__Damped oscillations__

The differential equation describing the displacement from equilibrium for
damped harmonic motion is

md^{2}x/dt^{2} + kx + cdx/dt = 0.

(a) State the conditions and find an expression for x(t) for underdamped,
critically damped, and overdamped motion.

(b) Show that for underdamped motion, the ration of two successive maxima in
the displacement x is constant.

Solution:

- Concepts:

The damped harmonic oscillator - Reasoning:

For underdamped motion we have oscillations with and exponentially decreasing amplitude. For critically and overdamped motion we have no oscillations and at most one zero crossing as the system returns to equilibrium. - Details of the calculation:

(a) Let x = Aexp(iΩt). (The real part is implied.) Then

-Ω^{2}+ k/m + iΩc/m = 0, Ω = ic/(2m) ± (ω_{0}^{2}– c^{2}/(2m)^{2})^{½}, with ω_{0}^{2}= k/m.

Let γ = c/(2m). Ω = iγ ± (ω_{0}^{2}– γ^{2})^{½}. Let ω^{2}= (ω_{0}^{2}– γ^{2}).

If ω_{0}^{2}> γ^{2}, or k/m > c^{2}/(2m)^{2}, then ω is real.

x = |A|exp(-γt)cos(ωt + δ), |A| and δ depend on the initial conditions.

We have underdamped motion.

If ω_{0}^{2}= γ^{2}, then ω is zero, we have critically damped motion

x = Aexp(-γt) is one solution to the differential equation d^{2}x/dt^{2}+ γ^{2}x + 2γdx/dt = 0,

x = Bt exp(-γt) is another solution.

The most general solution is x = Aexp(-γt) + Bt exp(-γt).

A and B depend on the initial conditions.

If ω_{0}^{2 }< γ^{2}, then ω^{ }is imaginary, ω =^{ }±i(γ^{2}– ω_{0}^{2})^{½}= ±iβ.

We have overdamped motion.

x = Aexp(-(γ + β)t) + Bexp(-(γ - β)t. A and B depend on the initial conditions.

(b) Maxima occur when dx/dt = 0, -γcos(ωt + δ) - ωsin(ωt + δ) = 0,

tan(ωt + δ - nπ) = -γ/ω,

ωt = nπ - δ - tan^{-1}(γ/ω) ∆t = (π/ω).

The ratio of successive maximum values of x(t) is R = exp(-γπ/ω).

A damped oscillator satisfies the equation d^{2}x/dt^{2} +
2γdx/dt + ω_{0}^{2}x = 0, where γ and ω_{0} are
positive constants with γ < ω_{0} (under-damping). Assume that the
equation of motion of a particle of mass m is given by this equation. At time t
= 0, the particle is released from rest at the point x = a.

(a) Find x as a function of time t and sketch of x as a function of t.

(b) Find all the turning points of the motion and determine the ratio of
successive maximum values of x(t).

(c) Re-do part( a) for the case of critical damping, i.e. when γ = ω_{0}.

Solution:

- Concepts:

The underdamped harmonic oscillator - Reasoning:

For underdamped motion we have oscillations with and exponentially decreasing amplitude. For critically and overdamped motion we have no oscillations and at most one zero crossing as the system returns to equilibrium. - Details of the calculation:

(a) d^{2}x/dt^{2}= -ω_{0}^{2}x - 2γdx/dt.

Then the solution for underdamped motion is

x(t) = a exp(-γt) cos(ω_{1}t + δ), with ω_{1}^{2}= ω_{0}^{2}– γ^{2}.

Here ω_{0}is the angular frequency of an undamped oscillator with the same spring constant and δ ~ 0, since γ < < ω_{0}, (tanδ = - γ/ω_{1}).

(b) Zeroth order approximations: γ < < ω

_{0}. The turning points of the motion occur when

ω_{1}t = (ω_{0}^{2}– γ^{2})^{½}t = nπ, cos(ω_{1}t) = ±1. Then t = (nπ/ω_{1}).

The ratio of successive maximum values of |x(t)| is R = exp(-γ(t + ∆t)/exp(-γt) = exp(-γ∆t),

where ∆t = (π/ω_{1}). R = exp(-γπ/ω_{1}). ω_{1}≈ ω_{0}(1 – ½γ^{2}/ω_{0}^{2}).

Higher order: dx/dt = 0. -γcos(ω_{1}t) - ω_{1}sin(ω_{1}t) = 0, tan(ω_{1}t - nπ) = -γ/ω_{1},

ω_{1}t = nπ - tan^{-1}(γ/ω_{1}) ∆t = (π/ω_{1}).

The turning points of the motion occur when t = nπ/ω_{1}- tan^{-1}(γ/ω_{1})/ω_{1}, earlier than t = nπ/ω_{1}.

The ratio of successive maximum values of |x(t)| stays R = exp(-γπ/ω_{1}).

(c) When γ = ω_{0}then ω_{1}= 0 and x(t) = (a + bω_{0}t) exp(-ω_{0}t).

Consider a damped harmonic oscillator. Let us define T_{1} as
the time between adjacent zero crossings, 2T_{1} as its “period”, and ω_{1}
= 2π/(2T_{1}) as its “angular frequency”. If the amplitude of the
damped oscillator decreases to 1/e of its initial value after n periods, show
that the frequency of the oscillator must be approximately [1 - (8π^{2}n^{2})^{-1}]
times the frequency of the corresponding undamped oscillator.

Solution:

- Concepts:

The underdamped harmonic oscillator - Reasoning:

The oscillator is underdamped, since it crosses the equilibrium position many times. - Details of the calculation:

The equation of motion for the damped harmonic oscillator is d^{2}x/dt^{2}+ 2βdx/dt + ω_{0}^{2}x = 0.

The solution for underdamped motion is x(t) = A exp(-βt) cos(ω_{1}t - δ), with ω_{1}^{2}= ω_{0}^{2}– β^{2}. Here ω_{0}is the angular frequency of an undamped oscillator with the same spring constant.

After n periods, nT = 2nT_{1}, the amplitude of the oscillator decreases to A/e.

So exp(-1) = exp(-βnT), βnT_{ }= 1, β = 1/nT.

With T = 2π/ω_{1}we have β = ω_{1}/(n2π).

ω_{1}^{2}= ω_{0}^{2}– (ω_{1}/n2π.)^{2}, ω_{1}^{2}+ ω_{1}^{2}/(4n^{2}π^{2}) = ω_{0}^{2}.

ω_{1}= ω_{0}[1 + 1/(4n^{2}π^{2})]^{–½}.

(1 + x)^{-½}= 1 – ½ x + … .

ω_{1}≈ ω_{0}[1 - 1/(8n^{2}π^{2})].

A mass on a spring moves in one dimension and is subject to a velocity-dependent force. The net force is F = -kx – bv. Assuming is small, what fraction of energy is dissipated in each cycle?

Solution:

- Concepts:

The underdamped harmonic oscillator - Reasoning:

The oscillator is underdamped, since it goes through many cycles. - Details of the calculations:

md^{2}x/dt^{2}= -kx - bdx/dt, d^{2}x/dt^{2}= -(k/m)x - (b/m)dx/dt.

Let ω_{0}^{2}= k/m, β = b/2m.

Then the solution for underdamped motion is x(t) = A exp(-βt) cos(ω_{1}t - δ),

with ω_{1}^{2}= ω_{0}^{2}– β^{2}.

Here ω_{0}is the angular frequency of an undamped oscillator with the same spring constant.

The energy is proportional to the square of the effective amplitude, (A exp(-βt))^{2}. After one cycle the amplitude decreases by a factor of exp(-βτ), where τ = 2π/ω_{1}. The energy decreases by a factor of exp(-2βτ).

If β is small, then exp(-2βτ) = 1 - 2βτ.

The fraction of the energy dissipated in each cycle is

2βτ = (b/m)(4πm)/(4k - b^{2})^{½}= (4πb)/(4k - b^{2})^{½}.

The terminal speed of a freely falling object is
v_{t} (assume the drag force is proportional to the speed of the
object). When the object is suspended by a spring, while still influenced by
the same drag force, the spring stretches by an amount x_{0}. Show that
the frequency of oscillation of the object (when it is suspended by the spring)
is ω, where ω^{2} = g/x_{0} – g^{2}/(4v_{t}^{2}).

Solution:

- Concepts:

The damped harmonic oscillator.. - Reasoning:

The equation of motion for the damped harmonic oscillator with the x-axis pointing up is md^{2}x/dt^{2}= -γdx/dt – kx – mg. Here x is the displacement of the free end of the spring from its equilibrium position. The terminal velocity v_{t}lets us calculate γ, and x_{0}lets us calculate the spring constant k. - Details of the calculation:

When the system is not accelerating: mg = γv_{t}, γ = mg/v_{t}. mg = kx_{0}, k = mg/x_{0}.

d^{2}x/dt^{2}= -(g/v_{t})dx/dt – (g/x_{0})x – mg.

Try a solution x = -x_{0}+ A(exp(ict) for the equation of motion. Here c = -iβ + ω

-mc^{2}A(exp(ict) = -iγc A(exp(ict) + kx_{0}- kA(exp(ict) – mg. c^{2}– i(γ/m)c – k/m = 0 .

c = -iγ/(2m) + (k/m – γ^{2}/(4m^{2}))^{½}. ω^{2}= k/m – γ^{2}/(4m^{2}) = g/x_{0}– g^{2}/(4v_{t}^{2}).

Consider a pendulum consisting of a rigid thin rod with length L and
negligible mass supporting a ball of mass M. The pendulum is immersed in a
viscous medium which causes a frictional force F whose magnitude is proportional
to the speed v of the ball, F = -μv. It swings in a vertical plane under
the influence of gravity.

(a) Derive the equation of motion of the pendulum,
allowing for arbitrary angles θ of deflection from the vertical axis.

(b)
Determine the fixed points for which d^{2}θ/dt^{2} = 0 when dθ/dt
= 0. Determine for each fixed point the critical value of the drag
coefficient μ above which there is no oscillation about the point for small
displacements.

Solution:

- Concepts:

Newton's 2^{nd}law - Reasoning:

Newton's second law gives the equation of motion. We are asked to find the equilibrium points, where the acceleration is zero. The equation of motion will tell us if small displacements away from an equilibrium point will lead to oscillations. - Details of the calculation:

F = Ma

(a) Equation of motion: MLd^{2}θ/dt^{2}= -Mgsinθ – μLdθ/dt

d^{2}θ/dt^{2}+ (μ/M)dθ/dt + (g/L)sinθ = 0

(b) Fixed points: θ = 0 and θ = π.

For small deviations from θ = 0 we have d^{2}θ/dt^{2}+ (μ/M)dθ/dt + (g/L)θ = 0.

θ = θ_{0}exp(iωt), ω^{2}- iωm/M - g/L = 0, ω = iμ/(2M) ± (-μ^{2}/(4M^{2}) + g/L)^{½}= iα ± β.

(The displacement as a function of time is given by the real part of this solution.)

If μ < 2M(g/L)^{½}, then β is real and θ = θ_{0}exp(-αt) exp(±iβt), we have oscillations.

If μ > 2M(g/L)^{½}, then β is imaginary and we have no oscillations.

For small deviations from θ = π we have d^{2}φ/dt^{2}+ (μ/M)dφ/dt - (g/L)φ = 0, where φ = π + θ.

φ = φ_{0}exp(λt), λ^{2}+ λm/M - g/L = 0, λ = -μ/(2M) ± (μ^{2}/(4M^{2}) + g/L)^{½}.

λ is always real, we have no oscillations for any value of μ.

A grandfather clock has a pendulum of length L = 1 m and a bob of mass m = 0.5
kg. A mass of 2 kg falls 0.7 m in seven days to keep the amplitude if the
pendulum oscillations steady at 0.03 rad.

(a) The quality factor Q of a damped oscillator is defined as

Q = 2π*(average energy)/(energy lost per cycle).

What is the Q of this system?

(b) With no energy input from the falling mass, the pendulum obeys the small
angle equation of motion

d^{2}θ/dt^{2} + 2bdθ/dt + ω_{0}^{2}θ = 0.

Find b and ω_{0}.

Solution:

- Concepts:

The driven, damped oscillator - Reasoning:

With no energy input, the oscillator is underdamped. The damping is weak. - Details of the calculation:

(a) The rate at which energy is dissipated is

dE/dt = -(2 kg*9.8 m/s^{2}*0.7 m)/(7*24*3600 s) = -2.3*10^{-5}J/s.

The average energy of an oscillator is E = ½kA^{2}. Here k = mg/L and A = Lθ.

E = ½*0.5 kg*9.8 m/s^{2}*1 m*(0.03)^{2}= 2.2*10^{-3}J.

Since the rate of energy dissipation is small, the period of the pendulum is nearly equal to its natural period T = 2π(L/g)^{½}= 2s.

Therefore Q = 2π*(2.2*10^{-3}J)/( 2.3*10^{-5}J/s*2s) = 300.

(b) ω_{0}= (g/L)^{½}= 3.13/s

Then the solution for underdamped motion is x(t) = A exp(-βt) cos(ω_{1}t + δ),

with ω_{1}^{2}= ω_{0}^{2}– β^{2}.

The energy is proportional to the square of the effective amplitude, (A exp(-βt))^{2}.

After one cycle the ration of the amplitudes is A(t + T)/A(t) = exp(-βT).

The ration of the energies is E(t + T)/E(t) = exp(-2βT).

If β is small, then exp(-2βT) = 1 - 2βT.

The fraction of the energy dissipated in each cycle is |(E(t + T) - E(t))/E(t) | = 2βT.

2βT = 2π/Q. β = π/(2s*300) = 5.2*10^{-3}/s.

A mass m fixed to a spring of spring constant k and
immersed in a fluid provides a model for a damped harmonic oscillator. A
circuit with inductance L, capacitance C and resistance R provides an electric
analog to such an oscillator.

(a) Write out the circuit equation for the LCR circuit and Newton's second law
of motion for the damped oscillator. What assumption must be made about the
dependence of the damping of the mass on velocity for the two equations to have
the same mathematical form?

(b) How are the constants m, k, and b (damping constant) related to the circuit
constants L, C and R? To what parameters of the electric circuit are the
mechanical quantities x (displacement) and v (velocity) related?

(c) Derive and expression for the displacement and velocity in the limit of
weak damping.

(d) What voltages measured in the circuit would give values proportional to the
displacement and velocity of the mechanical oscillator?

__Driven and damped oscillations__

Consider a damped harmonic oscillator. Let us define T_{1} as the
time between adjacent zero crossings, 2T_{1} as its “period”, and ω_{1}
= 2π/(2T_{1}) as its “angular frequency”. The damped harmonic
oscillator is characterized by the quality factor Q = ω_{1}/(2β), where
1/β is the relaxation time, i.e. the time in which the amplitude of the
oscillation is reduced by a factor of 1/e.

(a) After 8.6 seconds and 5 periods of oscillations, the amplitude of a
damped oscillator decreased to 17% of its originally set value. Determine
the quality factor Q of the system.

(b) A motor now supplies an external driving term M cosωt. Determine
the stationary solution for the now driven and damped oscillator.

(c) For a system with an equation of motion

d^{2}x/dt^{2} + 2β dx/dt + ω_{0}^{2}x = A cos(ωt)

the quality factor Q is defined as Q = ω_{R}/(2β), where ω_{R}
is the angular frequency for amplitude resonance. Compare the shape of a
plot of the amplitude of the oscillations versus the driving frequency Ω for
the above driven and damped oscillator with a Lorentzian atomic line shape
for Q = 4 and Q = 10^{7}.

Solution:

- Concepts

The underdamped harmonic oscillator, the driven oscillator - Reasoning:

The oscillator in part (a) is underdamped, since it crosses the equilibrium position many times. For part (b) a harmonic driving force is given. - Details of the calculations:

(a) The equation of motion for the damped harmonic oscillator is d^{2}x/dt^{2}+ 2βdx/dt + ω_{0}^{2}x = 0.

The solution for underdamped motion is x(t) = A exp(-βt) cos(ω_{1}t - δ), with ω_{1}^{2}= ω_{0}^{2}– β^{2}.

Here ω_{0}is the angular frequency of an undamped oscillator with the same spring constant.

Aexp(-β*8.6 s) = 0.17 A, -β*8.6 s = ln(0.17) , β = 0.206/s.

5(2T_{1}) = 8.6 s, ω_{1}= 2π/(2T_{1}) = 3.651, Q = ω_{1}/(2β) = 8.87.

(b) The equation of motion for the harmonically driven harmonic oscillator (with damping) is d^{2}x/dt^{2}+ 2βdx/dt + ω_{0}^{2}x = Acosωt, with A = M/m.

The stationary solution x(t) = Dcos(ωt - δ) with tanδ = 2ωβ/(ω_{0}^{2 }- ω^{2}) and D = A/[(ω_{0}^{2 }- ω^{2})^{2}+ 4ω^{2}β^{2}]^{½}.

[The expressions for tanδ and D are found by substituting x(t) = Dcos(ωt - δ) into the differential equation d^{2}x/dt^{2}+ 2βdx/dt + ω_{0}^{2}x = Acosωt.]

When ω_{ }= ω_{R }= [ω_{0}^{2}- 2β^{2}]^{½}we have amplitude resonance. [dD/dω = 0 when evaluated at ω_{R}.]

For an undamped oscillator ω_{R }= ω_{0}and Q = infinity. The amplitude increases without limit when the driving frequency ω approaches ω_{0}.

(c) When Q = 4, we have ω_{R}= 8β, so at amplitude resonance is D ≈ A/(2ω_{R}β) = 4A/ω_{R}^{2}.

When Q = 10^{7}, we have ω_{R}= 2 10^{7}β, so at amplitude resonance D ≈ A/(2ω_{R}β) = 10^{7}A/ω_{R}^{2}. The shape of the resonance curve approaches that of an undamped oscillator.

The figure below shows plots of amplitude versus driving frequency for different values of β. (A = 1, ω_{0 }= 10, β = 0.01 (blue), 0.1 (pink), 0.5 (orange).)

- Additional information:

For a lightly damped oscillator we have Q ≈ ω_{0}/Δω, where Δω represents points on the amplitude resonance curve that are 2^{-½}= 0.707 of the maximum amplitude.

Proof:

For a lightly damped oscillator ω_{R}≈ ω_{0 }and Q ≈ ω_{0}/(2β). The amplitude at resonance is D ≈ A/(2ω_{0}β). The driving frequency for which the amplitude is 2^{-½}of the maximum amplitude is found from A/(2^{-½}2ωβ) = A/[8ω^{2}β^{2}]^{½}= A/[(ω_{0}^{2}-ω^{2})^{2}+ 4ω^{2}β^{2}]^{½}.

8ω^{2}β^{2}≈ (ω_{0}^{2}-ω^{2})^{2}+ 4ω^{2}β^{2}, 4ω^{2}β^{2}= (ω_{0}^{2}-ω^{2})^{2}, 2ωβ = (ω_{0}^{2}-ω^{2}) = (ω_{0}+ω)(ω_{0}-ω).

2ωβ ≈ 2ωΔω/2, since (ω_{0}+ω) ≈ 2ω_{0}near resonance. This yields 2β ≈ Δω and Q ≈ ω_{0}/Δω.

Q becomes large when Δω becomes small compared to ω_{0}.

A 0.500 kg mass is attached to a spring of constant 150 N/m. A driving
force F(t) = (12.0 N) cos(ωt) is applied to the mass, and the damping coefficient
b is 6.00 Ns/m. What is the amplitude (in cm) of the steady-state motion if
ω is equal to half of the natural frequency ω_{0} of the system?

Solution:

- Concepts:

Forced oscillations - Reasoning:

The spring constant, driving force, and damping coefficient for a forced harmonic oscillator are given and we are asked to find the steady-state amplitude. - Details of the calculation:

The equation of motion for the harmonically driven harmonic oscillator [F(t) = F_{0}cos(ωt)] with damping coefficient b = 2βm is d^{2}x/dt^{2}+ 2βdx/dt + ω_{0}^{2}x = Acosωt, with A = F_{0}/m.

The stationary solution x(t) = Dcos(ωt - δ) with tanδ = 2ωb/(ω_{0}^{2}- ω^{2}) and D = A/[(ω_{0}^{2}- ω^{2})^{2}+ 4ω^{2}β^{2}]^{½}.

[The expressions for tanδ and D are found by substituting x(t) = Dcos(ωt - δ) into the differential equation d^{2}x/dt^{2}+ 2βdx/dt + ω_{0}^{2}x = Acosωt.}

In this problem

D = (F_{0}/m)/[(ω_{0}^{2 }- ω^{2})^{2 }+ (bω/m)^{2}]^{½},

ω_{0}^{2}= k/m = 300, ω = ω_{0}/2, ω^{2}= 300/4 = 75.

F_{0}/m = 12/0.5 = 24.

D = 24/[(300 - 75)^{2}+ 36*75/0.25]^{½}= 0.0968 m = 9.7 cm is the amplitude of the steady state motion.

A particle of mass m = 1 oscillates without friction attached to a spring with k = 4. The motion of the particle is driven by the external force F(t) = 3 t cos(t). Find the equation of motion and solve it. Discuss the physical meaning of the solution.

Solution:

- Concepts:

Driven oscillations - Reasoning:

The amplitude of the driving force increases linearly with time. For large t we expect a term x proportional to tcos(t) to dominate the solution. For small t, x = Ccos(2t + φ), the solution to

d^{2}x/dt^{2}+ 4x = 0, may dominate, depending on the initial conditions, which determine the constants C and φ. - Details of the calculation:

F = md^{2}x/dt^{2}= -kx + F(t), d^{2}x/dt^{2}+ 4x - 3t cos(t) = 0 is the equation of motion.

In complex notation: d^{2}x/dt^{2}+ 4x - 3t exp(it) = 0.

Try x = A(t) exp(it).

Then d^{2}A/dt^{2}+2idA/dt + 3A – 3t = 0.

Let A = t + f(t). (This eliminates the term proportional to t.)

Then d^{2}f/dt^{2}+ 2i(1 + df/dt) + 3f = 0.

Let f = -2i/3 + g(t). (This eliminates the constant term.)

Then d^{2}g/dt^{2}+ 2i dg/dt + 3g = 0.

For solutions with g not equal to zero, let g = Cexp(iat)

Then -a^{2}- 2a + 3 = 0, a = -1 ± 2.

Most general solution:

x = (t - 2i/3 + C_{1}exp(it) + C_{2}exp(-3it))exp(it)

x = (t - 2i/3)exp(it) + C_{1}exp(2it) + C_{2}exp(-2it)

x = texp(it) + ⅔exp(i(t - π/2)) + C_{1}exp(2it) + C_{2}exp(-2it)

Take the real part: x = tcos(t) + ⅔sin(t) + Ccos(2t + φ).

The constants C and φ are determined by the initial conditions.

A mass m hangs in equilibrium by a spring which exerts a force F = -K(x - l), where x is the length of the spring and l is its length when relaxed. At t = 0 the point of support P to which the upper end of the spring is attached begins to oscillate sinusoidally up and down with amplitude A, angular frequency ω as shown (x' = A sin(ωt)). Set up and solve the equation of motion for x(t).

Solution:

- Concepts:

Forced oscillations - Reasoning:

The problem is equivalent to a forced oscillation problem with no damping. - Details of the calculation:

Take the upper end of the spring, P, as the origin of the x coordinate of the mass m. Let the x-axis point downward.

T = ½m[(d/dt)(x + x']^{2}= ½m[(dx/dt)^{2}+ (dx'/dt)^{2}+ 2(dx/dt)(dx'/dt)].

U = -mg(x + x') + ½K(x - l)^{2}. L = T - U.

dL/dv_{x}= mdx/dt + mdx'/dt. d/dt(dL/dv_{x}) = md^{2}x/dt^{2}- mω^{2}Asin(ωt).

dL/dx = mg - K(x - l).

Equation of motion: md^{2}x/dt^{2}- mω^{2}Asin(ωt) = mg - K(x - l).

d^{2}x/dt^{2}+ (K/m)(x - l - mg/K) = ω^{2}Asin(ωt).

Let y = x - l - mg/K, ω_{0}^{2}= K/m. Then the equation of motion becomes d^{2}y/dt^{2}+ ω_{0}^{2}y = ω^{2}Asin(ωt).

Try a solution y = Bsin(ωt). This yields B = ω^{2}A/(ω_{0}^{2}- ω^{2}). To find the most general solution we add the homogeneous solution y = Csin(ω_{0}t + δ).

Most general solution: y(t) = Csin(ω_{0}t + δ) + [ω^{2}A/(ω_{0}^{2}- ω^{2})]sin(ωt).

Initial conditions: mg = K(x - l), x = mg/K + l, y = 0 and dy/dt = 0 at t = 0.

This implies δ = 0 and C = -ω^{3}A/(ω_{0}(ω_{0}^{2}- ω^{2})).

y(t) = -[ω^{3}A/(ω_{0}(ω_{0}^{2}- ω^{2}))]sin(ω_{0}t) + [ω^{2}A/(ω_{0}^{2}- ω^{2})]sin(ωt).

x(t) = -[ω^{2}A/(ω_{0}^{2}- ω^{2})][(ω/ω_{0})sin(ω_{0}t) + sin(ωt)] + mg/K + l.