A 2 kg particle undergoes SHM acc. to x=1.5sin[ (pi)t/4 +(pi)/6 ]. Find the min. time required to move from x=0.5m to x=-0.75?

rajan jha Grade: 12

        A 2 kg particle undergoes SHM acc. to x=1.5sin[ (pi)t/4 +(pi)/6 ]. Find the min.  time required to move from x=0.5m  to   x=-0.75?

8 years ago

Answers : (1)

vikas askiitian expert

509 Points

							let the particle is at x = +1.5   (extream position)
now time take by particle to reach x=.5 is t₁
from extream position we use x = 1.5cos[ pit/4+pi/6]            (coz phase difference of pi/2)
.5 = 1.5cos[pit/4+pi/6]
 pit/4 + pi/6 = cos^-11/3
                  =7pi/18 (approx)
         t/4 = 4pi/18
        t₁ = 16/18 sec .........1
now, let time taken by particle from extream position to  x=-.75 is t₂ then
- .75 = 1.5cos[pit/4+pi/6]
     pit/4+pi/6 = cos^-1 -(1/2)
                     =2pi/3
     t/4 = 1/2
      t₂ = 2sec    .......2
 the difference t₂ - t₁ will give the time taken by particle from x=.5 to x=-.75
    T = 2 - 16/18
       =1.11sec approx