A particle executes simple harmonic motion according to equation `4(d^(2)x)/(dt^(2))+320x=0`. Its time period of oscillation is :-

Options

A. `(2pi)/(5sqrt(3))s`
B. `(pi)/(3sqrt(2))s`
C. `(pi)/(2sqrt(5))s`
D. `(2pi)/(sqrt(3))s`

Answer

To solve the problem, we start with the given equation of motion for a particle executing simple harmonic motion:

\[ 4 \frac{d^2x}{dt^2} + 320x = 0 \]

### Step 1: Rewrite the equation in standard form

We can rewrite the equation in the standard form of simple harmonic motion, which is:

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

To do this, we divide the entire equation by 4:

\[ \frac{d^2x}{dt^2} + 80x = 0 \]

Here, we identify \( \omega^2 = 80 \).

### Step 2: Find the angular frequency \( \omega \)

To find \( \omega \), we take the square root of both sides:

\[ \omega = \sqrt{80} \]

### Step 3: Calculate the time period \( T \)

The time period \( T \) of simple harmonic motion is given by the formula:

\[ T = \frac{2\pi}{\omega} \]

Substituting the value of \( \omega \):

\[ T = \frac{2\pi}{\sqrt{80}} \]

### Step 4: Simplify the expression

We can simplify \( \sqrt{80} \):

\[ \sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5} \]

Now substituting this back into the equation for \( T \):

\[ T = \frac{2\pi}{4\sqrt{5}} \]

### Step 5: Final simplification

This simplifies to:

\[ T = \frac{\pi}{2\sqrt{5}} \]

Thus, the time period of oscillation is:

\[ T = \frac{\pi}{2\sqrt{5}} \text{ seconds} \]

### Final Answer

The time period of oscillation is \( \frac{\pi}{2\sqrt{5}} \) seconds.

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