A particle is executing a simple harmonic motion. Its maximum acceleration is alpha and maximum velocity is beta. Then, its time period of vibration will be………

A particle is executing a simple harmonic motion. Its maximum accelera

1.	A particle is executing a simple harmonic motion. Its maximum acceleration is alpha and maximum velocity is beta. Then, its time period of vibration will be………
Answer» `(2pi BETA)/(ALPHA)` `(beta^2)/(alpha^2)` `(alpha)/(beta)` `(beta^2)/(alpha)` Solution :MAXIMUM acceleration in SHM `a_("max")= A omega^(2)` `therefore alpha = A omega^(2)` and Maximum velocity `v_("max")= A omega` `therefore beta = A omega` `therefore (a_("max"))/(v_("max"))= (A omega^(2))/(A omega)` `(alpha)/(beta)= omega` `therefore (alpha)/(beta)= (2pi)/(T)` `therefore T= 2pi (beta)/(alpha)`.