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, we need to find the fraction of the time period \( T \) after which the velocity of a point performing simple harmonic motion (SHM) is equal to half of its maximum velocity. The equation of motion is given as:

\[ x = a \sin(\omega t + \frac{\pi}{6}) \]

### Step 1: Identify the maximum velocity

The maximum velocity \( V_{\text{max}} \) for a particle in SHM is given by:

\[ V_{\text{max}} = \omega a \]

### Step 2: Set up the equation for half of the maximum velocity

We need to find the time when the velocity \( V \) is half of the maximum velocity:

\[ V = \frac{1}{2} V_{\text{max}} = \frac{1}{2} \omega a \]

### Step 3: Use the velocity equation in SHM

The velocity \( V \) in SHM can also be expressed as:

\[ V = \omega \sqrt{a^2 - x^2} \]

### Step 4: Set the two expressions for velocity equal

Now, we can set the two expressions for velocity equal to each other:

\[ \omega \sqrt{a^2 - x^2} = \frac{1}{2} \omega a \]

### Step 5: Cancel \( \omega \) and solve for \( x \)

Assuming \( \omega \neq 0 \), we can cancel \( \omega \) from both sides:

\[ \sqrt{a^2 - x^2} = \frac{1}{2} a \]

Now, squaring both sides gives:

\[ a^2 - x^2 = \frac{1}{4} a^2 \]

### Step 6: Rearrange to find \( x^2 \)

Rearranging the equation, we have:

\[ x^2 = a^2 - \frac{1}{4} a^2 = \frac{3}{4} a^2 \]

### Step 7: Solve for \( x \)

Taking the square root gives:

\[ x = \sqrt{\frac{3}{4}} a = \frac{\sqrt{3}}{2} a \]

### Step 8: Substitute \( x \) back into the equation of motion

Now we substitute \( x = \frac{\sqrt{3}}{2} a \) into the equation of motion:

\[ \frac{\sqrt{3}}{2} a = a \sin(\omega t + \frac{\pi}{6}) \]

Dividing both sides by \( a \) (assuming \( a \neq 0 \)):

\[ \frac{\sqrt{3}}{2} = \sin(\omega t + \frac{\pi}{6}) \]

### Step 9: Find the angle corresponding to \( \sin^{-1}(\frac{\sqrt{3}}{2}) \)

The angle whose sine is \( \frac{\sqrt{3}}{2} \) is:

\[ \omega t + \frac{\pi}{6} = \frac{\pi}{3} \]

### Step 10: Solve for \( \omega t \)

Rearranging gives:

\[ \omega t = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \]

### Step 11: Substitute \( \omega \) in terms of \( T \)

Since \( \omega = \frac{2\pi}{T} \):

\[ \frac{2\pi}{T} t = \frac{\pi}{6} \]

### Step 12: Solve for \( t \)

Multiplying both sides by \( \frac{T}{2\pi} \):

\[ t = \frac{T}{12} \]

### Conclusion

The fraction of the time period after which the velocity of the point will be equal to half of its maximum velocity is:

\[ \frac{t}{T} = \frac{1}{12} \]

### Final Answer

The answer is \( \frac{1}{12} \).

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