Question

The displacement of the particle performing S.H.M. is given by x = 4 sin πt, where x is in cm and t is in second. The time taken by the particle in second to move from the equilibrium position to the position of half the maximum displacement, is ______.

`[sin30^circ=cos60^circ=0.5, cos30^circ=sin60^circ=sqrt3/2]`

Options

• `1/6`

• `1/2`

• `1/4`

• `1/3`

Answer 1

The displacement of the particle performing S.H.M. is given by x = 4 sin πt, where x is in cm and t is in second. The time taken by the particle in second to move from the equilibrium position to the position of half the maximum displacement, is `underline(1/6)`.

Explanation:

x = 4 sin πt

At t = 0, x = 0

when x = 2 cm we have

2 = 4 sin π t

`thereforesinpit=1/2`

`thereforepit=pi/6`

`thereforet=1/6s`