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 solve the problem of finding 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 Motion
The particle is executing simple harmonic motion with an amplitude \( A \) and a period \( T \). The displacement \( x \) of the particle at any time \( t \) can be expressed as:
\[ x = A \sin(\omega t) \]
where \( \omega \) is the angular frequency given by \( \omega = \frac{2\pi}{T} \).

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

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

### Step 4: Find the Angle
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:
\[ \omega t = \frac{\pi}{6} \]

### Step 5: Substitute for Angular Frequency
Now substituting \( \omega = \frac{2\pi}{T} \):
\[ \frac{2\pi}{T} t = \frac{\pi}{6} \]

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

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