Question

The displacement of the particle executing linear S.H.M. is x = 0.25 Sill ( 11 l + 0.5) m. The period of S.H.M. is ______.

(`pi = 22/7` )

Options

• `3/7`s

• `2/7`s

• `4/7`s

• `1/7`s

Answer 1

The displacement of the particle executing linear S.H.M. is x = 0.25 Sill ( 11 l + 0.5) m. The period of S.H.M. is `4/7`s.

Explanation:

x = 0.25 sin(11 t + 0.5)

Comparing with x =A sin `(omega"t" + phi)`

∴ `omega` = 11

∴ Time Period, T = `(2pi)/omega= (2 xx 22)/(7 xx 11) = 4/7`s