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

Options

A. `2pisqrt(alpha)`
B. `2pialpha`
C. `(2pi)/(sqrt(alpha))`
D. `(2pi)/(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:
\[ \frac{d^2x}{dt^2} + \omega^2 x = 0 \]
where \(\omega\) is the angular frequency.

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

3. **Find the Angular Frequency \(\omega\)**:
To find \(\omega\), we take the square root of both sides:
\[ \omega = \sqrt{\alpha} \]

4. **Relate Angular Frequency to Time Period**:
The relationship between angular frequency \(\omega\) and the time period \(T\) is given by:
\[ \omega = \frac{2\pi}{T} \]
Therefore, we can express the time period \(T\) in terms of \(\omega\):
\[ T = \frac{2\pi}{\omega} \]

5. **Substitute \(\omega\) into the Time Period Formula**:
Now substituting \(\omega = \sqrt{\alpha}\) into the time period formula:
\[ T = \frac{2\pi}{\sqrt{\alpha}} \]

6. **Conclusion**:
Thus, the time period \(T\) of the simple harmonic motion described by the given equation is:
\[ T = \frac{2\pi}{\sqrt{\alpha}} \]

### Final Answer:
The time period is \(T = \frac{2\pi}{\sqrt{\alpha}}\).

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