The equation of SHM of a particle is given as `2(d^(2)x)/(dt^(2))+32x=0` where x is the displacement from the mean position. The period of its oscillation ( in seconds) is -

Options

A. 4
B. `(pi)/(2)`
C. `(pi)/(2sqrt(2))`
D. `2pi`

Explanation

To find the period of oscillation for the given simple harmonic motion (SHM) equation, we can follow these steps:

### Step 1: Write down the given equation
The equation of SHM is given as:
\[ 2\frac{d^2x}{dt^2} + 32x = 0 \]

### Step 2: Rearrange the equation
We can rearrange the equation to isolate the second derivative term:
\[ \frac{d^2x}{dt^2} + 16x = 0 \]

### Step 3: Identify the standard form of SHM
The standard form of the SHM equation is:
\[ \frac{d^2x}{dt^2} + \omega^2 x = 0 \]
where \(\omega\) is the angular frequency.

### Step 4: Compare coefficients
From our rearranged equation \(\frac{d^2x}{dt^2} + 16x = 0\), we can see that:
\[ \omega^2 = 16 \]

### Step 5: Solve for \(\omega\)
Taking the square root of both sides gives us:
\[ \omega = \sqrt{16} = 4 \, \text{rad/s} \]

### Step 6: Calculate the period \(T\)
The relationship between angular frequency \(\omega\) and the period \(T\) is given by:
\[ \omega = \frac{2\pi}{T} \]
Rearranging this to solve for \(T\) gives:
\[ T = \frac{2\pi}{\omega} \]

### Step 7: Substitute the value of \(\omega\)
Substituting \(\omega = 4\) into the equation for \(T\):
\[ T = \frac{2\pi}{4} = \frac{\pi}{2} \, \text{seconds} \]

### Final Answer
The period of oscillation is:
\[ T = \frac{\pi}{2} \, \text{seconds} \]

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