The displacement of a particle executing simple harmonic motion is given by
`x=3sin(2pit+(pi)/(4))`
where x is in metres and t is in seconds. The amplitude and maximum speed of the particle is

Options

A. 3m, `2pims^(-1)`
B. 3m, `4pims^(-1)`
C. 3m, `6pims^(-1)`
D. 3m, `8pims^(-1)`

Explanation

To solve the problem, we need to find the amplitude and maximum speed of a particle executing simple harmonic motion (SHM) given by the equation:

\[ x = 3 \sin(2\pi t + \frac{\pi}{4}) \]

### Step 1: Identify the Amplitude
The general form of the equation for simple harmonic motion is:

\[ x = A \sin(\omega t + \phi) \]

Where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

From the given equation, we can directly identify the amplitude \( A \):

\[ A = 3 \, \text{m} \]

### Step 2: Find the Angular Frequency
Next, we identify the angular frequency \( \omega \) from the equation:

\[ \omega = 2\pi \, \text{rad/s} \]

### Step 3: Calculate the Maximum Speed
The maximum speed \( v_{\text{max}} \) in SHM can be calculated using the formula:

\[ v_{\text{max}} = A \omega \]

Substituting the values we found:

\[ v_{\text{max}} = 3 \, \text{m} \times 2\pi \, \text{rad/s} \]

Calculating this gives:

\[ v_{\text{max}} = 6\pi \, \text{m/s} \]

### Final Answers
- Amplitude \( A = 3 \, \text{m} \)
- Maximum Speed \( v_{\text{max}} = 6\pi \, \text{m/s} \)