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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?

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?

Grade:12

1 Answers

vikas askiitian expert
509 Points
10 years ago

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

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