# Establish the equation of motion of a damped oscillator and show that for a weakly damped oscillator, the displacement is given by x(t)=a0exp(-bt)cos(wt+¢)

1
by Nick16

2015-09-25T14:11:03+05:30

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I have done the solution in two ways.  One by deriving the equation and another by assuming the form of the solution.

We have a system oscillating in a SHM with an characteristic angular frequency of w₀.  In the system suppose there is a damping force externally acting on the mass m that oscillates.  The damping force is directly proportional to the velocity and opposite to it.

Suppose we have a spring with a constant k with a mass m.  It is present in a thick gas or  fluid, offering a small resistance to the movement.  Let b be the small damping factor.

Equation of motion then is :
m d² x / dt² = F = -  k x - b v = - k x - b dx/dt    --- (1)

Let us assume that the solution is in the form of  x = A₀ e^{p t } * Sin (wt + B).  This derivation and assumption is explained in the differential equations lessons.  A linear homogeneous differential equation has a solution in the form of exponential and sine wave product.  we need to find w and p.

x' =  dx/dt = A₀ e^{p t} [ w Cos (wt+B) + p Sin (wt+B) ]
x" = A₀ e^{p t} [ - w² Sin(wt+B) + pw Cos (wt+B) + pw Cos (wt+B) + p² Sin(pt+B) ]
= A₀ e^{p t} [ (p² - w²) Sin(wt+B) + 2 pw Cos (wt+B) ]

Substituting in eq (1) we get:
m e^{p t} [ (p² - w²) Sin(wt+B) + 2 pw Cos (wt+B) ]
=  - k e^{p t} Sin(wt+B) - b e^{p t} [ w Cos (wt+B) + p Sin (wt+B) ]

[ m (p² - w²) + k + b p ] Sin(wt+B) + (b + 2pm) w Cos(wt+B) = 0

Since the above is true for all x and t,  and Sine and Cosine are independent and orthogonal functions,  the coefficients must be 0 for the above to be true.
p = -b/2m
m b²/4m²  - mw² + k -b²/2m = 0       =>   w² =  k/m -  b²/4m²
w = √(w₀² - b²/4m²)     where  w₀ = √(k/m)

so we get   x = A₀ e^{-b t/2m} * Sin (w t + B)              ---- (2)

We apply the initial boundary conditions:
At t = 0,    x = A        =>    A =   A₀  Sin B     =>  ∠B =  Sin-1 A/A0.
also,  v = dx/dt = v₀  =>      A₀ [w cos B - b/2m * sin B] = v₀
A/Sin B  * [w cos B - b/2m * Sin B] = v₀
A w Cot B =  (v₀ + Ab/2m)      =>     Tan B = (A m w)/ (m v₀ + A b)
=>   Sin B =  Amw / √[(Amw)² + (m v₀+Ab)²]
A₀  =  A/SinB = √[ (Amw)²+ (m v₀+Ab)²]/mw = √[ A² + (v₀/w + Ab/mw)² ]

So we have found the constants in the eq (2).

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Another way of Derivation using linear homogeneous differential equations of 2nd order:

m d² x / dt² = Force on the body  = - k x - b v = - k x - b dx/dt    --- (1)
m x" + b x' + k x = 0

The auxiliary equation for this is:     m r² + b r + k  = 0
r₁ = [ -b/2m + root(b2/4m2 - k/m) ]   and   r₂ = [ -b/2m - root(b2/4m2 - k/m) ]

Here b is very small.   b2 << 4mk.  so the solutions are imaginary and are:

r₁ =   g + i h   and  r₂ =  g - i h  ,
where   g = -b/2m   and   h = √(k/m - b2/4m²)      --- (3)

we know that e^{(g + i h)t} = e^{g t} [ Cos ht + Sin ht ]

Here  cos h and sin h are orthogonal independent solutions.   So the general solution of differential equation of motion is:

x = A₀ e^{gt} [ c Cos ht + d Sin ht ]  = A₀ e^{gt} Cos (ht + Q)

We apply the initial boundary conditions for deciding constants c and d.
at  t = 0 ,   x = A        =>     A =  A₀ Cos Q
at  t = 0,   x' = v = v₀  =>  A₀ [g Cos(ht+Q) - h Sin (ht+Q)] = v₀
=>  A g - A h Tan Q = v₀   =>   Tan Q = (A g - v₀) / (A h)

Then Cos Q =  A h /√[ A²h² + (A g - v₀)²]
A₀ =  √[ A² + (A g/h - v₀/h)²]      where  g and h are given in  (3)

Thus we have found the equation of motion for displacement of SHM with a small damping factor b.
x =  A0 e^{gt} Cos (ht + Q)        where  Q, A0 and h are found as above.
A₀ =  √[ A² + (A g/h - v₀/h)²]
Tan Q = (A g - v₀) / (A h)
g = -b/2m        h = root(k/m - b²/4m²)         w = √(w₀² - g²)
w₀ = natural frequency of the system with out damping = √(k/m)
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