NEET Class neet Answered
A particle exciting SHM an amplitude of 20 cm & t.p of 12 sec.
Asked by brijk456 | 12 Aug, 2019, 06:46: PM
Expert Answer
Displacement x for SHM is given by, x = A sin(ωt) .....................(1)
where A is amplitude, ω is angular frequency and t is time period. we have ω = 2π/T , where T is time period.
for displacement +10 cm, we have, 10 = 20 sin[ (2π/12) t1 ] , hence phase angle (π/6)t1 = sin-1(1/2) ............(2)
to find the travel time between x=+10 to x=-10, we have to consider that the particle is at x=+10 after reaching the amplitude.
Hence eqn.(2) is written as, (π/6)t1 = sin-1(1/2) = π - (π/6) = 5π/6 or t1 = 5 s
for displacement -10 cm, we have, -10 = 20 sin[ (2π/12) t2 ] , hence phase angle (π/6)t2 = sin-1(-1/2) ............(3)
to find the travel time from x = +10 to x = -10, we have to consider that the
particle is at x = -10 after reaching the equlibrium position(i.e. after half period).
Hence eqn.(3) is written as, (π/6)t2 = sin-1(-1/2) = π + (π/6) = 7π/6 or t2 = 7 s
Hence time duration t to travel from x = +10 to x = -10 is given by, t = t2 - t1 = (7 - 5) s = 2 s
Answered by Thiyagarajan K | 12 Aug, 2019, 09:14: PM
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