A particle is executing a linear S.H.M. Its velocity at a distance x from the mean position is given by `v^2=144-9x^2`. The maximum velocity of the particle is

Options

A. 12 unit
B. 18 unit
C. 24 unit
D. 36 unit

Answer

To solve the problem, we need to find the maximum velocity of a particle executing simple harmonic motion (SHM), given the equation for its velocity at a distance \( x \) from the mean position:

\[ v^2 = 144 - 9x^2 \]

### Step-by-step Solution:

1. **Understanding the Equation**: The given equation relates the velocity \( v \) of the particle to its displacement \( x \) from the mean position. The maximum velocity occurs when the displacement \( x \) is zero (i.e., at the mean position).

2. **Substituting \( x = 0 \)**: To find the maximum velocity, we substitute \( x = 0 \) into the equation:
\[ v^2 = 144 - 9(0)^2 \]
Simplifying this gives:
\[ v^2 = 144 \]

3. **Calculating Maximum Velocity**: To find \( v \), we take the square root of both sides:
\[ v = \sqrt{144} = 12 \, \text{m/s} \]

4. **Conclusion**: The maximum velocity of the particle executing linear SHM is:
\[ v_{\text{max}} = 12 \, \text{m/s} \]