JEE MAIN - Physics (2005 - No. 42)
If a simple harmonic motion is represented by $${{{d^2}x} \over {d{t^2}}} + \alpha x = 0.$$ its time period is
$${{2\pi } \over {\sqrt \alpha }}$$
$${{2\pi } \over \alpha }$$
$$2\pi \sqrt \alpha $$
$$2\pi \alpha $$
توضیح
$${{{d^2}x} \over {d{t^2}}} = - \alpha x = - {\omega ^2}x$$
$$ \Rightarrow \omega = \sqrt \alpha $$ $$\,\,\,\,$$ or $$\,\,\,\,$$ $$T = {{2\pi } \over \omega } = {{2\pi } \over {\sqrt \alpha }}$$
$$ \Rightarrow \omega = \sqrt \alpha $$ $$\,\,\,\,$$ or $$\,\,\,\,$$ $$T = {{2\pi } \over \omega } = {{2\pi } \over {\sqrt \alpha }}$$