A particle is subjected to three SHM's in same direction simultaneously each having amplitude a and equal time period. The phase of the second motion is `30^(0)` ahead of the first and the phase of the third motion is `30^(0)` ahead of the second. Find the amplitude of the resultant motion.

Answer

To find the amplitude of the resultant motion when a particle is subjected to three simple harmonic motions (SHMs) with the same amplitude and equal time period, we can follow these steps:

### Step 1: Define the SHMs
Let the amplitude of each SHM be \( A \). The phases of the three SHMs are:
- First SHM: \( \phi_1 = 0^\circ \)
- Second SHM: \( \phi_2 = 30^\circ \)
- Third SHM: \( \phi_3 = 60^\circ \)

### Step 2: Write the equations for each SHM
The displacement for each SHM can be expressed in terms of their respective amplitudes and phases:
- For the first SHM:
\[ x_1 = A \cos(0^\circ) = A \]
- For the second SHM:
\[ x_2 = A \cos(30^\circ) = A \cdot \frac{\sqrt{3}}{2} \]
- For the third SHM:
\[ x_3 = A \cos(60^\circ) = A \cdot \frac{1}{2} \]

### Step 3: Calculate the resultant displacement
The resultant displacement \( R \) can be found by summing the contributions from all three SHMs:
\[ R = x_1 + x_2 + x_3 = A + A \cdot \frac{\sqrt{3}}{2} + A \cdot \frac{1}{2} \]
Factoring out \( A \):
\[ R = A \left( 1 + \frac{\sqrt{3}}{2} + \frac{1}{2} \right) \]

### Step 4: Simplify the expression
Combine the terms inside the parentheses:
\[ R = A \left( 1 + \frac{\sqrt{3} + 1}{2} \right) = A \left( \frac{2 + \sqrt{3} + 1}{2} \right) = A \left( \frac{3 + \sqrt{3}}{2} \right) \]

### Step 5: Calculate the amplitude of the resultant motion
The amplitude of the resultant motion \( A_R \) is given by the magnitude of the resultant displacement:
\[ A_R = \sqrt{R_x^2 + R_y^2} \]
Where:
- \( R_x = A \left( 1 + \frac{\sqrt{3}}{2} + \frac{1}{2} \right) \)
- \( R_y = A \left( \frac{1}{2} + \frac{\sqrt{3}}{2} \right) \)

Calculating \( R_x \) and \( R_y \):
\[ R_x = A \left( \frac{3 + \sqrt{3}}{2} \right) \]
\[ R_y = A \left( \frac{1 + \sqrt{3}}{2} \right) \]

### Step 6: Final expression for amplitude
Now substituting back into the amplitude formula:
\[ A_R = \sqrt{\left(A \cdot \frac{3 + \sqrt{3}}{2}\right)^2 + \left(A \cdot \frac{1 + \sqrt{3}}{2}\right)^2} \]
\[ = A \cdot \frac{1}{2} \sqrt{(3 + \sqrt{3})^2 + (1 + \sqrt{3})^2} \]

Calculating the squares:
\[ (3 + \sqrt{3})^2 = 9 + 6\sqrt{3} + 3 = 12 + 6\sqrt{3} \]
\[ (1 + \sqrt{3})^2 = 1 + 2\sqrt{3} + 3 = 4 + 2\sqrt{3} \]
Adding these:
\[ = (12 + 6\sqrt{3}) + (4 + 2\sqrt{3}) = 16 + 8\sqrt{3} \]

Thus, the final amplitude of the resultant motion is:
\[ A_R = A \cdot \frac{1}{2} \sqrt{16 + 8\sqrt{3}} \]