A particle moves along x-axis according to relation `x= 1+2 sin omegat`. The amplitude of S.H.M. is

Options

A. 2
B. 1
C. `sqrt(5)`
D. 3

Answer

To find the amplitude of the simple harmonic motion (S.H.M.) for the given equation \( x = 1 + 2 \sin(\omega t) \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the General Form of S.H.M.**:
The general form of the equation for simple harmonic motion is \( x = A \sin(\omega t + \phi) + C \), where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant,
- \( C \) is the vertical shift (mean position).

2. **Analyze the Given Equation**:
The equation given is \( x = 1 + 2 \sin(\omega t) \). Here, we can see that:
- The term \( 2 \sin(\omega t) \) indicates that the amplitude \( A \) is 2.
- The constant term \( 1 \) indicates that the mean position (equilibrium position) is at \( x = 1 \).

3. **Determine the Extreme Positions**:
- The maximum value of \( \sin(\omega t) \) is 1, which gives the maximum displacement:
\[ x_{\text{max}} = 1 + 2(1) = 3 \]
- The minimum value of \( \sin(\omega t) \) is -1, which gives the minimum displacement:
\[ x_{\text{min}} = 1 + 2(-1) = -1 \]

4. **Calculate the Distance Between Extreme Positions**:
- The distance \( d \) between the maximum and minimum positions is:
\[ d = x_{\text{max}} - x_{\text{min}} = 3 - (-1) = 3 + 1 = 4 \]

5. **Relate the Distance to Amplitude**:
- The distance between the two extreme positions is given by \( d = 2A \), where \( A \) is the amplitude.
- Therefore, we can set up the equation:
\[ 2A = 4 \]

6. **Solve for Amplitude**:
- Dividing both sides by 2 gives:
\[ A = \frac{4}{2} = 2 \]

### Conclusion:
The amplitude of the simple harmonic motion is \( 2 \).