Find time period of the function, `y=sin omega t + sin 2omega t + sin 3omega t`

Explanation

To find the time period of the function \( y = \sin(\omega t) + \sin(2\omega t) + \sin(3\omega t) \), we will follow these steps:

### Step 1: Identify the individual time periods
The time period \( T \) of a sine function \( \sin(\omega t) \) is given by the formula:
\[ T = \frac{2\pi}{\omega} \]
For each term in the function, we can calculate the time periods as follows:

1. For \( y_1 = \sin(\omega t) \):
\[ T_1 = \frac{2\pi}{\omega} \]

2. For \( y_2 = \sin(2\omega t) \):
\[ T_2 = \frac{2\pi}{2\omega} = \frac{\pi}{\omega} \]

3. For \( y_3 = \sin(3\omega t) \):
\[ T_3 = \frac{2\pi}{3\omega} \]

### Step 2: Find the overall time period
The overall time period of the function \( y \) will be the least common multiple (LCM) of the individual time periods \( T_1, T_2, \) and \( T_3 \).

We have:
- \( T_1 = \frac{2\pi}{\omega} \)
- \( T_2 = \frac{\pi}{\omega} \)
- \( T_3 = \frac{2\pi}{3\omega} \)

### Step 3: Calculate the LCM
To find the LCM of the three time periods, we will first express them in a common format:

- \( T_1 = \frac{2\pi}{\omega} \)
- \( T_2 = \frac{\pi}{\omega} \)
- \( T_3 = \frac{2\pi}{3\omega} \)

The LCM of the numerators \( 2\pi, \pi, 2\pi \) is \( 2\pi \) (since \( 2\pi \) is the largest).

The LCF of the denominators \( \omega, \omega, 3\omega \) is \( \omega \).

Thus, the overall time period \( T \) is given by:
\[ T = \frac{\text{LCM of numerators}}{\text{LCF of denominators}} = \frac{2\pi}{\omega} \]

### Final Result
The time period of the function \( y = \sin(\omega t) + \sin(2\omega t) + \sin(3\omega t) \) is:
\[ T = \frac{2\pi}{\omega} \]