5: Relate amplitude and maximum speed
The maximum speed (v_max) of the particle in SHM is given by:
\[ v_{\text{max}} = A \cdot \omega \]

### Step 6: Substitute v_max into the average speed formula
Now, we can express the average speed in terms of the maximum speed:
\[ v_{\text{avg}} = \frac{2A \cdot \omega}{\pi} = \frac{2}{\pi} \cdot (A \cdot \omega) = \frac{2}{\pi} \cdot v_{\text{max}} \]

### Final Answer
Thus, the average speed of the particle in one time period is:
\[ v_{\text{avg}} = \frac{2}{\pi} v_{\text{max}} \]