Physics · 2015
NEET 2015 · 25 July · Code C · Q45
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 particle is executing a simple harmonic motion. Its maximum acceleration is $\displaystyle \alpha$ and maximum velocity is $\displaystyle \beta$. Then, its time period of vibration will be :
Official answer
From NTA’s final answer key for this paper.
(2)
$\displaystyle \frac{2 \pi \beta}{\alpha}$
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