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/8
B. T/12
C. T/2
D. T/4

Answer

To solve the problem of finding the minimum time taken by a particle executing simple harmonic motion to travel half of the amplitude from the equilibrium position, we can follow these steps:

### Step 1: Understand the parameters of SHM
The particle oscillates with:
- Amplitude \( A \)
- Period \( T \)

### Step 2: Write the equation of motion
The displacement \( x \) of a particle in simple harmonic motion can be expressed as:
\[ x = A \sin(\omega t) \]
where \( \omega \) is the angular frequency.

### Step 3: Relate angular frequency to period
The angular frequency \( \omega \) is related to the period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]

### Step 4: Set up the equation for half the amplitude
We want to find the time taken to travel half the amplitude, which means we set:
\[ x = \frac{A}{2} \]
Substituting this into the equation of motion gives:
\[ \frac{A}{2} = A \sin(\omega t) \]

### Step 5: Simplify the equation
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{1}{2} = \sin(\omega t) \]

### Step 6: Solve for \( \omega t \)
To find \( \omega t \), we take the inverse sine:
\[ \omega t = \sin^{-1}\left(\frac{1}{2}\right) \]
The value of \( \sin^{-1}\left(\frac{1}{2}\right) \) is:
\[ \omega t = \frac{\pi}{6} \quad \text{(or 30 degrees)} \]

### Step 7: Substitute \( \omega \) back into the equation
Now we substitute \( \omega \) back in:
\[ \frac{2\pi}{T} t = \frac{\pi}{6} \]

### Step 8: Solve for \( t \)
To find \( t \), we rearrange the equation:
\[ t = \frac{\pi}{6} \cdot \frac{T}{2\pi} = \frac{T}{12} \]

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