# 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+¢)

by Nick16 22.09.2015

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by Nick16 22.09.2015

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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).

===============================================

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. s*o* 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)

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 :

Let us assume that the solution is in the form of

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

so we get

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) =>

=> Sin B = Amw / √[(Amw)² + (m v₀+Ab)²]

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

===============================================

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

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. s

r₁ = g + i h and r₂ = g - i h ,

where

we know that e^{(g + i h)t} = e^

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 ] =

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 =>

Then Cos Q = A h /√[ A²h² + (A g - v₀)²]

Thus we have found the equation of motion for displacement of SHM with a small damping factor b.

g = -b/2m h = root(k/m - b²/4m²)