Flag Wave Motion> Lead 3...
question mark

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 , 14 Years ago
Grade 12
anser 1 Answers
vikas askiitian expert

Last Activity: 14 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

star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments