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JEE PHYSICS MCQ: OSCILLATIONS

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If a simple harmonic is represented by d2xdt2\frac{{{d^2}x}}{{d{t^2}}}+αx=0+ \alpha x = 0,its time period is
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Correct option: B
2πα\frac{{2\pi }}{{\sqrt \alpha }}
Option B is correct because the time period of a simple harmonic motion is given by T=2πωT = \frac{{2\pi}}{{\omega}}, where ω=α\omega = \sqrt{{\alpha}} in this case.

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