`a - x` equation of a body in SHM is `a + 16 x = 0`. Here, `x` is in `cm` and a in `cm//s^(2)`. Find time period of oscillations.

Explanation

To solve the problem step by step, we will analyze the given equation of motion for a body in simple harmonic motion (SHM) and derive the time period of oscillation.

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

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

### Step 3: Compare the two equations
From the equations derived in Steps 1 and 2, we can equate the expressions for acceleration:
\[ -16x = -\omega^2 x \]
This implies:
\[ \omega^2 = 16 \]

### Step 4: Solve for angular frequency \( \omega \)
Taking the square root of both sides gives:
\[ \omega = \sqrt{16} = 4 \, \text{radians/second} \]

### Step 5: Calculate the time period \( T \)
The time period \( T \) of oscillation is related to the angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{4} = \frac{\pi}{2} \]

### Step 6: Convert to decimal form
Calculating the numerical value:
\[ T \approx 1.57 \, \text{seconds} \]

### Final Answer
The time period of oscillation is approximately:
\[ T \approx 1.57 \, \text{seconds} \]

---