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.
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.
\[ x=\frac{x_{0}}{2}=0.02\;\text{m}\quad\Longrightarrow\quad v=\pm\omega\sqrt{x_{0}^{2}-x^{2}} =\pm10\sqrt{(0.04)^{2}-(0.02)^{2}} =\pm10\sqrt{1.2\times10^{-3}} =\pm0.35\;\text{m s}^{-1}. \]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 \(x_{0}\).
\[ E=\frac12 m\omega^{2}x_{0}^{2} \;\Longrightarrow\; x_{0}= \sqrt{\frac{2E}{m\omega^{2}}} =\sqrt{\frac{2\times0.45}{0.25\times36}} =\sqrt{0.10}=0.316\;\text{m}. \]| 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.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.
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