Question

In a certain oscillatory system, the amplitude of
motion is 5 m and the time period is 4 s. The time
taken by the particle for passing between points
which are at distances of 4 m and 2 m from the
centre and on the same side of it will be
(a) 0.30 s (b) 0.32 s (c) 0.33 s (d) 0.35 s​

Answer 1

Answer:

C)0.33s

Explanation:

The equation for position of particle at time 't' is given by x=(A)sin(ωt)

ω=2π/T ,where T is time period

  =2π/4

  =π/2

  4=5sin(πt/2)

=>t=(53⁰)2/π=(0.92)(2)/π=0.58s

similarly;

   2=5sin(πt/2)

   t=(0.41)(2)/π =0.26s

the particle is at 4m and 2m when t₁=0.58s and t₂=0.26s

∴ time interval is given by t₁-t₂=0.3248=0.33 (approx)

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