The amplitide of a particle due to superposition of following S.H.Ms. Along the same line is
`{:(X_(1)=2 sin 50 pi t,,X_(2)=10 sin(50 pi t +37^(@))),(X_(3)=-4 sin 50 pi t,,X_(4)=-12 cos 50 pi t):}`

Options

A. `4 sqrt(2)`
B. `4`
C. `6 sqrt(2)`
D. none of these

Explanation

To find the resultant amplitude of the superposition of the given simple harmonic motions (S.H.Ms), we will follow these steps:

### Step 1: Identify the Amplitudes and Phases
We have four S.H.Ms given as:
1. \( X_1 = 2 \sin(50 \pi t) \)
2. \( X_2 = 10 \sin(50 \pi t + 37^\circ) \)
3. \( X_3 = -4 \sin(50 \pi t) \)
4. \( X_4 = -12 \cos(50 \pi t) \)

From these equations, we can identify the amplitudes and the phases:
- For \( X_1 \): Amplitude = 2, Phase = 0
- For \( X_2 \): Amplitude = 10, Phase = 37°
- For \( X_3 \): Amplitude = 4, Phase = 180° (since \( -4 \sin(50 \pi t) = 4 \sin(50 \pi t + 180^\circ) \))
- For \( X_4 \): Amplitude = 12, Phase = 90° (since \( -12 \cos(50 \pi t) = 12 \sin(50 \pi t + 90^\circ) \))

### Step 2: Resolve Each Component into X and Y Directions
We will resolve the amplitudes into their x and y components using trigonometric identities.

1. **For \( X_1 \)**:
- \( A_{x1} = 2 \)
- \( A_{y1} = 0 \)

2. **For \( X_2 \)**:
- \( A_{x2} = 10 \cos(37^\circ) \)
- \( A_{y2} = 10 \sin(37^\circ) \)

3. **For \( X_3 \)**:
- \( A_{x3} = -4 \)
- \( A_{y3} = 0 \)

4. **For \( X_4 \)**:
- \( A_{x4} = 0 \)
- \( A_{y4} = -12 \)

### Step 3: Calculate the Total Amplitudes in X and Y Directions
Now, we can sum the components in the x and y directions.

**Total Amplitude in X Direction**:
\[ A_x = A_{x1} + A_{x2} + A_{x3} + A_{x4} = 2 + 10 \cos(37^\circ) - 4 + 0 \]
Using \( \cos(37^\circ) \approx \frac{4}{5} \):
\[ A_x = 2 + 10 \cdot \frac{4}{5} - 4 = 2 + 8 - 4 = 6 \]

**Total Amplitude in Y Direction**:
\[ A_y = A_{y1} + A_{y2} + A_{y3} + A_{y4} = 0 + 10 \sin(37^\circ) + 0 - 12 \]
Using \( \sin(37^\circ) \approx \frac{3}{5} \):
\[ A_y = 0 + 10 \cdot \frac{3}{5} - 12 = 0 + 6 - 12 = -6 \]

### Step 4: Calculate the Resultant Amplitude
The resultant amplitude \( A \) can be calculated using the Pythagorean theorem:
\[ A = \sqrt{A_x^2 + A_y^2} = \sqrt{6^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \]

### Final Result
The resultant amplitude of the particle due to the superposition of the given S.H.Ms is \( 6\sqrt{2} \).

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