Q143 / 180 · COMEDK 2021 · Physics
The function $y=\log \omega t$ can represent
- Aa periodic function
- Correct answerBnon-periodic function
- Coscillatory function
- Dcirculatory motion
Solution
The function $y = \log(\omega t)$ is defined for $\omega t > 0$.
A function $f(t)$ is periodic if there exists a positive constant $T$ such that $f(t + T) = f(t)$ for all $t$.
For the given function, $\log(\omega(t + T)) = \log(\omega t + \omega T) = \log(\omega t) + \log(1 + \dfrac{T}{t})$.
Since $\log(1 + \dfrac{T}{t}) \neq 0$ for any $T > 0$ and $t > 0$, the condition $f(t + T) = f(t)$ cannot be satisfied for any constant $T$.
Therefore, the function $y = \log(\omega t)$ is a non-periodic function.
Answer: non-periodic function
A function $f(t)$ is periodic if there exists a positive constant $T$ such that $f(t + T) = f(t)$ for all $t$.
For the given function, $\log(\omega(t + T)) = \log(\omega t + \omega T) = \log(\omega t) + \log(1 + \dfrac{T}{t})$.
Since $\log(1 + \dfrac{T}{t}) \neq 0$ for any $T > 0$ and $t > 0$, the condition $f(t + T) = f(t)$ cannot be satisfied for any constant $T$.
Therefore, the function $y = \log(\omega t)$ is a non-periodic function.
Answer: non-periodic function
Hence, the correct answer is option: 'B'.