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 determine the amplitude and maximum speed of a particle executing simple harmonic motion (SHM) given its displacement equation:

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

### Step 1: Identify the Amplitude
The standard form of the equation of SHM 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 see that:
- The amplitude \( a = 3 \) meters.

### Step 2: Identify the Angular Frequency
From the equation, we can also identify the angular frequency \( \omega \):
- Comparing \( 2\pi t \) with \( \omega t \), we find that \( \omega = 2\pi \) radians per second.

### Step 3: Calculate the Maximum Speed
The maximum speed \( v_{\text{max}} \) in SHM can be calculated using the formula:
\[ v_{\text{max}} = \omega a \]
Substituting the values we found:
- \( \omega = 2\pi \) radians/second,
- \( a = 3 \) meters.

Thus, we calculate:
\[ v_{\text{max}} = (2\pi)(3) = 6\pi \text{ meters/second} \]

### Summary of Results
- **Amplitude**: \( 3 \) meters
- **Maximum Speed**: \( 6\pi \) meters/second

### Final Answer
- Amplitude: \( 3 \) meters
- Maximum Speed: \( 6\pi \) meters/second

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