Passage XI) The differential equation of a particle undergoing SHM is given by `a(d^(2)x)/(dt^(2))`+bx = 0. The particle starts from the extreme position.
The time period of osciallation is given by

Options

A. `(2pi)/(b)`
B. `(2pi)sqrt(b)`
C. `(2pi)sqrt(b/a)`
D. `(2pi(sqrt(a/b))`

Answer

To find the time period of oscillation for a particle undergoing Simple Harmonic Motion (SHM) described by the differential equation \( a \frac{d^2x}{dt^2} + bx = 0 \), we can follow these steps:

### Step 1: Rewrite the Differential Equation
The given differential equation is:
\[ a \frac{d^2x}{dt^2} + bx = 0 \]
We can rearrange this to express the acceleration in terms of displacement:
\[ \frac{d^2x}{dt^2} = -\frac{b}{a} x \]

### Step 2: Identify the Form of the Equation
The equation \( \frac{d^2x}{dt^2} = -\omega^2 x \) is the standard form of SHM, where \( \omega^2 = \frac{b}{a} \). Here, \( \omega \) is the angular frequency.

### Step 3: Solve for Angular Frequency
From the comparison, we have:
\[ \omega^2 = \frac{b}{a} \]
Taking the square root gives us:
\[ \omega = \sqrt{\frac{b}{a}} \]

### Step 4: Relate Angular Frequency to Time Period
The relationship between angular frequency \( \omega \) and the time period \( T \) is given by:
\[ \omega = \frac{2\pi}{T} \]
Substituting for \( \omega \):
\[ \sqrt{\frac{b}{a}} = \frac{2\pi}{T} \]

### Step 5: Solve for Time Period
Rearranging the equation to solve for \( T \):
\[ T = \frac{2\pi}{\sqrt{\frac{b}{a}}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{a}{b}} \]

### Conclusion
Thus, the time period of oscillation is given by:
\[ T = 2\pi \sqrt{\frac{a}{b}} \]