Three simple harmonic motion of equal amplitudes `A` and equal time periods in the same direction combine. The phase of the second motion is `60^(@)` ahead of the first and the phase of the third motion is `60^(@)` ahead of the second. Find the amplitude of the resultant motion.

Explanation

To find the amplitude of the resultant motion when three simple harmonic motions (SHM) combine, we can follow these steps:

### Step 1: Define the SHM equations
Let the amplitude of each SHM be \( A \) and the time period be the same. The equations for the three SHMs can be defined as follows:

1. First SHM:
\[ \vec{R_1} = A \hat{i} \]

2. Second SHM (60 degrees ahead of the first):
\[ \vec{R_2} = A \cos(60^\circ) \hat{i} + A \sin(60^\circ) \hat{j} \]

3. Third SHM (60 degrees ahead of the second, which makes it 120 degrees ahead of the first):
\[ \vec{R_3} = A \cos(120^\circ) \hat{i} + A \sin(120^\circ) \hat{j} \]

### Step 2: Calculate the components of each SHM
Now, we need to calculate the components of the second and third SHMs using the trigonometric values:

- For \( \vec{R_2} \):
\[ \cos(60^\circ) = \frac{1}{2}, \quad \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
Thus,
\[ \vec{R_2} = A \left(\frac{1}{2}\right) \hat{i} + A \left(\frac{\sqrt{3}}{2}\right) \hat{j} = \frac{A}{2} \hat{i} + \frac{A\sqrt{3}}{2} \hat{j} \]

- For \( \vec{R_3} \):
\[ \cos(120^\circ) = -\frac{1}{2}, \quad \sin(120^\circ) = \frac{\sqrt{3}}{2} \]
Thus,
\[ \vec{R_3} = A \left(-\frac{1}{2}\right) \hat{i} + A \left(\frac{\sqrt{3}}{2}\right) \hat{j} = -\frac{A}{2} \hat{i} + \frac{A\sqrt{3}}{2} \hat{j} \]

### Step 3: Add the vectors
Now, we can add the three vectors together to find the resultant vector \( \vec{R} \):
\[ \vec{R} = \vec{R_1} + \vec{R_2} + \vec{R_3} \]
Substituting the values:
\[ \vec{R} = A \hat{i} + \left(\frac{A}{2} - \frac{A}{2}\right) \hat{i} + \left(\frac{A\sqrt{3}}{2} + \frac{A\sqrt{3}}{2}\right) \hat{j} \]
This simplifies to:
\[ \vec{R} = A \hat{i} + 0 \hat{i} + A\sqrt{3} \hat{j} = A \hat{i} + A\sqrt{3} \hat{j} \]

### Step 4: Calculate the magnitude of the resultant vector
To find the amplitude of the resultant motion, we calculate the magnitude of \( \vec{R} \):
\[ |\vec{R}| = \sqrt{(A)^2 + (A\sqrt{3})^2} = \sqrt{A^2 + 3A^2} = \sqrt{4A^2} = 2A \]

### Final Result
Thus, the amplitude of the resultant motion is:
\[ \text{Amplitude of resultant motion} = 2A \]

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