A particle is executing S.H.M. from mean position at 5 cm distance, acceleration is `20 cm//sec^(2)` then value of angular velocity will be

Options

A. 2 rad/sec
B. 4 rad/sec
C. 10 rad/sec
D. 15 rad/sec

Explanation

To solve the problem step by step, we will use the relationship between acceleration, angular velocity, and displacement in Simple Harmonic Motion (SHM).

### Step 1: Understand the relationship
In SHM, the acceleration \( a \) of a particle is given by the formula:
\[ a = -\omega^2 x \]
where:
- \( a \) is the acceleration,
- \( \omega \) is the angular velocity,
- \( x \) is the displacement from the mean position.

### Step 2: Identify the given values
From the problem, we have:
- Displacement \( x = 5 \, \text{cm} \)
- Acceleration \( a = 20 \, \text{cm/s}^2 \)

### Step 3: Substitute the values into the equation
We can ignore the negative sign for now since we are interested in the magnitude of acceleration. Thus, we can write:
\[ a = \omega^2 x \]
Substituting the known values:
\[ 20 = \omega^2 \cdot 5 \]

### Step 4: Solve for \( \omega^2 \)
Rearranging the equation to solve for \( \omega^2 \):
\[ \omega^2 = \frac{20}{5} = 4 \, \text{(s}^{-2}\text{)} \]

### Step 5: Calculate \( \omega \)
Now, taking the square root to find \( \omega \):
\[ \omega = \sqrt{4} = 2 \, \text{radians/s} \]

### Final Answer
The value of angular velocity \( \omega \) is:
\[ \omega = 2 \, \text{radians/s} \]