The periodic time of a particle executing S.H.M. is12 second.After how much intervalfrom`t=0` will its displacement be half of its amplitude ?

Options

A. 1s
B. 2s
C. 6s
D. 3s

Answer

To solve the problem, we need to find the time at which the displacement of a particle executing Simple Harmonic Motion (SHM) is half of its amplitude. Here’s a step-by-step solution:

### Step 1: Understand the parameters
Given:
- The periodic time \( T = 12 \) seconds.
- The amplitude \( A \) (not given numerically but is a variable).

### Step 2: Write the equation of motion for SHM
The displacement \( x \) of a particle in SHM can be described by the equation:
\[ x(t) = A \sin(\omega t + \phi) \]
Where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

Since the particle starts from the mean position, we can take \( \phi = 0 \). Thus, the equation simplifies to:
\[ x(t) = A \sin(\omega t) \]

### Step 3: Find the angular frequency \( \omega \)
The angular frequency \( \omega \) is given by:
\[ \omega = \frac{2\pi}{T} \]
Substituting \( T = 12 \) seconds:
\[ \omega = \frac{2\pi}{12} = \frac{\pi}{6} \text{ rad/s} \]

### Step 4: Set the displacement to half of the amplitude
We want to find the time \( t \) when the displacement \( x \) is half of the amplitude:
\[ x = \frac{A}{2} \]
Substituting into the equation:
\[ \frac{A}{2} = A \sin(\omega t) \]
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{1}{2} = \sin(\omega t) \]

### Step 5: Solve for \( \omega t \)
The equation \( \sin(\omega t) = \frac{1}{2} \) corresponds to:
\[ \omega t = \frac{\pi}{6} \quad \text{or} \quad \omega t = \frac{5\pi}{6} \quad \text{(since sine is positive in the first and second quadrants)} \]

### Step 6: Calculate \( t \)
Using \( \omega = \frac{\pi}{6} \):
1. For \( \omega t = \frac{\pi}{6} \):
\[ t = \frac{\frac{\pi}{6}}{\frac{\pi}{6}} = 1 \text{ second} \]
2. For \( \omega t = \frac{5\pi}{6} \):
\[ t = \frac{\frac{5\pi}{6}}{\frac{\pi}{6}} = 5 \text{ seconds} \]

### Step 7: Conclusion
The first time \( t \) when the displacement is half of the amplitude is:
\[ t = 1 \text{ second} \]
The second time is \( t = 5 \text{ seconds} \). However, since the question asks for the first occurrence, the answer is:
\[ \text{The answer is } 1 \text{ second.} \]