If the magnitude of displacement is numerically equal to that of acceleration, then the time period is

Options

A. 1 second
B. `pi` second
C. `2pi` second
D. `4pi` second

Explanation

To solve the problem, we need to establish the relationship between displacement and acceleration in Simple Harmonic Motion (SHM) and find the time period when the magnitude of displacement is equal to the magnitude of acceleration.

### Step-by-Step Solution:

1. **Understanding the Relationship in SHM**:
In SHM, the acceleration \( a \) is given by the formula:
\[ a = -\omega^2 x \]
where \( \omega \) is the angular frequency and \( x \) is the displacement from the mean position.

2. **Magnitude of Acceleration**:
The magnitude of acceleration can be expressed as:
\[ |a| = \omega^2 |x| \]

3. **Given Condition**:
According to the problem, the magnitude of displacement is numerically equal to that of acceleration:
\[ |x| = |a| \]
Substituting the expression for acceleration:
\[ |x| = \omega^2 |x| \]

4. **Simplifying the Equation**:
We can simplify this equation. Since \( |x| \) is not zero (as we are looking for a non-trivial solution), we can divide both sides by \( |x| \):
\[ 1 = \omega^2 \]

5. **Finding Angular Frequency**:
From the equation \( \omega^2 = 1 \), we find:
\[ \omega = 1 \text{ rad/s} \]

6. **Calculating the Time Period**:
The time period \( T \) of oscillation is related to the angular frequency by the formula:
\[ T = \frac{2\pi}{\omega} \]
Substituting \( \omega = 1 \):
\[ T = \frac{2\pi}{1} = 2\pi \text{ seconds} \]

### Final Answer:
The time period of the oscillation is \( 2\pi \) seconds.

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