Three simple harmonic motions 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 solve the problem of finding the amplitude of the resultant motion when three simple harmonic motions (SHMs) combine, we can follow these steps:

### Step 1: Understand the Components of SHM
Each SHM can be represented as a vector in the complex plane. The first SHM can be represented as:
\[ y_1 = A \sin(\omega t) \]
The second SHM, which is \(60^\circ\) ahead of the first, can be represented as:
\[ y_2 = A \sin(\omega t + 60^\circ) \]
The third SHM, which is \(60^\circ\) ahead of the second, can be represented as:
\[ y_3 = A \sin(\omega t + 120^\circ) \]

### Step 2: Convert Sine Functions to Cosine
Using the sine addition formula, we can express these SHMs in terms of cosine:
\[ y_1 = A \sin(\omega t) \]
\[ y_2 = A \left( \sin(\omega t) \cos(60^\circ) + \cos(\omega t) \sin(60^\circ) \right) \]
\[ y_3 = A \left( \sin(\omega t) \cos(120^\circ) + \cos(\omega t) \sin(120^\circ) \right) \]

### Step 3: Calculate the Components
Using the values of \( \cos(60^\circ) = \frac{1}{2} \) and \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \):
- For \( y_2 \):
\[ y_2 = A \left( \sin(\omega t) \cdot \frac{1}{2} + \cos(\omega t) \cdot \frac{\sqrt{3}}{2} \right) \]
- For \( y_3 \):
Using \( \cos(120^\circ) = -\frac{1}{2} \) and \( \sin(120^\circ) = \frac{\sqrt{3}}{2} \):
\[ y_3 = A \left( \sin(\omega t) \cdot -\frac{1}{2} + \cos(\omega t) \cdot \frac{\sqrt{3}}{2} \right) \]

### Step 4: Sum the Components
Now, we can sum the three SHMs:
\[ y_{total} = y_1 + y_2 + y_3 \]
This gives us:
\[ y_{total} = A \sin(\omega t) + A \left( \frac{1}{2} \sin(\omega t) + \frac{\sqrt{3}}{2} \cos(\omega t) \right) + A \left( -\frac{1}{2} \sin(\omega t) + \frac{\sqrt{3}}{2} \cos(\omega t) \right) \]

### Step 5: Combine Like Terms
Combining the sine and cosine terms:
- Sine terms:
\[ A \sin(\omega t) + \frac{1}{2}A \sin(\omega t) - \frac{1}{2}A \sin(\omega t) = A \sin(\omega t) \]
- Cosine terms:
\[ \frac{\sqrt{3}}{2} A \cos(\omega t) + \frac{\sqrt{3}}{2} A \cos(\omega t) = \sqrt{3} A \cos(\omega t) \]

### Step 6: Resultant Amplitude
The resultant amplitude \( R \) can be found using the Pythagorean theorem:
\[ R = \sqrt{(A)^2 + (\sqrt{3}A)^2} \]
\[ R = \sqrt{A^2 + 3A^2} = \sqrt{4A^2} = 2A \]

### Final Result
The amplitude of the resultant motion is:
\[ \boxed{2A} \]