Question

A particle starting from mean position oscillates simple harmonically with period 4 s. After what time will its kinetic energy be 75% of the total energy?

`( cos 30° = sqrt(3/2))`

विकल्प

• `1/4` s

• `1/3` s

• `1/2` s

• `1/5` s

Answer 1

`1/3`s

Explanation:

T = 4s

TE = `1/2 "m" omega^2"A"^2   "KE" = 1/2 "m" omega^2 ("A"^2 - "X"^2)`

`3/4 xx 1/2 "m"omega^2 "A"^2 = 1/2 "m" omega^2 ("A"^2 - omega^2)`

` 3/4 "A"^2 = "A"^2 -"X"^2`

`therefore "X"^2 = "A"^2 - 3/4 "A"^2 = 1/4 "A"^2`

`"X" ="A"/2`

X = A sin `pi/2` t

sin `pi/6` = sin `(pi"t")/2`       `therefore pi/6 = pi"t"/2 therefore "t" = 1/3s`