The motion of a particle executing S.H.M. is given by `x= 0.01 sin 100 pi (t+.05)` , where x is in metres and time is in seconds. The time period is

Options

A. 0.01 sec
B. 0.02 sec
C. 0.1 sec
D. 0.2 sec

Answer

To find the time period of the particle executing simple harmonic motion (S.H.M.) given by the equation \( x = 0.01 \sin(100\pi(t + 0.05)) \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given equation**:
The equation of motion is given as:
\[ x = 0.01 \sin(100\pi(t + 0.05)) \]

2. **Rewrite the equation**:
We can express the equation in the standard form of S.H.M.:
\[ x = a \sin(\omega t + \phi) \]
Here, \( a = 0.01 \), \( \omega = 100\pi \), and \( \phi = 100\pi \times 0.05 = 5\pi \).

3. **Extract the angular frequency (\(\omega\))**:
From the equation, we see that:
\[ \omega = 100\pi \]

4. **Relate angular frequency to the time period**:
The relationship between angular frequency and time period \( T \) is given by:
\[ T = \frac{2\pi}{\omega} \]

5. **Substitute the value of \(\omega\)**:
Now, substituting the value of \(\omega\) into the formula for time period:
\[ T = \frac{2\pi}{100\pi} \]

6. **Simplify the expression**:
Simplifying the equation gives:
\[ T = \frac{2}{100} = 0.02 \text{ seconds} \]

7. **Final answer**:
Therefore, the time period \( T \) of the particle is:
\[ T = 0.02 \text{ seconds} \]

### Conclusion:
The time period of the particle executing simple harmonic motion is \( 0.02 \) seconds.