question_answer

A simple harmonic motion is represented by \[\frac{{{d}^{2}}x}{d{{t}^{2}}}+\alpha x=0.\] What is its time period?

Answer:

                Clearly, \[\frac{{{d}^{2}}x}{d{{t}^{2}}}=-\alpha x\] or  \[a=-ax\]                 Time period, \[T=2\pi \sqrt{\frac{x}{a}}=2\pi \sqrt{\frac{x}{\alpha x}}=\frac{2\pi }{\sqrt{\alpha }}\]