A particle executes shm in a line 10 cm long. When it passes through the mean position its velocity is `15 cms^(-1)` . Find the period.

Explanation

To solve the problem step by step, we will follow the concepts of Simple Harmonic Motion (SHM).

### Step 1: Understand the problem
We have a particle executing SHM along a line of length 10 cm. The mean position is at the center (5 cm), and the amplitude (A) is half of the total length, which is 5 cm. The maximum velocity (Vmax) at the mean position is given as 15 cm/s.

### Step 2: Identify the amplitude
The amplitude (A) of the motion is the maximum displacement from the mean position. Since the total length is 10 cm, the amplitude is:
\[ A = \frac{10 \text{ cm}}{2} = 5 \text{ cm} \]

### Step 3: Use the formula for maximum velocity
In SHM, the maximum velocity (Vmax) is given by the formula:
\[ V_{max} = A \omega \]
Where:
- \( V_{max} \) = maximum velocity
- \( A \) = amplitude
- \( \omega \) = angular frequency

Given that \( V_{max} = 15 \text{ cm/s} \) and \( A = 5 \text{ cm} \), we can substitute these values into the formula:
\[ 15 \text{ cm/s} = 5 \text{ cm} \cdot \omega \]

### Step 4: Solve for angular frequency (ω)
Rearranging the equation to find \( \omega \):
\[ \omega = \frac{15 \text{ cm/s}}{5 \text{ cm}} = 3 \text{ rad/s} \]

### Step 5: Find the time period (T)
The relationship between angular frequency and time period is given by:
\[ \omega = \frac{2\pi}{T} \]
Rearranging this equation to solve for T:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{3 \text{ rad/s}} \]

### Step 6: Calculate the time period
Now we can calculate the time period:
\[ T \approx \frac{2 \times 3.14}{3} \approx \frac{6.28}{3} \approx 2.09 \text{ seconds} \]

### Final Answer
The time period of the particle executing SHM is approximately \( 2.09 \text{ seconds} \).

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