Exercises, examples & solutions - Classical Mechanics & Special Relativity for Starters

Updated: 18 Jun 2026

Here are some examples and exercises that deal with oscillations.

Worked examples¶

Particle on a spring¶

A particle of mass $m$ is fixed to a spring (with spring constant $k$ and rest length $l_0$ ). The spring rests on a horizontal table, see figure. At $t=0$ , the spring has its rest length. The particle is released with zero velocity. What is the motion of the particle?

a point mass fixed on top of a vertical positioned spring that rests on a table.

Interpret the problem

Develop the solution

Evaluate the problem

Assess the problem

First we finish the sketch by drawing what is relevant for this problem.

The motion of the particle is governed by gravity and the force exerted by the spring. Both act along the $z$ -axis, making this a 1-dimensional problem. We have a particle that is initially at rest and will start moving when it is released.

Exercises¶

Exercise 1 (Particle oscillating in viscous oil)

A particle is hanging on a spring (spring constant $k$ ). The spring is attached to the ceiling. The particle is a sphere of mass $m$ and radius $R$ . It is submersed in a viscous oil with viscosity $\mu$ and density $\rho_{oil}$ .

The particle has initial velocity $v_0$ . Its position is such that the spring is at its rest length $l_0$ .

a. Determine the equilibrium position of $m$ .

b. Set up N2 for this situation. The oil exerts a friction force of the form $F = - 6\pi\mu Rv$ , with $\mu$ the oil viscosity and $R$ the sphere radius (this is Stokes law for the drag force from a fluid at low velocities). Note that the oil exerts a buoyancy force on the mass $m$ .

c. Give the general solution for the motion of the particle.

Exercise 3 (Charged particle)

A point particle, $m$ , with a positive charge, $q$ is placed in between two positive chrges $Q$ . These charges are fixed on the $x$ -axis. The distance between both charges Q is $2L$ . The mass $m$ can only moving over de $x$ -axis.

Initially, $m$ is at rest at a position exactly in between both charges. At time $t=0$ , the mass gets displaced by a small distance $\Delta x$ from its equilibrium position ( $\Delta x \ll L$ ). Gravity should be ignored in this exercise.

a. Make a sketch of the situation after the displacement. Draw all relevant quantities. Give the forces acting on $m$ ; give a mathematical expression for each of them. Determine the equilibrium position of $m$ (it should not be based on symmetry arguments).

b. Discuss what kind of motion $m$ will exhibit (after being released with zero initial velocity from position $\Delta x$ . Take the origin in the middle of the charges $Q$ .

c. Set up the equation of motion for $m$ .

d. Expand the force acting on $m$ using a Taylor series for positions of $m$ around the equilibrium position.

e. Show that for small deviations from the equilibrium position $m$ will move like a harmonic oscillator.

f. Determine the frequency of the harmonic oscillation.

Exercise 4 (Electronic Circuit)

Consider the electronic circuit shown in the drawing. It contains a resistor ( $V_R = IR$ ), a capacitor ( $V_C = \frac{1}{C}\int Idt$ ) and a coil ( $V_L = L\frac{dI}{dt}$ ).

According to Kirchhoff’s Law is the sum of all Voltage drops in the circuit equal to zero.

a. Apply Kirchhoff’s law on the circuit in the drawing and cast it in a form that it describes the evolution of the current $I(t)$ as a function of time.

b. Substitute in this equation that the charge $q$ is equal to the integral of the current and eliminate $I$ from the equation. Compare your equation for $q$ with the equation of motion of a damped harmonic oscillator. Which quantity plays the role of mass, of the spring and of friction?

c. The situation changes because an AC voltage source is added to the circuit. Determine now the ‘equation of motion’ and provide the general solution.

d. For $R = 10$ Ohm, a coil of $L = 0.1$ H and a capacitor of $C = 1.0 \cdot 10^{-5}$ F, find the resonance frequency of this oscillator.

Answers¶

Solution to Exercise 1

a. equilibrium: $\sum F = 0$

0 = -mg + V_p \rho_{oil} g - kz_{eq} \rightarrow z_{eq} = \frac{V_p (\rho_p - \rho_{oil} ) g}{k}

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with $V_p$ the particle volume and $\rho_p$ the density of the particle.

b. N2: now we have to take the frictional force by the oil into account:

m\frac{d^2z}{dt^2} = -kz - V_p (\rho_p - \rho_{oil} ) g - b\frac{dz}{dt}

(13)

c. Rewrite this as

m\ddot{z} + b \dot{z} + kz = - V_p (\rho_p - \rho_{oil} ) g

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This equation describes a damped harmonic oscillator. It will oscillate sinusoidal with angular frequency $\omega_\pm = \frac{-3\pi\mu R \pm \sqrt{9\pi^2\mu^2 R^2 - mk}}{m}$

The general solution is:

z(t) = z_{eq} + A \sin \omega t + B \cos \omega t

(15)

Solution to Exercise 2

b. To find the equilibrium positions, we need to find the points where the derivative of the potential energy is zero:

\frac{dV}{dx} = \frac{c(a^2 - x^2)}{(x^2 + a^2)^2} = 0 \rightarrow x = \pm a

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The stability follows from inspecting

\frac{d^2V}{dx^2} = -2cx \frac{-x^2 + 3a^2}{x^2 + a^2}^3

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x=-a \rightarrow \frac{d^2V}{dx^2} = \frac{c}{2a^3} > 0 \rightarrow \text{stable}

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x=a \rightarrow \frac{d^2V}{dx^2} = -\frac{c}{2a^3} < 0 \rightarrow \text{unstable}

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The same conclusion can also be drawn by inspecting the graph of $V(x)$ .

c. Make a Taylor expansion around $x=-a$ :

V(x) = V(-a) + \frac{1}{1!} V'(-a) (x+a) +\frac{1}{2!} V"(-a) (x+a)^2 + h.o.t.

(20)

with $V'(-a)=0$ and $V"(-a)=\frac{c}{2a^3}$ . Thus, we can write the energy for a small interval around $x=-a$ as:

\frac{1}{2}mv^2 + \frac{1}{2}\frac{c}{2a^3} (x+a)^2 -\frac{c}{2a} = E_0

(21)

This is the energy of a harmonic oscillator with angular frequency $\omega = \sqrt{\frac{c}{2a^3 m}}$ and period $T = 2\pi \sqrt{\frac{2a^3 m}{c}}$ .

Solution to Exercise 3

2 forces, both Coulomb type:
left charge $F_L$ , right charge $F_R$
position: $\Delta x = x$
left charge: $F_L = \frac{1}{4\pi\epsilon_0}\frac{Qq}{(L+ x )^2}$
right charge: $F_R = -\frac{1}{4\pi\epsilon_0}\frac{Qq}{(L- x )^2}$

equilibrium:

\begin{split} \sum F_i = 0 \rightarrow F_L + F_R = 0 \rightarrow \\ \frac{1}{4\pi\epsilon_0}\frac{Qq}{(L+ x )^2} - \frac{1}{4\pi\epsilon_0}\frac{Qq}{(L- x )^2} = 0 \rightarrow x = 0 \rightarrow\\ (L-x)^2 = (L+x)^2 \rightarrow x_{eq} = 0 \end{split}

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b. Energy conservation!! Coulomb force → potential exists: same relation with distance as gravity

F_L = \frac{1}{4\pi\epsilon_0}\frac{Qq}{(L+ x )^2} \rightarrow V_L(x) = -\int_\infty^x \frac{1}{4\pi\epsilon_0}\frac{Qq}{(L+ x )^2} dx = \frac{1}{4\pi\epsilon_0}\frac{Qq}{L+ x}

(23)

F_R = -\frac{1}{4\pi\epsilon_0}\frac{Qq}{(L- x )^2} \rightarrow V_R(x) = -\int_\infty^x -\frac{1}{4\pi\epsilon_0}\frac{Qq}{(L- x )^2} dx = \frac{1}{4\pi\epsilon_0}\frac{Qq}{L- x}

(24)

So, energy conservation:

\begin{split} \frac{1}{2}mv^2 +V_L(x) + V_R(x) = E_0 \rightarrow \\ \frac{1}{2}mv^2 + \frac{1}{4\pi\epsilon_0}\frac{Qq}{L+ x} + \frac{1}{4\pi\epsilon_0}\frac{Qq}{L- x} =E_0 \end{split}

(25)

From a plot of the potential energy we see: the motion of $m$ is confined! Moreover, $m$ will oscillate, but probably not in a harmonic way!

c. Equation of motion:

m\frac{d^2x}{dt^2} = \frac{1}{4\pi\epsilon_0}\frac{Qq}{(L+ x )^2} - \frac{1}{4\pi\epsilon_0}\frac{Qq}{(L- x )^2}

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d. Rewrite N2 using $s=x/L$ :

mL\frac{d^2s}{dt^2} = \frac{Qq}{4\pi\epsilon_0 L^2} \left (\frac{1}{(1+s)^2} - \frac{1}{(1-s)^2} \right )

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e. Make a Taylor series for the forces around $s=0$ :

\begin{split} \frac{1}{1+s^2} \approx 1 -2s + 3s^2 \\ \frac{1}{1-s^2} \approx 1 +2s + 3s^2 \end{split}

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Thus, for $s \ll 1$ $ we find in good approximation:

\begin{split} mL\frac{d^2s}{dt^2} = \frac{Qq}{4\pi\epsilon_0 L^2} \cdot -4s \\ \frac{d^2s}{dt^2} + \frac{Qq}{\pi\epsilon_0 mL^3} s = 0 \end{split}

(29)

harmonic oscillator!!

f. with frequency

\omega_0^2 = \frac{Qq}{\pi\epsilon_0 mL^3}

(30)

Solution to Exercise 4

a. Kirchhoff: $V_L + V_R + V_C = 0 \rightarrow L\frac{dI}{dt} + IR + \frac{1}{C}\int Idt = 0$

b. relation charge and current: $q = \int Idt \leftrightarrow I = \frac{dq}{dt} \rightarrow \frac{dI}{dt} = \frac{d^2q}{dt^2}$

Thus, we can write for the charge:

L\frac{d^2q}{dt} + R\frac{dq}{dt} +\frac{1}{C}q = 0

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Damped harmonic oscillator!
‘mass’: $L$ , thus the coil is the mass=inertia
spring constant: $\frac{1}{C}$ , hence the capacitor is the spring, that generates the restoring force
friction: $R$ , the resistor is the damping.
$q(t)$ acts as ‘position coordinate’

c. with driving voltage source:

V_L + V_R +V_C = V_0 \sin \nu t \rightarrow L\frac{d^2q}{dt} + R\frac{dq}{dt} +\frac{1}{C}q = V_0 \sin \nu t

(32)

This describes a forced, damped harmonic oscillator with general solution:

q(t) = A e^{-\frac{R}{2L}t} \sin \left (\sqrt{\frac{1}{LC} - \frac{R^2}{4L^2}}t + \epsilon \right ) + q_{max} \sin (\nu t + \alpha )

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d. with

q_{max} = \frac{V_0 / L}{\sqrt{(\omega_0^2 - \nu^2)^2 + \frac{R^2}{L^2} \nu^2}}

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which has a maximum if the denominator is minimal, i.e. $\nu_{res}^2 = \omega_0^2 - \frac{R^2}{L^2}$ and $\omega_0^2 = \frac{1}{LC}$

Thus, for $R=10$ Ohm, $L=0.1$ H and $C=1.0 \cdot 10^{-5}$ F: $\nu_{res}$ = 994 rad/s

6.1 Exercises, examples & solutions

Worked examples¶

Particle on a spring¶

Exercises¶

Answers¶