Question

a) A simple harmonic motion is represented by x(t) = a cosωt Obtain expressions for velocity and acceleration of the oscillator. Also, plot the time variation of displacement, velocity and acceleration of the oscillator. b) The time period of a simple pendulum, called ‘seconds pendulum’, is 2 s. Calculate the length, angular frequency and frequency of the pendulum. What is the difference between a simple pendulum and a compound pendulum? (6+4) c) Two collinear harmonic oscillations x1 = 8 sin (100 πt) and x2 = 12 sin (96 πt) are superposed. Calculate the values of time when the amplitude of the resultant oscillation will be (i) maximum and (ii) minimum. (5+5) d) For a damped harmonic oscillator, the equation of motion is ( / ) ( / ) 0 2 2 m d x dt + γ dx dt + kx = with m = 0.50 kg, γ = 0.70 kgs−1 and k = 70 Nm−1. Calculate (i) the period of motion, (ii) number of oscillations in which its amplitude will become half of its initial value, (iii) the number of oscillations in which its mechanical energy will drop to half of its initial value, (iv) its relaxation time, and (v) quality factor. (4+4+4+4+4) e) Establish the equation of motion of a weakly damped forced oscillator explaining the significance of each term. Differentiate between transient and steady state of the oscillator.

Answer 1

x(t)=a cos wt
velocity = -sin wt
Acceleration = DV/DT
= -sin wt
= cos wt

Answer 2

A damped harmonic oscillator:  Equation of motion :
   x = displacement,   t = time,    k = constant (like spring constant)
   ¥ = damping factor (drag coefficient) = 0.70 N/m,
   m = mass of particle executing Periodic/oscillations = 0.50 kg

   m d²x/dt²  + ¥  dx/dt  + k x = 0
   dx²/dt² + 1.40 dx/dt + 140 x = 0
           std form of ODE:  d²x/dt² + p(t) dx/dt + q(t) x = 0
 
   Using the method of solving Ordinary differential equations for second degree equations:
   Let  x(t) = u(t) v(t)
   p(t) = 1.40,  v(t) = exp(-0.70 t),  q(t) = 140
   Q(t) = v''(t) + p(t) v'(t) + q(t) v(t) = 140 exp(- 0.70 t)
   Normal form of ODE:   v(t) u''(t) + Q(t) u(t) = 0
                                        u''(t) +140 u(t) = 0
        This is the equation of motion for a SHM.
      ω² = 140, ω = 2√35 rad/s
      u(t) = A Sin ωt 

Solution :  x(t) = A exp(- 0.70 t) Cos ωt
    If we are given x(t=0), then we can know the value of A.

    Period = 2π/ω =π/√35  Sec = 0.531 sec
 
    Amplitude = A exp(- 0.70 t)
   Time duration for amplitude to be come 1/2 of initial value
         =>   exp(- 0.70 t) = 1/2  
         =>  t = Ln 2 / 0.70 = 0.99 sec
    Number of oscillations = 0.99 /0.531 = 1.86 

  mechanical energy of the oscillator 
     = 1/2 k A² = 1/2 * 70 * A² * exp(- 1.40 t)
  It becomes half in t =  Ln 2 /1.40 = 0.495 sec
      ie., in 0.93 oscillations.

Relaxation time period of a damped oscillator is the time duration for its amplitude become 1/e of its initial value:
      So relaxation time =>   0.70 t = 1,   t = 1.43 sec

Quality factor = Energy stored in the oscillator / Energy lost during one oscillation
           = energy stored in the oscillator /energy lost during one radian of oscillation
 Q = ω/p(t) = 2√35 / 1.40 = 8.45

Here  p(t) = gamma is less than  2 ω.  So it is underdamping.