If the displacement, velocity and acceleration of a particle in SHM are 1 cm, 1cm/sec, 1cm/`sec^2` respectively its time period will be (in seconds) :

Options

A. `pi`
B. `0.5 pi`
C. `2 pi`
D. `1. 5 pi`

Answer

To find the time period of a particle in Simple Harmonic Motion (SHM) given its displacement, velocity, and acceleration, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Displacement (x) = 1 cm
- Velocity (v) = 1 cm/sec
- Acceleration (a) = 1 cm/sec²

2. **Use the Relation Between Acceleration and Displacement:**
In SHM, the acceleration (a) is related to the displacement (x) by the formula:
\[ a = -\omega^2 x \]
Here, \(\omega\) is the angular frequency.

3. **Calculate \(\omega^2\):**
Since we are only interested in the magnitudes, we can rewrite the equation as:
\[ a = \omega^2 |x| \]
Substituting the known values:
\[ 1 \, \text{cm/sec}^2 = \omega^2 \times 1 \, \text{cm} \]
This simplifies to:
\[ \omega^2 = 1 \, \text{sec}^{-2} \]

4. **Find \(\omega\):**
Taking the square root of both sides gives:
\[ \omega = 1 \, \text{rad/sec} \]

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

6. **Final Answer:**
The time period of the particle in SHM is:
\[ T = 2\pi \, \text{seconds} \approx 6.28 \, \text{seconds} \]