The displacement of a body executing SHM is given by `x=Asin(2pit+pi//6)`. The first time from t = 0 when the velocity is maximum is :

Options

A. 0.33 sec
B. 0.16 sec
C. 0.25 sec
D. 0.5 sec

Explanation

To solve the problem, we need to find the first time from \( t = 0 \) when the velocity of the body executing simple harmonic motion (SHM) is maximum. The displacement is given by:

\[ x = A \sin(2\pi t + \frac{\pi}{6}) \]

### Step 1: Understand the relationship between displacement and velocity in SHM
In SHM, the velocity \( v \) is given by the derivative of displacement \( x \) with respect to time \( t \):

\[ v = \frac{dx}{dt} \]

### Step 2: Differentiate the displacement equation
We differentiate the displacement equation to find the velocity:

\[ v = \frac{d}{dt}(A \sin(2\pi t + \frac{\pi}{6})) = A \cdot 2\pi \cos(2\pi t + \frac{\pi}{6}) \]

### Step 3: Determine when velocity is maximum
The velocity \( v \) is maximum when \( \cos(2\pi t + \frac{\pi}{6}) = \pm 1 \). This occurs when the argument of the cosine function is an integer multiple of \( \pi \):

\[ 2\pi t + \frac{\pi}{6} = n\pi \]

where \( n \) is an integer.

### Step 4: Solve for \( t \)
Rearranging the equation gives:

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

Dividing by \( 2\pi \):

\[ t = \frac{n}{2} - \frac{1}{12} \]

### Step 5: Find the first positive time from \( t = 0 \)
To find the first positive time, we start with \( n = 1 \):

\[ t = \frac{1}{2} - \frac{1}{12} \]

Finding a common denominator (which is 12):

\[ t = \frac{6}{12} - \frac{1}{12} = \frac{5}{12} \]

### Step 6: Convert to decimal
Calculating \( \frac{5}{12} \):

\[ t = 0.4167 \text{ seconds (approximately)} \]

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

\[ t \approx 0.4167 \text{ seconds} \]

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