The equation of a simple harmonic motion of a particle is `(d^(2)x)/(dt^(2)) + 0.2 (dx)/(dt) + 36x = 0`. Its time period is approximately

Options

A. `(pi)/(2) sec`
B. `(pi)/(4) sec`
C. `(pi)/(3) sec`
D. `(pi)/(6)sec`

Answer

To find the time period of the simple harmonic motion described by the equation

\[ \frac{d^2x}{dt^2} + 0.2 \frac{dx}{dt} + 36x = 0, \]

we can follow these steps:

### Step 1: Identify the coefficients
The given equation can be compared to the standard form of a damped harmonic oscillator:

\[ \frac{d^2x}{dt^2} + \frac{y}{m} \frac{dx}{dt} + \frac{k}{m} x = 0. \]

From the equation, we can identify:
- \(y = 0.2\)
- \(k = 36\)

### Step 2: Calculate \(\omega^2\) and \(\beta\)
From the standard form, we know:
- \(\omega^2 = \frac{k}{m}\)
- \(\beta = \frac{y}{2m}\)

We need to find \(\omega\) and \(\beta\). However, we don't have \(m\) explicitly, but we can express \(\omega\) in terms of \(k\) and \(m\).

Given:
\[ \omega^2 = 36 \implies \omega = \sqrt{36} = 6. \]

Next, we can find \(\beta\):
\[ \beta = \frac{0.2}{2m}. \]

### Step 3: Determine the time period
The time period \(T\) of the oscillation is given by the formula:
\[ T = \frac{2\pi}{\omega}. \]

Substituting the value of \(\omega\):
\[ T = \frac{2\pi}{6} = \frac{\pi}{3}. \]

### Conclusion
Thus, the time period of the simple harmonic motion is approximately:

\[ T \approx \frac{\pi}{3} \text{ seconds}. \]