The displacement of an object attached to a spring and executing simple harmonic motion is given by `x=2xx10^(-2) cos(pit)` `m`. The time at which the maximum speed first occurs is.

Options

A. 0.75 s
B. 0.125 s
C. 0.25 s
D. 0.5 s

Explanation

To solve the problem, we need to find the time at which the maximum speed occurs for an object attached to a spring executing simple harmonic motion. The displacement is given by:

\[ x = 2 \times 10^{-2} \cos(\pi t) \, \text{m} \]

### Step 1: Find the velocity function
The velocity \( V \) is the derivative of the displacement \( x \) with respect to time \( t \). Thus, we differentiate \( x \):

\[ V = \frac{dx}{dt} = \frac{d}{dt}(2 \times 10^{-2} \cos(\pi t)) \]

Using the chain rule, we have:

\[ V = 2 \times 10^{-2} \cdot (-\sin(\pi t)) \cdot \frac{d}{dt}(\pi t) = -2 \times 10^{-2} \pi \sin(\pi t) \]

So, the velocity function is:

\[ V = -2\pi \times 10^{-2} \sin(\pi t) \, \text{m/s} \]

### Step 2: Determine when the velocity is maximum
The velocity will be maximum when the absolute value of \( \sin(\pi t) \) is maximum, which occurs when:

\[ \sin(\pi t) = 1 \]

### Step 3: Solve for time
The sine function equals 1 at:

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

Solving for \( t \):

\[ t = \frac{1}{2} \, \text{s} \]

### Conclusion
Thus, the time at which the maximum speed first occurs is:

\[ \boxed{0.5 \, \text{s}} \]