The displacement of a body executing SHM is given by `x=lambdasin(2pit+pi//3)`. 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 find the first time from \( t = 0 \) when the velocity of a body executing Simple Harmonic Motion (SHM) is maximum, we start with the given displacement equation:

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

### Step 1: Understand the condition for maximum velocity
The velocity \( v \) in SHM is given by the derivative of displacement \( x \) with respect to time \( t \). The velocity is maximum when the particle is at the mean position, which occurs when \( x = 0 \).

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

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

Since \( \lambda \) cannot be zero, we need:

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

### Step 3: Solve for the angle where sine is zero
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 us:

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

Dividing through by \( 2\pi \):

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

### Step 5: Find the first positive time
We need to find the smallest non-negative \( t \).

1. For \( n = 0 \):

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

2. For \( n = 1 \):

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

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

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

Thus, the correct answer is \( 0.33 \) seconds.

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