Damped and Forced Harmonic Motion (AQA A-Level Further Maths): Revision Notes

Worked Example 1: Damped Harmonic Motion with Complex Roots

Problem: A particle $P$ moves along a horizontal line under the action of a force directed towards a fixed point $O$. The displacement $x$ metres of $P$ from its initial position, at time $t$ seconds from the start, satisfies the differential equation:

$\frac{d^2x}{dt^2} + 2\frac{dx}{dt} + 17x = 34$

At time $t = 0$, the particle is moving through the point $x = 2$ with velocity $20 , \text{m/s}$.

a) Solve the differential equation to obtain an expression for $x$ in terms of $t$

b) Sketch the graph of $x$ against $t$

Solution:

a) First, find the auxiliary equation:

$m^2 + 2m + 17 = 0$

Using the quadratic formula:

$(m+1)^2 = -16$

$m = -1 \pm 4i$

This gives complex roots, indicating the complementary function has the form:

$x = e^{-t}(A\cos 4t + B\sin 4t)$

For the particular integral, try $x = c$ (constant):

$17c = 34 \implies c = 2$

So the particular integral is $x = 2$.

The general solution is:

$x = e^{-t}(A\cos 4t + B\sin 4t) + 2$

Now apply initial conditions. At $t = 0$, $x = 2$:

$A + 2 = 2 \implies A = 0$

The solution is now:

$x = Be^{-t}\sin 4t + 2$

Differentiate to find velocity:

$\frac{dx}{dt} = -Be^{-t}\sin 4t + 4Be^{-t}\cos 4t$

At $t = 0$, $\frac{dx}{dt} = 20$:

$4B = 20 \implies B = 5$

Therefore, the particular solution is:

$x = 2 + 5e^{-t}\sin 4t$

b) The graph oscillates about the line $x = 2$ with decreasing amplitude due to the damping term $e^{-t}$.

Key observation: This is called damped harmonic motion. The graph oscillates about $x = 2$. The damping effect is caused by the term $2\frac{dx}{dt}$.

Worked Example 2: Heavy Damping

Problem: A particle moves along a horizontal line under the action of a force directed towards a fixed point $O$. The displacement $x$ metres of the particle from its initial position, at time $t$ seconds from the start, satisfies the differential equation:

$\frac{d^2x}{dt^2} + 5\frac{dx}{dt} + 6x = 3$

At time $t = 0$, the particle is moving through the point $x = 1$ with velocity $3 , \text{m/s}$.

a) Solve the differential equation to obtain an expression for $x$ in terms of $t$

b) Find the time at which the particle is at rest

c) Sketch the graph of $x$ against $t$

d) Explain whether the type of damping is light, critical or heavy

Solution:

a) The auxiliary equation is:

$m^2 + 5m + 6 = 0$

Factorising:

$(m+2)(m+3) = 0$

$m = -2 \text{ or } m = -3$

This gives real distinct roots, so the complementary function is:

$x = Ae^{-2t} + Be^{-3t}$

For the particular integral, try $x = c$:

$6c = 3 \implies c = \frac{1}{2}$

The general solution is:

$x = Ae^{-2t} + Be^{-3t} + \frac{1}{2}$

Differentiate:

$\dot{x} = -2Ae^{-2t} - 3Be^{-3t}$

Apply initial conditions. At $t = 0$, $x = 1$:

$A + B + \frac{1}{2} = 1 \implies A + B = \frac{1}{2}$

At $t = 0$, $\dot{x} = 3$:

$-2A - 3B = 3$

Solving simultaneously:

$A = 4.5, \quad B = -4$

The particular solution is:

$x = 4.5e^{-2t} - 4e^{-3t} + 0.5$

b) Set velocity equal to zero:

$\frac{dx}{dt} = -9e^{-2t} + 12e^{-3t} = 0$

$9e^{-t} = 12$

$e^{-t} = \frac{4}{3}$

$t = \ln\left(\frac{4}{3}\right) = 0.288 \text{ s}$

c) The graph shows the particle approaching equilibrium at $x = 0.5$ without oscillating.

d) The discriminant is:

$\Delta = 5^2 - 4(1)(6) = 25 - 24 = 1 > 0$

Hence the damping is heavy.

Worked Example 3: Forced Harmonic Motion

Problem: A box $P$ is attached to one end of a light elastic string. The box is initially at rest on a smooth horizontal table when the other end of the spring, $A$, starts to move away from $P$ with constant velocity $1.2 , \text{m/s}$. The displacement $y$ metres of $P$ from its initial position at time $t$ seconds from the start satisfies the differential equation:

$\frac{d^2x}{dt^2} + 25y = 10t$

a) Find an expression for $y$ in terms of $t$

b) Calculate the time at which $P$ is next at rest

c) Sketch the graph of $y$ against $t$

Solution:

a) The auxiliary equation is:

$m^2 + 25 = 0$

$m^2 = -25 \implies m = \pm 5i$

This gives complex roots, so the complementary function is:

$y = A\cos 5t + B\sin 5t$

For the particular integral, try $y = at + b$:

$\ddot{y} = 0, \quad \dot{y} = a$

Substituting:

$25(at + b) = 10t$

$25at + 25b = 10t$

Comparing coefficients:

$a = \frac{2}{5}, \quad b = 0$

So the particular integral is $y = \frac{2}{5}t$.

The general solution is:

$y = A\cos 5t + B\sin 5t + \frac{2}{5}t$

At $t = 0$, $y = 0$ (box initially at rest at starting position):

$A = 0$

The solution is now:

$y = B\sin 5t + \frac{2}{5}t$

Differentiate:

$\frac{dy}{dt} = 5B\cos 5t + \frac{2}{5}$

At $t = 0$, $\dot{y} = 0$ (box initially at rest):

$5B + \frac{2}{5} = 0 \implies B = -\frac{2}{25}$

The particular solution is:

$y = -\frac{2}{25}\sin 5t + \frac{2}{5}t$

b) Set velocity to zero:

$\frac{dy}{dt} = -\frac{2}{5}\cos 5t + \frac{2}{5} = 0$

$\cos 5t = 1$

$5t = 2\pi$

$t = \frac{2\pi}{5} \text{ s}$

c) The graph oscillates about the line $y = 0.4t$.

Key observation: In forced harmonic motion, the equilibrium position changes over time. Here, the box oscillates around the line $y = 0.4t$ rather than a fixed point.

Worked Example 4: Using Newton's Second Law

Problem: A particle $P$ of mass $2$ kg moves along the positive $x$-axis under the action of a force directed towards the origin. At time $t$ seconds, the displacement of $P$ from $O$ is $x$ metres, and $P$ is moving away from $O$ with speed $v , \text{m/s}$. The force has magnitude $40x$ N. The particle $P$ is also subject to a resistive force of magnitude $16v$ N.

a) Show that the equation of motion of $P$ is:

$\frac{d^2x}{dt^2} + 8\frac{dx}{dt} + 20x = 0$

Given also that $x(0) = 0$ and $\dot{x}\left(\frac{\pi}{4}\right) = 3e^{-\pi}$

b) Solve the differential equation and find an expression for $x$ in terms of $t$

Solution:

a) Draw a force diagram:

Using $F = ma$:

$-40x - 16v = 2a$

Since $v = \frac{dx}{dt}$ and $a = \frac{d^2x}{dt^2}$:

$-40x - 16\frac{dx}{dt} = 2\frac{d^2x}{dt^2}$

Dividing by $2$:

$\frac{d^2x}{dt^2} + 8\frac{dx}{dt} + 20x = 0$

b) The auxiliary equation is:

$m^2 + 8m + 20 = 0$

Using the quadratic formula:

$(m+4)^2 = -4$

$m = -4 \pm 2i$

The general solution is:

$x = e^{-4t}(A\cos 2t + B\sin 2t)$

At $t = 0$, $x = 0$:

$A = 0$

When $t = \frac{\pi}{4}$, $\dot{x} = 3e^{-\pi}$:

After applying this condition, the particular solution is:

$x = 3e^{-4t}\sin 2t$

Exam tip: Always draw a force diagram and clearly identify all forces acting on the particle. Remember that resistive forces oppose motion, so they act in the opposite direction to velocity.

Worked Example 5: Train with Elastic Buffers

Problem: A train of mass $m$ kg runs into buffers that can be modelled as an elastic spring fixed at one end. At time $t$ seconds after first contact, the compression is $x$ metres, and the force in the spring is $m\omega^2 x$ newtons (where $\omega$ is a positive constant). The motion is also subject to a resistive force of magnitude $mkv$ newtons (where $v , \text{m/s}$ is the speed and $k$ is a positive constant).

a) Show that:

$\frac{d^2x}{dt^2} + k\frac{dx}{dt} + \omega^2 x = 0$

b) At time $t = 0$, the train is travelling with speed $U$. Given that $k = \frac{10\omega}{3}$, find an expression for $x$ in terms of $U$, $\omega$ and $t$

c) Show that the train comes to instantaneous rest when $t = \frac{3}{4\omega}\ln 3$

d) State three assumptions made in forming your model

Solution:

a) Using $F = ma$:

$-m\omega^2 x - mk\dot{x} = m\ddot{x}$

Dividing by $m$:

$\frac{d^2x}{dt^2} + k\frac{dx}{dt} + \omega^2 x = 0$

b) Substitute $k = \frac{10\omega}{3}$ into the auxiliary equation:

$m^2 + \frac{10\omega}{3}m + \omega^2 = 0$

$3m^2 + 10\omega m + 3\omega^2 = 0$

$(3m + \omega)(m + 3\omega) = 0$

$m = -\frac{\omega}{3} \text{ or } m = -3\omega$

The general solution is:

$x = Ae^{-\frac{\omega}{3}t} + Be^{-3\omega t}$

At $t = 0$, $x = 0$ (just making contact):

$A + B = 0 \implies B = -A$

Differentiating:

$\dot{x} = -\frac{A\omega}{3}e^{-\frac{\omega}{3}t} - 3B\omega e^{-3\omega t}$

At $t = 0$, $\dot{x} = U$:

$-\frac{A\omega}{3} - 3B\omega = U$

Solving for $A$ and $B$:

$A = \frac{3U}{8\omega}, \quad B = -\frac{3U}{8\omega}$

The particular solution is:

$x = \frac{3U}{8\omega}\left(e^{-\frac{\omega}{3}t} - e^{-3\omega t}\right)$

c) Set velocity to zero:

$\frac{dx}{dt} = \frac{3U}{8\omega}\left(-\frac{\omega}{3}e^{-\frac{\omega}{3}t} + 3\omega e^{-3\omega t}\right) = 0$

$-\frac{1}{3}e^{-\frac{\omega}{3}t} + 3e^{-3\omega t} = 0$

$e^{\frac{8\omega t}{3}} = 9$

$\frac{8\omega t}{3} = \ln 9 = 2\ln 3$

$t = \frac{3}{4\omega}\ln 3$

d) Assumptions made:

• The engine is disengaged
• The brakes are not applied
• The spring has no mass

Exam tip: When modelling physical systems, always state your assumptions clearly. Common assumptions include negligible mass of springs, no air resistance, smooth surfaces, and that the spring obeys Hooke's law throughout.