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

d

Explanation

Here amplitude of particle A= 25 cm and time period T= 3s
if the paritcle at t=0 s is at mean position its displacement equation will be
`x=A sin omega t`
if it takes time `t_(1)` to move a distance x= 12.5 cm to one side of its mena position then
`12.5 =25 sin (2pit_(1))/(3)`
`1/2=-sin(2pit_(1))/(3) ro sin (pi)/(6)=sin (2pit_(1))/(3)`
`therefore (pi)/(6)=(2pit_(1))/(3)rarr t_(1)=1/4` s
The same will be the time to move 12.5 cm to the other side of its mean position therefore total time
`t=t_(1)+t_(2)=1/4+1/4=1/2=0.5 `s