A tiny mass performs S.H.M along a straight line with a time period of T=0.60sec and amplitude A=10.0cm. Calculate the mean velocity (in m/sec) in the time to displace by `A/2`.

Explanation

To solve the problem of calculating the mean velocity of a tiny mass performing Simple Harmonic Motion (SHM) as it displaces by \( A/2 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Time period, \( T = 0.60 \) sec
- Amplitude, \( A = 10.0 \) cm

2. **Convert Amplitude to Meters:**
- Since we need the final answer in meters per second, convert the amplitude from centimeters to meters:
\[ A = 10.0 \, \text{cm} = 0.10 \, \text{m} \]

3. **Determine the Time to Displace by \( A/2 \):**
- The displacement \( A/2 = 0.10 \, \text{m} / 2 = 0.05 \, \text{m} \).
- From the properties of SHM, the time taken to go from the mean position to \( A/2 \) is \( \frac{T}{12} \).
- Calculate \( \frac{T}{12} \):
\[ \text{Time} = \frac{0.60 \, \text{sec}}{12} = 0.05 \, \text{sec} \]

4. **Calculate the Average Velocity:**
- The average velocity \( v_{avg} \) is defined as the total displacement divided by the total time taken:
\[ v_{avg} = \frac{\text{Displacement}}{\text{Time}} = \frac{A/2}{T/12} \]
- Substitute the values:
\[ v_{avg} = \frac{0.05 \, \text{m}}{0.05 \, \text{sec}} = 1 \, \text{m/sec} \]

5. **Final Answer:**
- The mean velocity in the time to displace by \( A/2 \) is:
\[ v_{avg} = 1 \, \text{m/sec} \]