A point performs simple harmonic oscillation of period T and the equation of motion is given by `x = a sin (omega t + (pi)/(6))`. After the elapse of what fraction of the time period, the velocity of the point will be equal to half of its maximum velocity ?

Options

A. `(T)/(8)`
B. `(T)/(6)`
C. `(T)/(3)`
D. `(T)/(12)`

Answer

To solve the problem step by step, we will follow these instructions:

### Step 1: Understand the equation of motion
The equation of motion for the simple harmonic oscillator is given as:
\[ x = a \sin(\omega t + \frac{\pi}{6}) \]
where:
- \( a \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( t \) is the time.

### Step 2: Find the velocity
The velocity \( v \) of the point can be found by differentiating the position \( x \) with respect to time \( t \):
\[ v = \frac{dx}{dt} = a \omega \cos(\omega t + \frac{\pi}{6}) \]

### Step 3: Determine the maximum velocity
The maximum velocity \( v_{max} \) occurs when \( \cos(\omega t + \frac{\pi}{6}) = 1 \):
\[ v_{max} = a \omega \]

### Step 4: Set up the equation for half of the maximum velocity
We want to find the time when the velocity \( v \) is equal to half of its maximum velocity:
\[ v = \frac{1}{2} v_{max} = \frac{1}{2} (a \omega) \]
Thus, we have:
\[ a \omega \cos(\omega t + \frac{\pi}{6}) = \frac{1}{2} (a \omega) \]

### Step 5: Simplify the equation
Dividing both sides by \( a \omega \) (assuming \( a \) and \( \omega \) are not zero):
\[ \cos(\omega t + \frac{\pi}{6}) = \frac{1}{2} \]

### Step 6: Solve for \( \omega t + \frac{\pi}{6} \)
The cosine function equals \( \frac{1}{2} \) at specific angles:
\[ \omega t + \frac{\pi}{6} = \frac{\pi}{3} \quad \text{or} \quad \omega t + \frac{\pi}{6} = \frac{5\pi}{3} \]

### Step 7: Solve for \( \omega t \)
From the first equation:
\[ \omega t = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \]
From the second equation:
\[ \omega t = \frac{5\pi}{3} - \frac{\pi}{6} = \frac{10\pi}{6} - \frac{\pi}{6} = \frac{9\pi}{6} = \frac{3\pi}{2} \]

### Step 8: Relate \( \omega \) to the time period \( T \)
The angular frequency \( \omega \) is related to the time period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]
Substituting this into the equation:
\[ \frac{2\pi}{T} t = \frac{\pi}{6} \]
Solving for \( t \):
\[ t = \frac{T}{12} \]

### Step 9: Conclusion
The fraction of the time period \( T \) after which the velocity of the point will be equal to half of its maximum velocity is:
\[ \frac{t}{T} = \frac{1}{12} \]