The displacement of a body executing `SHM` is given by `x=A sin (2pi t+pi//3)`. The first time from `t=0` when the velocity is maximum is

Options

A. `(1)/(12)`s
B. 0.16 s
C. 0.25 s
D. 0.33 s

Explanation

To find the first time from \( t = 0 \) when the velocity of the body executing simple harmonic motion (SHM) is maximum, we start with the given displacement equation:

\[ x = A \sin(2\pi t + \frac{\pi}{3}) \]

### Step 1: Understand the conditions for maximum velocity
In SHM, the velocity is maximum when the displacement \( x \) is zero. This is because the velocity is given by the derivative of the displacement with respect to time, and it reaches its maximum value when the object crosses the mean position (where \( x = 0 \)).

### Step 2: Set the displacement equation to zero
To find the times when the velocity is maximum, we set the displacement \( x \) to zero:

\[ 0 = A \sin(2\pi t + \frac{\pi}{3}) \]

Since \( A \) is not zero, we can simplify this to:

\[ \sin(2\pi t + \frac{\pi}{3}) = 0 \]

### Step 3: Solve the sine equation
The sine function is zero at integer multiples of \( \pi \):

\[ 2\pi t + \frac{\pi}{3} = n\pi \quad (n \in \mathbb{Z}) \]

### Step 4: Rearrange the equation
Rearranging gives:

\[ 2\pi t = n\pi - \frac{\pi}{3} \]

Dividing through by \( 2\pi \):

\[ t = \frac{n}{2} - \frac{1}{6} \]

### Step 5: Find the first positive time
Now, we will find the first positive time by substituting different integer values for \( n \):

1. For \( n = 0 \):
\[ t = \frac{0}{2} - \frac{1}{6} = -\frac{1}{6} \quad (\text{not valid}) \]

2. For \( n = 1 \):
\[ t = \frac{1}{2} - \frac{1}{6} = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \quad (\text{valid}) \]

3. For \( n = 2 \):
\[ t = \frac{2}{2} - \frac{1}{6} = 1 - \frac{1}{6} = \frac{6}{6} - \frac{1}{6} = \frac{5}{6} \quad (\text{valid but not first}) \]

### Conclusion
The first time from \( t = 0 \) when the velocity is maximum is:

\[ t = \frac{1}{3} \text{ seconds} \quad \text{or} \quad 0.33 \text{ seconds} \]