Four particles of masses m, 2m, 3m and 4m are arranged at the corners of a parallelogram with each side equal to a and one of the angle between two adjacent sides is `60^(@)`. The parallelogram lies in the x-y plane with mass m at the origin and 4 m on the x-axis. The centre of mass of the arrangement will be located at

Options

A. `(sqrt3/2a, 0.95a)`
B. `(0.95a, sqrt3/4a)`
C. `((3a)/4,a/2)`
D. `(a/2,(3a)/4)`

Answer

To find the center of mass of the four particles arranged at the corners of a parallelogram, we will follow these steps:

### Step 1: Define the Positions of the Masses
We have four masses located at the corners of a parallelogram. The masses and their positions are as follows:
- Mass \( m \) is at the origin: \( (0, 0) \)
- Mass \( 4m \) is on the x-axis: \( (a, 0) \)
- Mass \( 2m \) is at an angle of \( 60^\circ \) from the x-axis:
- \( x_2 = a \cos(60^\circ) = a \cdot \frac{1}{2} = \frac{a}{2} \)
- \( y_2 = a \sin(60^\circ) = a \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}a}{2} \)
- Mass \( 3m \) is at the opposite corner:
- \( x_3 = a + a \cos(60^\circ) = a + \frac{a}{2} = \frac{3a}{2} \)
- \( y_3 = a \sin(60^\circ) = \frac{\sqrt{3}a}{2} \)

### Step 2: List the Coordinates
Now we can summarize the coordinates:
- \( (x_1, y_1) = (0, 0) \) for mass \( m \)
- \( (x_2, y_2) = \left(\frac{a}{2}, \frac{\sqrt{3}a}{2}\right) \) for mass \( 2m \)
- \( (x_3, y_3) = \left(\frac{3a}{2}, \frac{\sqrt{3}a}{2}\right) \) for mass \( 3m \)
- \( (x_4, y_4) = (a, 0) \) for mass \( 4m \)

### Step 3: Calculate the Center of Mass
The center of mass \( (X_{cm}, Y_{cm}) \) can be calculated using the formulas:
\[ X_{cm} = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3 + m_4 x_4}{m_1 + m_2 + m_3 + m_4} \]
\[ Y_{cm} = \frac{m_1 y_1 + m_2 y_2 + m_3 y_3 + m_4 y_4}{m_1 + m_2 + m_3 + m_4} \]

### Step 4: Substitute the Values
Substituting the values into the equations:
1. For \( X_{cm} \):
\[ X_{cm} = \frac{m \cdot 0 + 2m \cdot \frac{a}{2} + 3m \cdot \frac{3a}{2} + 4m \cdot a}{m + 2m + 3m + 4m} \]
\[ = \frac{0 + m a + \frac{9ma}{2} + 4ma}{10m} \]
\[ = \frac{0 + ma + 4.5ma + 4ma}{10m} = \frac{9.5ma}{10m} = \frac{19a}{20} \]

2. For \( Y_{cm} \):
\[ Y_{cm} = \frac{m \cdot 0 + 2m \cdot \frac{\sqrt{3}a}{2} + 3m \cdot \frac{\sqrt{3}a}{2} + 4m \cdot 0}{10m} \]
\[ = \frac{0 + m\sqrt{3}a + \frac{3\sqrt{3}ma}{2}}{10m} \]
\[ = \frac{0 + \sqrt{3}a + 1.5\sqrt{3}a}{10} = \frac{2.5\sqrt{3}a}{10} = \frac{\sqrt{3}a}{4} \]

### Final Result
Thus, the center of mass of the arrangement is located at:
\[ \left( \frac{19a}{20}, \frac{\sqrt{3}a}{4} \right) \]