The differential equation of a particle performing a S.H.M. is `(d^(2)x)/(dt^(2))+ 64x=0`. The period of oscillation of the particle is

Options

A. 10 sec
B. 5 sec
C. `(pi)/(3)` sec
D. `(pi)/(4)` sec

Explanation

To find the period of oscillation of a particle performing simple harmonic motion (SHM) given the differential equation:

\[ \frac{d^2x}{dt^2} + 64x = 0 \]

we can follow these steps:

### Step 1: Rewrite the differential equation
We start by rewriting the given differential equation in a more standard form. We can express it as:

\[ \frac{d^2x}{dt^2} = -64x \]

### Step 2: Identify the angular frequency (ω)
The standard form of the SHM equation is:

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

By comparing this with our rewritten equation, we can see that:

\[ \omega^2 = 64 \]

To find ω, we take the square root:

\[ \omega = \sqrt{64} = 8 \text{ radians/second} \]

### Step 3: Calculate the period (T)
The period of oscillation (T) is related to the angular frequency (ω) by the formula:

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

Substituting the value of ω we found:

\[ T = \frac{2\pi}{8} = \frac{\pi}{4} \text{ seconds} \]

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

\[ \boxed{\frac{\pi}{4} \text{ seconds}} \]

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