Question

A particle executing SHM has angular frequency of $pi rad s^{-1}$ and amplitude of $5 m$. Deduce(i) time period(ii) maximum velocity(iii) maximum acceleration(iv) velocity when the displacement is 3 m.

Answer 1

in SHM, the displacement of a particle from mean equilibrium position is given by
    
  x  = A  Sin (ω t + Ф)

 given  ω = angular freq = π rad/ sec
           A = amplitude = 5 m

 Time period = T = 2π /ω = 2 Seconds.
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instantaneous velocity of the particle:  v = dx/dt = A ω Cos (ω t + Ф)
  maximum velocity = A ω = 5 m * π rad/sec = 5 π m/sec.
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acceleration =  dv/dt = - A ω² Sin (ωt +Ф)
maximum acceleration = A ω² = 5 * π²  m/sec²

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displacement    x =  5 Sin (ωt + Ф) = 3 meters
 =>  Sin (ωt + Ф) = 3/5
=>  Cos (ωt + Ф) = √(5²-3²) /5 = 4/5
 
velocity = v = A ω Cos (ω t + Ф) = 5 * π * 4/5 = 4 π    m/sec