The equation of motion of a particle executing simple harmonic motion is `a+16pi^(2)x = 0` In this equation, a is the linear acceleration in `m//s^(2)` of the particle at a displacement x in meter. The time period in simple harmonic motion is

Options

A. `(3)/(4)sec`
B. `(1)/(2)sec`
C. `1sec`
D. `2 sec`

Explanation

To find the time period of the simple harmonic motion (SHM) given the equation of motion \( a + 16\pi^2 x = 0 \), we can follow these steps:

### Step 1: Identify the equation of motion
The given equation is:
\[ a + 16\pi^2 x = 0 \]
This can be rearranged to express acceleration \( a \) in terms of displacement \( x \):
\[ a = -16\pi^2 x \]

### Step 2: Relate acceleration to angular frequency
In SHM, the relationship between acceleration \( a \) and displacement \( x \) is given by:
\[ a = -\omega^2 x \]
where \( \omega \) is the angular frequency.

### Step 3: Compare the two equations
From our rearranged equation, we have:
\[ -16\pi^2 x = -\omega^2 x \]
By comparing the coefficients of \( x \) from both equations, we find:
\[ \omega^2 = 16\pi^2 \]

### Step 4: Solve for \( \omega \)
Taking the square root of both sides gives us:
\[ \omega = 4\pi \]

### Step 5: Relate angular frequency to time period
The angular frequency \( \omega \) is related to the time period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting \( \omega = 4\pi \) into this equation:
\[ 4\pi = \frac{2\pi}{T} \]

### Step 6: Solve for the time period \( T \)
To find \( T \), we rearrange the equation:
\[ T = \frac{2\pi}{4\pi} = \frac{1}{2} \text{ seconds} \]

### Conclusion
Thus, the time period \( T \) of the simple harmonic motion is:
\[ \boxed{\frac{1}{2} \text{ seconds}} \]

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