A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is…….

A particle executes linear simple harmonic motion with an amplitude of

1.	A particle executes linear simple harmonic motion with an amplitude of 3 cm. When the particle is at 2 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is…….
Answer» SOLUTION :Velocity in SHM `v= pm OMEGA SQRT(A^(2)-X^(2))` and Acceleration `a= -omega^(2)x` but `\|v\|=\|a\|` `therefore omega sqrt(A^(2)-x^(2))= omega^(2)x` `sqrt(A^(2)-x^(2))= omega x` `therefore A^(2)- x^(2) = omega^(2) x^(2)` `therefore omega^(2) = (A^(2)-x^(2))/(x^(2))` `= ((3)^(2)-(2)^(2))/((2)^(2)` `= (9-4)/(4)` `=(5)/(4)` `therefore omega = (sqrt(5))/(2)` Now`omega = (2pi)/(T)"" therefore T= (2pi)/(omega)` `therefore T= (2pi)/((sqrt(5))/(2))= (4pi)/(sqrt(5))`.