A particle executes simple harmonic oscillation with an amplitudes a. The period of oscillation is T. The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is

Options

A. `(T)/(4)`
B. `(T)/(8)`
C. `(T)/(12)`
D. `(T)/(2)`

Answer

To find the minimum time taken by a particle executing simple harmonic motion (SHM) to travel half of the amplitude from the equilibrium position, we can follow these steps:

### Step 1: Understand the SHM Equation
The position \( x \) of a particle in SHM can be described by the equation:
\[ x(t) = A \sin(\omega t) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency, given by \( \omega = \frac{2\pi}{T} \),
- \( T \) is the period of oscillation.

### Step 2: Set Up the Equation for Half Amplitude
We want to find the time taken to travel half of the amplitude from the equilibrium position. Half of the amplitude is \( \frac{A}{2} \). Therefore, we set up the equation:
\[ \frac{A}{2} = A \sin(\omega t) \]

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

### Step 4: Solve for \( \omega t \)
To find \( \omega t \), we take the inverse sine:
\[ \omega t = \sin^{-1}\left(\frac{1}{2}\right) \]
We know that:
\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]
Thus:
\[ \omega t = \frac{\pi}{6} \]

### Step 5: Substitute \( \omega \)
Now, substitute \( \omega \) with \( \frac{2\pi}{T} \):
\[ \frac{2\pi}{T} t = \frac{\pi}{6} \]

### Step 6: Solve for \( t \)
Now, solve for \( t \):
\[ t = \frac{\pi}{6} \cdot \frac{T}{2\pi} = \frac{T}{12} \]

### Final Answer
The minimum time taken by the particle to travel half of the amplitude from the equilibrium position is:
\[ t = \frac{T}{12} \]

---