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)/(5sqrt3) s`
B. `(pi)/(3sqrt2)s`
C. `(pi)/(2sqrt5)s`
D. `(2pi)/(sqrt3)s`

Answer

To find the time period of oscillation for a particle executing simple harmonic motion (SHM) described by the equation:

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

we can follow these steps:

### Step 1: Rewrite the equation
First, we can rewrite the given equation in a standard form for SHM. We can divide the entire equation by 4:

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

### Step 2: Identify the form of SHM
The standard form of the SHM equation is:

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

From our equation, we can identify that:

\[ \omega^2 = 80 \]

### Step 3: Calculate omega
To find \(\omega\), we take the square root of both sides:

\[ \omega = \sqrt{80} = 4\sqrt{5} \]

### Step 4: Calculate the time period
The time period \(T\) of SHM is given by the formula:

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

Substituting the value of \(\omega\):

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

### Step 5: Simplify the time period
We can also express this in a different form:

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

### Conclusion
Thus, the time period of oscillation is:

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

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