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 formula for acceleration in simple harmonic motion (S.H.M.) and the given values.

### Step 1: Understand the given values
- The particle is executing S.H.M. from the mean position at a distance \( x = 5 \, \text{cm} \).
- The acceleration at this position is given as \( a = 20 \, \text{cm/s}^2 \).

### Step 2: Recall the formula for acceleration in S.H.M.
The acceleration \( a \) of a particle in S.H.M. is given by the formula:
\[ a = -\omega^2 x \]
where:
- \( \omega \) is the angular velocity (in radians per second),
- \( x \) is the displacement from the mean position.

### Step 3: Substitute the known values into the formula
We can rearrange the formula to solve for \( \omega^2 \):
\[ \omega^2 = -\frac{a}{x} \]
Substituting the values \( a = 20 \, \text{cm/s}^2 \) and \( x = 5 \, \text{cm} \):
\[ 20 = -\omega^2 (5) \]
This simplifies to:
\[ 20 = -5\omega^2 \]

### Step 4: Solve for \( \omega^2 \)
Rearranging the equation gives:
\[ \omega^2 = -\frac{20}{5} = -4 \]
However, since we are dealing with magnitudes in this context, we can ignore the negative sign:
\[ \omega^2 = 4 \]

### Step 5: Calculate \( \omega \)
Now, take the square root of both sides to find \( \omega \):
\[ \omega = \sqrt{4} = 2 \, \text{radians/second} \]

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

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