Q. A particle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is β. Then, its time period of vibration will be :
A) $\frac{2\pi\beta}{\alpha}$
B) $\frac{\beta^2}{\alpha^2}$
C) $\frac{\alpha}{\beta}$
D) $\frac{\beta^2}{\alpha}$
Explanation
[Solution] $\omega^2A = \alpha$, $\omega A = \beta \Rightarrow \omega = \alpha/\beta$, so $T = \frac{2\pi\beta}{\alpha}$. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.