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QAIPM-2015-Ph-1243: NEET 2015 Physics - Class 11 AIPM-2015-Ph-1243 Solution: A particle is executing a simple harmonic motion. Its maximum…

A particle is executing a simple harmonic motion. Its maximum acceleration is α\alpha and maximum velocity is β\beta. Then its time period of vibration will be βα\frac{\beta}{\alpha} 2παβ\frac{2\pi\alpha}{\beta} αβ\frac{\alpha}{\beta} β2α2\frac{\beta^2}{\alpha^2}

Multiple Choice Options

  • A β2α{{{\beta ^2}} \over \alpha } _ _ 2 _
  • B 2πβα{{2\pi \beta } \over \alpha } _ 2 � _ _
  • C β2α2{{{\beta ^2}} \over {{\alpha ^2}}} _ 2 _ 2 _
  • D αβ{\alpha \over \beta } _ _ _

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