Identify function which represent periodic motion

Options

A. `e^(omega t)`
B. `log_e(omega t)`
C. ` sin (omega t) + cos (omega t)`
D. `e^(-omega t)`

Answer

To identify which function represents periodic motion, we need to analyze each of the given functions one by one. A function is considered periodic if it repeats its values at regular intervals, known as the period.

### Step-by-Step Solution:

1. **Analyze the first function: \( e^{\omega t} \)**
- The function \( e^{\omega t} \) is an exponential function.
- As \( t \) increases, \( e^{\omega t} \) continuously increases and does not return to any previous value.
- Therefore, \( e^{\omega t} \) is **not periodic**.

2. **Analyze the second function: \( \log(e^{\omega t}) \)**
- The function can be simplified to \( \omega t \) since \( \log(e^x) = x \).
- This function is a linear function that continuously increases without repeating any values.
- Thus, \( \log(e^{\omega t}) \) is **not periodic**.

3. **Analyze the third function: \( \sin(\omega t) + \cos(\omega t) \)**
- Both \( \sin(\omega t) \) and \( \cos(\omega t) \) are well-known periodic functions with a period of \( 2\pi \).
- The sum of two periodic functions (with the same period) is also periodic.
- Therefore, \( \sin(\omega t) + \cos(\omega t) \) is **periodic**.

4. **Analyze the fourth function: \( e^{-\omega t} \)**
- Similar to the first function, \( e^{-\omega t} \) is also an exponential function.
- As \( t \) increases, \( e^{-\omega t} \) continuously decreases and does not return to any previous value.
- Hence, \( e^{-\omega t} \) is **not periodic**.

### Conclusion:
From the analysis, the only function that represents periodic motion is:
- \( \sin(\omega t) + \cos(\omega t) \)