Question

The velocities of a particle executing SHM are 4 cm s^-1 and3 cm s^-1 , when its distance from the mean position is 2 cm and 3 cmrespectively. Calculate its amplitude and time period.

Answer 1

x(t) =  A Sin (w t)      let us assume that  initial phase angle is 0.
 
v(t) = A w  Cos wt    
 
  =>  v²  = A² w² Cos² wt = w² (A² - x²)
        4²  = w² (A² -  2²)
       3²  = w² (A² - 3²)
 
Solving the above by dividing one equation by another:
       16 (A² - 3²) = 9 (A² - 2²)
         A² = 108/7
         A = √(108/7) cm
     w  = √7/5 cm

time period =  2π/w =  2π √(5/7)  sec