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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?
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
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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