A particle executes S.H.M. of amplitude 25 cm and time period 3 s. What is the minimum time required for the particle to move between two points 12.5 cm on either side of the mean position ?

Options

A. 0.6 s
B. 0.5 s
C. 0.4 s
D. 0.2 s

Answer

B

Explanation

To get a visual picture, we will consider the oscillation of a simple pendulum, when the bob starts from the extreme left and goes to the right.
When the particel starts from the extreme position, we
use `x=A "cos" omegat= A "cos" ((2pi)/(T))t`
In this case, x=12.5 cm, A=25 cm and T=3 s
`therefore 12.5=25 "cos" ((2pi t)/(3))`
`therefore (1)/(2)=""cos" "(pi)/(3)="cos"((2pit )/(3))`
`therefore (pi)/(3)=(2pi t)/(3) therefore 2t =1 or t =0.5 s`

The time required to travel the dis"tan"ce AO
`=(T)/(4)=(3)/(4)=0.75 s`
`therefore` Time required to go from A' to O=0.75-0.5=0.25s
Similary the time required to go from O to B' =0.25s.
`therefore` Time required to go from A' to B'=0.25+0.25=0.5s