Question

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:

Answer 1

For a particle executing SHM, maximum acceleration is

α = Aw^2

A = Maximum amplitude

w = angular velocity.

Maximum Velocity is

β = Aw.

α/β = Aw^2/Aw = w = 2π/T

T - Time period of vibration.

=> T = 2πβ / α  

Answer 2

Answer:

open the picture to get ur ans.... hope it will hlp uh... mark it as brainlistttt