These notes cover the full syllabus requirements for sections 17.1 – 17.3, with explicit statements, derivations, quantitative examples and guidance on the sketches that are required in the examinations.
\[
x(t)=x_{0}\sin(\omega t+\phi)
\]
\[
v(t)=\frac{dx}{dt}=x{0}\omega\cos(\omega t+\phi)=\omega x{0}\cos(\omega t+\phi)
\]
\[
a(t)=\frac{d^{2}x}{dt^{2}}=-x_{0}\omega^{2}\sin(\omega t+\phi)=-\omega^{2}x(t)
\]
For many exam questions the time origin is chosen so that the phase constant is zero. In that case the three equations become
\[
\boxed{x=x_{0}\sin\omega t},\qquad
\boxed{v=v_{0}\cos\omega t},\qquad
\boxed{a=-\omega^{2}x},
\]
where the maximum speed is \(v{0}=\omega x{0}\).
Using the conservation of mechanical energy (see §2) we obtain
\[
\boxed{v=\pm\,\omega\sqrt{x_{0}^{2}-x^{2}}}
\]
The “\(\pm\)” sign indicates the direction of motion: “+’’ when the particle moves away from the equilibrium position, “–’’ when it moves back towards equilibrium.
A mass‑spring system has \(\omega=10\;\text{rad s}^{-1}\) and amplitude \(x_{0}=0.04\;\text{m}\). Find the speed when the mass is halfway between the centre and the extreme position.
\[
x=\frac{x_{0}}{2}=0.02\;\text{m}\quad\Longrightarrow\quad
v=\pm\omega\sqrt{x_{0}^{2}-x^{2}}
=\pm10\sqrt{(0.04)^{2}-(0.02)^{2}}
=\pm10\sqrt{1.2\times10^{-3}}
=\pm0.35\;\text{m s}^{-1}.
\]
\[
E=K+U=\frac12 m\omega^{2}x_{0}^{2}
\]
A 0.25 kg mass on a spring oscillates with angular frequency \(\omega=6\;\text{rad s}^{-1}\). The total mechanical energy is \(E=0.45\;\text{J}\). Find the amplitude \(x_{0}\).
\[
E=\frac12 m\omega^{2}x_{0}^{2}
\;\Longrightarrow\;
x_{0}= \sqrt{\frac{2E}{m\omega^{2}}}
=\sqrt{\frac{2\times0.45}{0.25\times36}}
=\sqrt{0.10}=0.316\;\text{m}.
\]
\[
\boxed{\ddot x+2\beta\dot x+\omega_{0}^{2}x=0},
\qquad\beta\;( \text{s}^{-1})\text{ is the damping coefficient, }\;
\omega_{0}=\sqrt{k/m}.
\]
| Damping type | Condition | Displacement solution | Physical interpretation |
|---|---|---|---|
| Underdamped | \(\beta<\omega_{0}\) | \(x(t)=A\,e^{-\beta t}\sin(\omega_{d}t+\phi)\) | Oscillations persist but the amplitude decays exponentially with time‑constant \(1/\beta\). \(\omega{d}=\sqrt{\omega{0}^{2}-\beta^{2}}\) is the damped angular frequency. |
| Critically damped | \(\beta=\omega_{0}\) | \(x(t)=(A+Bt)\,e^{-\omega_{0}t}\) | The system returns to equilibrium as quickly as possible without overshooting; no oscillation occurs. |
| Over‑damped | \(\beta>\omega_{0}\) | \(x(t)=A\,e^{-(\beta+\sqrt{\beta^{2}-\omega{0}^{2}})t}+B\,e^{-(\beta-\sqrt{\beta^{2}-\omega{0}^{2}})t}\) | Two exponential terms with different decay rates; motion is non‑oscillatory and slower than the critically damped case. |
Because the damping force is proportional to velocity (\(F_{\text{d}}=-2\beta m v\)), the mechanical energy decays as
\[
E(t)=E_{0}\,e^{-2\beta t}.
\]
A mass‑spring oscillator has \(\beta=0.8\;\text{s}^{-1}\) and initial amplitude \(A_{0}=5.0\;\text{cm}\). Find the amplitude and the mechanical energy after 3 s.
\[
A(t)=A_{0}e^{-\beta t}=5.0\,e^{-0.8\times3}=5.0\,e^{-2.4}=0.45\;\text{cm}.
\]
\[
E(t)=E{0}e^{-2\beta t}=E{0}e^{-1.6\times3}=E{0}e^{-4.8}=0.008\,E{0}.
\]
Thus after 3 s the amplitude has fallen to less than 1 % of its original value and the mechanical energy is reduced to about 0.8 % of the initial energy.
\[
\boxed{\ddot x+2\beta\dot x+\omega{0}^{2}x=\frac{F{0}}{m}\cos(\omega_{d}t)},
\]
where \(\omega{d}\) is the driving (angular) frequency and \(F{0}\) the amplitude of the driving force.
\[
x(t)=X(\omega{d})\cos\!\bigl(\omega{d}t-\delta\bigr),
\]
\[
X(\omega{d})=\frac{F{0}/m}{\sqrt{(\omega{0}^{2}-\omega{d}^{2})^{2}+(2\beta\omega_{d})^{2}}},\qquad
\tan\delta=\frac{2\beta\omega{d}}{\;\omega{0}^{2}-\omega_{d}^{2}\;}.
\]
\[
Q=\frac{\omega_{0}}{2\beta}.
\]
\[
\Delta\omega=\frac{\omega_{0}}{Q}=2\beta.
\]
The amplitude‑frequency curve is a Lorentzian centred at \(\omega{0}\) with half‑maximum points at \(\omega{0}\pm\beta\).
Draw a smooth curve of \(X(\omega{d})\) versus \(\omega{d}\):
Series RLC circuit: \(L=0.12\;\text{H}\), \(C=80\;\mu\text{F}\), \(R=6\;\Omega\).
\[
\omega_{0}=\frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{0.12\times80\times10^{-6}}}=288\;\text{rad s}^{-1}.
\]
\[
\beta=\frac{R}{2L}= \frac{6}{0.24}=25\;\text{s}^{-1}.
\]
\[
Q=\frac{\omega_{0}}{2\beta}= \frac{288}{50}=5.8.
\]
\[
\Delta\omega=2\beta=50\;\text{rad s}^{-1}.
\]
Hence the resonance curve is centred at \(288\;\text{rad s}^{-1}\) with half‑maximum points at \(263\) and \(313\;\text{rad s}^{-1}\).
| Quantity | Expression | Maximum value | Value when the other quantity is zero |
|---|---|---|---|
| Displacement | \(x=x{0}\sin(\omega t+\phi)\) | \(|x|{\max}=x{0}\) | \(x=0\) when \(v=\pm v{0}\) |
| Velocity | \(v=\omega x{0}\cos(\omega t+\phi)\) | \(|v|{\max}=v{0}=\omega x{0}\) | \(v=0\) when \(|x|=x_{0}\) |
| Acceleration | \(a=-\omega^{2}x{0}\sin(\omega t+\phi)=-\omega^{2}x\) | \(|a|{\max}=a{0}=\omega^{2}x{0}\) | \(a=0\) when \(x=0\) |
Thus the three quantities are all sinusoidal and are displaced by \(\pi/2\) rad (90°) with respect to each other.
These sketches are explicitly required in the Cambridge exam for “graphical analysis of SHM”.
A transverse wave on a stretched string can be regarded as a series of coupled mass‑spring oscillators. For a small element of mass \(\Delta m\) the restoring force is provided by the tension \(T\), leading to the wave speed
\[
v_{\text{wave}}=\sqrt{\frac{T}{\mu}},\qquad \mu=\frac{\Delta m}{\Delta x}.
\]
This result follows directly from the SHM equation \(\ddot y = -(T/\mu) \, \partial^{2}y/\partial x^{2}\).
In a lossless LC circuit the charge \(q\) on the capacitor obeys
\[
L\ddot q + \frac{q}{C}=0,
\]
which is mathematically identical to \(\ddot x +\omega{0}^{2}x=0\) with \(\omega{0}=1/\sqrt{LC}\). Hence the charge and the current oscillate with the same angular frequency as a mass‑spring system, and the concepts of resonance and quality factor are directly transferable.
The potential energy of a quantum harmonic oscillator is \(U(x)=\tfrac12 kx^{2}\), exactly the same as the classical SHM. The Schrödinger equation then yields discrete energy levels
\(E_{n}=\hbar\omega\left(n+\tfrac12\right)\). The classical expressions for \(\omega\) and for the shape of the potential remain valid, providing the conceptual bridge between the two topics.
| Symbol | Quantity | Units | Typical expression in SHM |
|---|---|---|---|
| \(x\) | Displacement from equilibrium | m | \(x=x_{0}\sin(\omega t+\phi)\) |
| \(x_{0}\) | Amplitude | m | Maximum \(|x|\) |
| \(v\) | Velocity | m s\(^{-1}\) | \(v=\omega x_{0}\cos(\omega t+\phi)\) |
| \(v{0}\) | Maximum speed | m s\(^{-1}\) | \(v{0}=\omega x_{0}\) |
| \(a\) | Acceleration | m s\(^{-2}\) | \(a=-\omega^{2}x\) |
| \(\omega\) | Angular frequency | rad s\(^{-1}\) | \(\omega=2\pi f=\sqrt{k/m}\) |
| \(f\) | Frequency | Hz | \(f=\omega/2\pi\) |
| \(T\) | Period | s | \(T=2\pi/\omega\) |
| \(k\) | Force constant (spring) | N m\(^{-1}\) | \(k=m\omega^{2}\) |
| \(\beta\) | Damping coefficient | s\(^{-1}\) | Appears in \(\ddot x+2\beta\dot x+\omega_{0}^{2}x=0\) |
| \(Q\) | Quality factor | – | \(Q=\omega_{0}/2\beta\) |
| \(F_{0}\) | Driving‑force amplitude | N | RHS of forced‑oscillation equation |
| \(E\) | Total mechanical energy | J | \(E=\tfrac12 m\omega^{2}x_{0}^{2}\) |
A 0.50 kg mass attached to a spring oscillates with \(\omega=12\;\text{rad s}^{-1}\). At the instant the displacement is \(x=0.03\;\text{m}\) the speed is measured to be \(0.42\;\text{m s}^{-1}\). Determine the amplitude \(x_{0}\).
\[
E=\tfrac12 m\omega^{2}x_{0}^{2}
=\tfrac12 mv^{2}+\tfrac12 kx^{2}
=\tfrac12 m\bigl(v^{2}+\omega^{2}x^{2}\bigr).
\]
\[
x_{0}= \sqrt{\frac{v^{2}+\omega^{2}x^{2}}{\omega^{2}}}
=\sqrt{\frac{(0.42)^{2}+12^{2}(0.03)^{2}}{12^{2}}}
=0.043\;\text{m}.
\]
A light‑damped oscillator has \(\beta=0.5\;\text{s}^{-1}\) and initial amplitude \(A_{0}=2.0\;\text{cm}\). Find (i) the amplitude after 4 s, and (ii) the fraction of the original mechanical energy remaining.
(i) \(A= A_{0}e^{-\beta t}=2.0\,e^{-0.5\times4}=2.0\,e^{-2}=0.27\;\text{cm}.\)
(ii) \(E=E{0}e^{-2\beta t}=E{0}e^{-1.0\times4}=E{0}e^{-4}=0.018\,E{0}.\)
A series RLC circuit has \(L=0.05\;\text{H}\), \(C=200\;\mu\text{F}\) and \(R=4\;\Omega\). Calculate:
Solution:
\[
\omega_{0}= \frac{1}{\sqrt{LC}}=\frac{1}{\sqrt{0.05\times200\times10^{-6}}}=316\;\text{rad s}^{-1},
\qquad
\beta=\frac{R}{2L}= \frac{4}{0.10}=40\;\text{s}^{-1}.
\]
\[
Q=\frac{\omega_{0}}{2\beta}= \frac{316}{80}=3.95,\qquad
\Delta\omega=2\beta=80\;\text{rad s}^{-1}.
\]
The half‑maximum occurs when the denominator of \(X(\omega{d})\) is \(\sqrt{2}\) times its minimum value, giving \(|\omega{d}-\omega{0}|=\beta\). Hence the required driving frequencies are \(\omega{d}= \omega_{0}\pm\beta = 276\) rad s\(^{-1}\) and \(356\) rad s\(^{-1}\).
With these statements, derivations, examples and sketch‑guidelines you will have full coverage of the Cambridge AS & A‑Level Physics requirements for Simple Harmonic Motion.
Your generous donation helps us continue providing free Cambridge IGCSE & A-Level resources, past papers, syllabus notes, revision questions, and high-quality online tutoring to students across Kenya.