If a SHM is represented by the equation `x=10 sin(pit+(pi)/(6))` in Si units, then determine its amplitude, time period and maximum uelocity `upsilon_(max)` ?

Explanation

To solve the problem step by step, we will analyze the given equation of simple harmonic motion (SHM) and extract the required parameters: amplitude, time period, and maximum velocity.

### Step 1: Identify the Amplitude
The given SHM equation is:
\[ x = 10 \sin\left(\pi t + \frac{\pi}{6}\right) \]

In the standard form of SHM:
\[ x = a \sin(\omega t + \phi) \]
where:
- \( a \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

By comparing the two equations, we can see that:
- The amplitude \( a = 10 \) meters.

**Hint:** The amplitude is the coefficient of the sine function in the SHM equation.

### Step 2: Calculate the Time Period
The angular frequency \( \omega \) can be identified from the equation:
\[ \omega = \pi \, \text{rad/s} \]

The time period \( T \) of SHM is given by the formula:
\[ T = \frac{2\pi}{\omega} \]

Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{\pi} = 2 \, \text{seconds} \]

**Hint:** The time period is the reciprocal of the frequency and can be calculated using the angular frequency.

### Step 3: Calculate the Maximum Velocity
The maximum velocity \( v_{\text{max}} \) in SHM can be calculated using the formula:
\[ v_{\text{max}} = a \omega \]

Substituting the values of \( a \) and \( \omega \):
\[ v_{\text{max}} = 10 \, \text{m} \cdot \pi \, \text{rad/s} = 10\pi \, \text{m/s} \]

**Hint:** The maximum velocity occurs when the displacement is zero, and it is the product of amplitude and angular frequency.

### Final Results
- **Amplitude**: \( 10 \, \text{meters} \)
- **Time Period**: \( 2 \, \text{seconds} \)
- **Maximum Velocity**: \( 10\pi \, \text{m/s} \)