1.

The equation of SHM of a particle is given as `2(d^(2)x)/(dt^(2))+32x=0` where x is the displacement from the mean position. The period of its oscillation ( in seconds) is -A. zeroB. `(pi)/(2)`C. `pi`D. `2 pi`

Answer» Correct Answer - B
`(d^(2)x)/(dt^(2))+16x = 0`
`therefore" "omega^(2)=16 rArr omega = 4 and T = (2pi)/(omega)=(2pi)/(4)=(pi)/(2)`


Discussion

No Comment Found

Related InterviewSolutions