For a damped harmonic oscillator, the equation of motion is ( / ) ( / ) 02 2m 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
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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
k = 70 N/m
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.
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
k = 70 N/m
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.
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