A particle executing of a simple harmonic motion covers a distance equal to half its amplitude in on e second . Then the time period of the simple harmonic motion is

Options

A. 4 s
B. 6 s
C. 8 s
D. 12 s

Explanation

To solve the problem, we need to determine the time period of a particle executing simple harmonic motion (SHM) given that it covers a distance equal to half its amplitude in one second. Let's break it down step by step.

### Step 1: Understanding the Motion
The particle is executing simple harmonic motion, which can be described by the equation:
\[ x(t) = a \sin(\omega t) \]
where:
- \( x(t) \) is the displacement from the mean position at time \( t \),
- \( a \) is the amplitude,
- \( \omega \) is the angular frequency.

### Step 2: Setting Up the Equation
We know that at \( t = 1 \) second, the particle is at a distance equal to half its amplitude:
\[ x(1) = \frac{a}{2} \]
Substituting this into the SHM equation gives:
\[ \frac{a}{2} = a \sin(\omega \cdot 1) \]

### Step 3: Simplifying the Equation
Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ \frac{1}{2} = \sin(\omega) \]

### Step 4: Finding the Value of \( \omega \)
From the equation \( \sin(\omega) = \frac{1}{2} \), we know that:
\[ \omega = \frac{\pi}{6} \text{ radians} \]
This corresponds to an angle of 30 degrees.

### Step 5: Calculating the Time Period
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 \( \omega \):
\[ T = \frac{2\pi}{\frac{\pi}{6}} \]

### Step 6: Simplifying the Time Period
Calculating the above expression:
\[ T = 2\pi \cdot \frac{6}{\pi} = 12 \text{ seconds} \]

### Conclusion
The time period of the simple harmonic motion is:
\[ T = 12 \text{ seconds} \]

### Final Answer
The correct option is 12 seconds.

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