A body is vibrating in simple harmonic motion . If its acceleration in `12 cms^(-2)` at a displacement 3 cm from the mean position, then time period is

Options

A. 6.28 s
B. 3.14 s
C. 1.57 s
D. 2.57 s

Explanation

To solve the problem step by step, we will follow the principles of simple harmonic motion (SHM) and use the given information effectively.

### Step 1: Understand the relationship between acceleration, displacement, and angular frequency.
In simple harmonic motion, the acceleration \( a \) at a displacement \( x \) from the mean position is given by the formula:
\[ a = -\omega^2 x \]
where \( \omega \) is the angular frequency.

### Step 2: Substitute the given values.
We are given:
- Acceleration \( a = 12 \, \text{cm/s}^2 \)
- Displacement \( x = 3 \, \text{cm} \)

Since we are only interested in magnitudes, we can ignore the negative sign for now. Thus, we can write:
\[ 12 = \omega^2 \cdot 3 \]

### Step 3: Solve for \( \omega^2 \).
Rearranging the equation gives:
\[ \omega^2 = \frac{12}{3} = 4 \]

### Step 4: Find \( \omega \).
Taking the square root of both sides, we find:
\[ \omega = \sqrt{4} = 2 \, \text{rad/s} \]

### Step 5: Calculate the time period \( T \).
The time period \( T \) of simple harmonic motion is related to angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{2} = \pi \, \text{seconds} \]

### Step 6: Provide the numerical value of \( T \).
Using the approximate value of \( \pi \):
\[ T \approx 3.14 \, \text{seconds} \]

### Final Answer:
The time period of the body vibrating in simple harmonic motion is approximately \( 3.14 \, \text{seconds} \).

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