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6.1 Exercises, examples & solutions

Updated: 17 Jul 2026

Here are some examples and exercises that deal with oscillations.

Worked examples

Particle on a spring

A particle of mass mm is fixed to a spring (with spring constant kk and rest length l0l_0). The spring rests on a horizontal table, see figure. At t=0t=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.

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

The motion of the particle is governed by gravity and the force exerted by the spring. Both act along the zz-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

Answers

Solution to Exercise 1

a. equilibrium: ∑F=0\sum F = 0

0=−mg+Vpρoilg−kzeq→zeq=Vp(ρp−ρoil)gk0 = -mg + V_p \rho_{oil} g - kz_{eq} \rightarrow z_{eq} = \frac{V_p (\rho_p - \rho_{oil} ) g}{k}
(13)

with VpV_p the particle volume and ρp\rho_p the density of the particle.

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

md2zdt2=−kz−Vp(ρp−ρoil)g−bdzdtm\frac{d^2z}{dt^2} = -kz - V_p (\rho_p - \rho_{oil} ) g - b\frac{dz}{dt}
(14)

c. Rewrite this as

mz¨+bz˙+kz=−Vp(ρp−ρoil)gm\ddot{z} + b \dot{z} + kz = - V_p (\rho_p - \rho_{oil} ) g
(15)

This equation describes a damped harmonic oscillator. It will oscillate sinusoidal with angular frequency ω±=−3πμR±9π2μ2R2−mkm\omega_\pm = \frac{-3\pi\mu R \pm \sqrt{9\pi^2\mu^2 R^2 - mk}}{m}

The general solution is:

z(t)=zeq+Asin⁡ωt+Bcos⁡ωtz(t) = z_{eq} + A \sin \omega t + B \cos \omega t
(16)
Solution to Exercise 2

a.

Plot of the potential energy.

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

dVdx=c(a2−x2)(x2+a2)2=0→x=±a\frac{dV}{dx} = \frac{c(a^2 - x^2)}{(x^2 + a^2)^2} = 0 \rightarrow x = \pm a
(17)

The stability follows from inspecting

d2Vdx2=−2cx−x2+3a2x2+a23\frac{d^2V}{dx^2} = -2cx \frac{-x^2 + 3a^2}{x^2 + a^2}^3
(18)
x=−a→d2Vdx2=c2a3>0→stablex=-a \rightarrow \frac{d^2V}{dx^2} = \frac{c}{2a^3} > 0 \rightarrow \text{stable}
(19)
x=a→d2Vdx2=−c2a3<0→unstablex=a \rightarrow \frac{d^2V}{dx^2} = -\frac{c}{2a^3} < 0 \rightarrow \text{unstable}
(20)

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

c. Make a Taylor expansion around x=−ax=-a:

V(x)=V(−a)+11!V′(−a)(x+a)+12!V"(−a)(x+a)2+h.o.t.V(x) = V(-a) + \frac{1}{1!} V'(-a) (x+a) +\frac{1}{2!} V"(-a) (x+a)^2 + h.o.t.
(21)

with V′(−a)=0V'(-a)=0 and V"(−a)=c2a3V"(-a)=\frac{c}{2a^3}. Thus, we can write the energy for a small interval around x=−ax=-a as:

12mv2+12c2a3(x+a)2−c2a=E0\frac{1}{2}mv^2 + \frac{1}{2}\frac{c}{2a^3} (x+a)^2 -\frac{c}{2a} = E_0
(22)

This is the energy of a harmonic oscillator with angular frequency ω=c2a3m\omega = \sqrt{\frac{c}{2a^3 m}} and period T=2π2a3mcT = 2\pi \sqrt{\frac{2a^3 m}{c}}.

Solution to Exercise 3

a.

Plot of forces acting on m.

2 forces, both Coulomb type:
left charge FLF_L, right charge FRF_R
position: Δx=x\Delta x = x
left charge: FL=14πϵ0Qq(L+x)2F_L = \frac{1}{4\pi\epsilon_0}\frac{Qq}{(L+ x )^2}
right charge: FR=−14πϵ0Qq(L−x)2F_R = -\frac{1}{4\pi\epsilon_0}\frac{Qq}{(L- x )^2}

equilibrium:

∑Fi=0→FL+FR=0→14πϵ0Qq(L+x)2−14πϵ0Qq(L−x)2=0→x=0→(L−x)2=(L+x)2→xeq=0\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}
(23)

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

FL=14πϵ0Qq(L+x)2→VL(x)=−∫∞x14πϵ0Qq(L+x)2dx=14πϵ0QqL+xF_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}
(24)

FR=−14πϵ0Qq(L−x)2→VR(x)=−∫∞x−14πϵ0Qq(L−x)2dx=14πϵ0QqL−xF_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}
(25)

So, energy conservation:

12mv2+VL(x)+VR(x)=E0→12mv2+14πϵ0QqL+x+14πϵ0QqL−x=E0\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}
(26)
Plot of potential energy.

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

c. Equation of motion:

md2xdt2=14πϵ0Qq(L+x)2−14πϵ0Qq(L−x)2m\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}
(27)

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

mLd2sdt2=Qq4πϵ0L2(1(1+s)2−1(1−s)2)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 )
(28)

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

11+s2≈1−2s+3s211−s2≈1+2s+3s2\begin{split} \frac{1}{1+s^2} \approx 1 -2s + 3s^2 \\ \frac{1}{1-s^2} \approx 1 +2s + 3s^2 \end{split}
(29)

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

mLd2sdt2=Qq4πϵ0L2⋅−4sd2sdt2+Qqπϵ0mL3s=0\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}
(30)

harmonic oscillator!!

f. with frequency

ω02=Qqπϵ0mL3\omega_0^2 = \frac{Qq}{\pi\epsilon_0 mL^3}
(31)
Solution to Exercise 4

a. Kirchhoff: VL+VR+VC=0→LdIdt+IR+1C∫Idt=0V_L + V_R + V_C = 0 \rightarrow L\frac{dI}{dt} + IR + \frac{1}{C}\int Idt = 0

b. relation charge and current: q=∫Idt↔I=dqdt→dIdt=d2qdt2 q = \int Idt \leftrightarrow I = \frac{dq}{dt} \rightarrow \frac{dI}{dt} = \frac{d^2q}{dt^2}

Thus, we can write for the charge:

Ld2qdt+Rdqdt+1Cq=0L\frac{d^2q}{dt} + R\frac{dq}{dt} +\frac{1}{C}q = 0
(32)

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

c. with driving voltage source:

VL+VR+VC=V0sin⁡νt→Ld2qdt+Rdqdt+1Cq=V0sin⁡νtV_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
(33)

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

q(t)=Ae−R2Ltsin⁡(1LC−R24L2t+ϵ)+qmaxsin⁡(νt+α)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 )
(34)

d. with

qmax=V0/L(ω02−ν2)2+R2L2ν2q_{max} = \frac{V_0 / L}{\sqrt{(\omega_0^2 - \nu^2)^2 + \frac{R^2}{L^2} \nu^2}}
(35)

which has a maximum if the denominator is minimal, i.e. νres2=ω02−R2L2\nu_{res}^2 = \omega_0^2 - \frac{R^2}{L^2} and ω02=1LC\omega_0^2 = \frac{1}{LC}

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

Solution to Exercise 5
Classical Mechanics & Special Relativity for Starters
Oscillations
Classical Mechanics & Special Relativity for Starters
Collisions