The differential equation for a freely vibrating particle is `(d^2 x)/(dt^2)+ alpha x=0`. The angular frequency of particle will be

Options

A. `alpha`
B. `sqrt alpha`
C. `alpha^2`
D. 0

Answer

To find the angular frequency of a freely vibrating particle described by the differential equation

\[ \frac{d^2 x}{dt^2} + \alpha x = 0, \]

we can follow these steps:

### Step 1: Identify the form of the differential equation
The given equation is a second-order linear differential equation. It can be recognized as the standard form of the equation for simple harmonic motion (SHM).

### Step 2: Relate acceleration to displacement
In SHM, the acceleration \(a\) is related to the displacement \(x\) by the equation:

\[ a = -\omega^2 x, \]

where \(\omega\) is the angular frequency. The negative sign indicates that the acceleration is directed opposite to the displacement.

### Step 3: Substitute acceleration in the differential equation
We know that the acceleration can also be expressed as:

\[ \frac{d^2 x}{dt^2} = a. \]

Substituting this into the original differential equation gives:

\[ -\omega^2 x + \alpha x = 0. \]

### Step 4: Rearrange the equation
Rearranging the equation leads to:

\[ (-\omega^2 + \alpha)x = 0. \]

### Step 5: Solve for \(\omega^2\)
Since \(x\) cannot be zero (we are considering non-trivial solutions), we can set the term in parentheses to zero:

\[ -\omega^2 + \alpha = 0. \]

This simplifies to:

\[ \omega^2 = \alpha. \]

### Step 6: Find \(\omega\)
Taking the square root of both sides gives us the angular frequency:

\[ \omega = \sqrt{\alpha}. \]

### Conclusion
Thus, the angular frequency of the particle is

\[ \omega = \sqrt{\alpha}. \]