The maximum velocity ad maximum acceleration of a particle performing a linear S.H.M. are `alpha " and " beta` respectively. Then the path length of the particle is

Options

A. `(alpha(2))/(beta)`
B. `(beta)/(2alpha^(2))`
C. `(2alpha^(2))/(beta)`
D. `(2beta)/(alpha^(2))`

Answer

C

Explanation

`alpha=V_("max")=A omega and a_("max")=Aomega^(2)=beta`
`therefore (alpha^(2))/(beta)=(A^(2)omega^(2))/(A omega^(2))=A`
`therefore " Path length " =(2alpha^(2))/(beta)`