If a simple harmonic motion is represented by `(d^(2)x)/(dt^(2)) + alphax = 0`, its time period is :

Options

A. `(2pi)/(alpha)`
B. `(2pi)/(sqrt(alpha))`
C. `2pialpha`
D. `2pisqrt(alpha)`

Explanation

To solve the problem, we need to determine the time period of a simple harmonic motion (SHM) represented by the equation:

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

### Step-by-Step Solution:

1. **Identify the Standard Form of SHM**:
The standard form of the equation for simple harmonic motion is given by:
\[ \frac{d^2x}{dt^2} + \omega^2 x = 0 \]
where \(\omega\) is the angular frequency.

2. **Compare the Given Equation with the Standard Form**:
We can rewrite the given equation:
\[ \frac{d^2x}{dt^2} + \alpha x = 0 \]
By comparing this with the standard form, we can identify that:
\[ \omega^2 = \alpha \]

3. **Find the Angular Frequency**:
From the comparison, we can find \(\omega\):
\[ \omega = \sqrt{\alpha} \]

4. **Determine the Time Period**:
The time period \(T\) of SHM is related to the angular frequency by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting \(\omega = \sqrt{\alpha}\) into the formula gives:
\[ T = \frac{2\pi}{\sqrt{\alpha}} \]

5. **Final Result**:
Thus, the time period of the simple harmonic motion is:
\[ T = \frac{2\pi}{\sqrt{\alpha}} \]