The motion of a particle executing simple harmonic motion is given by `X = 0.01 sin 100 pi (t + 0.05)`, where X is in metres andt in second. The time period is second is

Options

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

Explanation

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

### Step 1: Identify the angular frequency (\( \omega \))
The general form of the equation for SHM is:
\[ X = A \sin(\omega t + \phi) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.

From the given equation:
\[ X = 0.01 \sin(100 \pi (t + 0.05)) \]
we can see that:
\[ \omega = 100 \pi \]

### Step 2: 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 \( \omega \):
\[ T = \frac{2\pi}{100 \pi} \]

### Step 3: Simplify the expression
Now, we can simplify the expression:
\[ T = \frac{2\pi}{100 \pi} = \frac{2}{100} = \frac{1}{50} \]

### Step 4: Convert to decimal
Now, convert \( \frac{1}{50} \) to decimal form:
\[ T = 0.02 \text{ seconds} \]

### Conclusion
Thus, the time period of the particle executing simple harmonic motion is:
\[ T = 0.02 \text{ seconds} \]

### Final Answer
The time period is \( 0.02 \) seconds.

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