The equation of motion of a particle of mass `1g` is `(d^(2)x)/(dt^(2)) + pi^(2)x = 0`, where `x` is displacement (in m) from mean position. The frequency of oscillation is (in Hz)

Options

A. `1//2`
B. `2`
C. `5sqrt(10)`
D. `1//5 sqrt(10)`

Explanation

To solve the problem, we need to find the frequency of oscillation for the given equation of motion:

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

### Step-by-Step Solution:

**Step 1: Identify the form of the equation.**
The given equation is a second-order linear differential equation that can be recognized as the equation of motion for simple harmonic motion (SHM).

**Step 2: Compare with the standard SHM equation.**
The standard form of the SHM equation is:

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

where \(\omega\) is the angular frequency.

**Step 3: Identify \(\omega^2\).**
From the given equation, we can see that:

\[ \omega^2 = \pi^2 \]

**Step 4: Solve for \(\omega\).**
Taking the square root of both sides gives:

\[ \omega = \sqrt{\pi^2} = \pi \, \text{(radians per second)} \]

**Step 5: Calculate the time period \(T\).**
The time period \(T\) of SHM is related to the angular frequency by the formula:

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

Substituting \(\omega = \pi\):

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

**Step 6: Calculate the frequency \(f\).**
The frequency \(f\) is the reciprocal of the time period:

\[ f = \frac{1}{T} = \frac{1}{2} \, \text{Hz} \]

### Final Answer:
The frequency of oscillation is \( \frac{1}{2} \, \text{Hz} \).

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