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 amplitude of a particle due to the superposition of the given simple harmonic motions (SHMs), we will follow these steps:

### Step 1: Identify the given SHMs
The SHMs are:
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. \( X_4 = -12 \cos(50 \pi t) \)

### Step 2: Convert all SHMs to a common form
We can express \( X_4 \) in terms of sine:
\[ X_4 = -12 \cos(50 \pi t) = -12 \sin\left(50 \pi t + 90^\circ\right) \]

### Step 3: Break down the components of each SHM
- For \( X_1 \):
- Amplitude = 2, Angle = 0°

- For \( X_2 \):
- Amplitude = 10, Angle = 37°

- For \( X_3 \):
- Amplitude = -4, Angle = 0° (which means it acts in the opposite direction)

- For \( X_4 \):
- Amplitude = -12, Angle = 90° (which means it acts in the opposite direction)

### Step 4: Calculate the x and y components of each SHM
Using the angles, we can find the components:
1. \( X_1 \):
- \( X_{1x} = 2 \cos(0) = 2 \)
- \( X_{1y} = 2 \sin(0) = 0 \)

2. \( X_2 \):
- \( X_{2x} = 10 \cos(37^\circ) = 10 \cdot \frac{4}{5} = 8 \)
- \( X_{2y} = 10 \sin(37^\circ) = 10 \cdot \frac{3}{5} = 6 \)

3. \( X_3 \):
- \( X_{3x} = -4 \cos(0) = -4 \)
- \( X_{3y} = -4 \sin(0) = 0 \)

4. \( X_4 \):
- \( X_{4x} = -12 \cos(90^\circ) = 0 \)
- \( X_{4y} = -12 \sin(90^\circ) = -12 \)

### Step 5: Sum the components
Now, we sum the x and y components:
- Total x-component:
\[ X_{total x} = X_{1x} + X_{2x} + X_{3x} + X_{4x} = 2 + 8 - 4 + 0 = 6 \]

- Total y-component:
\[ X_{total y} = X_{1y} + X_{2y} + X_{3y} + X_{4y} = 0 + 6 + 0 - 12 = -6 \]

### Step 6: Calculate the resultant amplitude
The resultant amplitude \( R \) can be calculated using the Pythagorean theorem:
\[ R = \sqrt{(X_{total x})^2 + (X_{total y})^2} = \sqrt{(6)^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \]

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

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