Oscillations & Waves

Question 1

(a) [4 marks]

• Newton: $ma = F = -kx \Rightarrow m\ddot{x} = -kx$, so $\ddot{x} + \tfrac{k}{m}x = 0$. (1)
• Define $\omega^2 = k/m$. (1)
• Substitute trial $x = A\cos(\omega t+\phi)$: $\ddot{x} = -\omega^2 A\cos(\omega t+\phi) = -\omega^2 x$. (1)
• This satisfies $\ddot{x}+\omega^2 x = 0$, confirming solution; $\omega = \sqrt{k/m}$. (1)

(b) [3 marks]

• $x = A\cos\theta$, $v=\dot x = -A\omega\sin(\omega t+\phi)$. (1)
• $\sin\theta = \pm\sqrt{1-\cos^2\theta} = \pm\sqrt{1-(x/A)^2}$. (1)
• $v = -A\omega(\pm\sqrt{1-x^2/A^2}) = \pm\omega\sqrt{A^2-x^2}$. (1)

(c) [3 marks]

• $KE = \tfrac12 m v^2 = \tfrac12 m\omega^2(A^2-x^2)= \tfrac12 k(A^2-x^2)$ (using $m\omega^2=k$). (1)
• $PE = \tfrac12 kx^2$. (1)
• $E = KE+PE = \tfrac12 k A^2$, independent of $x$ ⇒ constant. (1)

(d) [2 marks]

• $\omega=\sqrt{100/0.25}=20\ \text{rad s}^{-1}$; $v_{max}=\omega A = 20(0.04)=0.8\ \text{m s}^{-1}$. (1)
• $E=\tfrac12 kA^2 = \tfrac12(100)(0.04)^2 = 0.08\ \text{J}$. (1)

Question 2

(a) [4 marks]

• Restoring tangential force: $F = -mg\sin\theta$. (1)
• Torque/arc: $mL\ddot\theta = -mg\sin\theta$. (1)
• Small angle: $\sin\theta\approx\theta$ (in radians) ⇒ $\ddot\theta + \tfrac{g}{L}\theta=0$; needed to obtain linear SHM form. (1)
• $\omega=\sqrt{g/L}$, $T=2\pi\sqrt{L/g}$. (1)

(b) [4 marks]

• Torque about pivot: $\tau = -Mgd\sin\theta$. (1)
• $I\ddot\theta = -Mgd\sin\theta$. (1)
• Small angle: $\ddot\theta + \tfrac{Mgd}{I}\theta=0$ ⇒ $\omega=\sqrt{Mgd/I}$. (1)
• $T=2\pi\sqrt{I/(Mgd)}$. (1)

(c) [2 marks]

• $d=L/2$, $I=\tfrac13 ML^2$: $T=2\pi\sqrt{\tfrac13 ML^2/(Mg\cdot L/2)}=2\pi\sqrt{\tfrac{2L}{3g}}$. (1)
• Ratio to simple pendulum $2\pi\sqrt{L/g}$: factor $\sqrt{2/3}\approx0.816$. (1)

Question 3

(a) [5 marks]

• $m\ddot x + b\dot x + kx = 0$. (2)
• Discriminant of characteristic eqn via $\omega_0=\sqrt{k/m}$, $\gamma=b/2m$:
  • Underdamped: $\gamma<\omega_0$ — oscillates with decaying amplitude. (1)
  • Critical: $\gamma=\omega_0$ — returns to equilibrium fastest without oscillating. (1)
  • Overdamped: $\gamma>\omega_0$ — returns slowly, no oscillation. (1)

(b) [3 marks]

• High Q ⇒ small energy loss per cycle (Q = $2\pi\times$ energy stored / energy lost per cycle). (1.5)
• High Q ⇒ narrow, sharp resonance peak (small bandwidth $\Delta\omega=\omega_0/Q$). (1.5)

(c) [2 marks]

• $Q=\omega_0 m/b = 50(0.2)/0.4 = 25$. (2)

Question 4

(a) [6 marks]

• Element from $x$ to $x+dx$, mass $\mu\,dx$; tension $T$ acts along tangents at both ends. (1)
• Net vertical force $=T\sin\theta_2 - T\sin\theta_1$. (1)
• Small angle: $\sin\theta\approx\tan\theta = \partial y/\partial x$. (1)
• Net force $= T[(\partial y/\partial x)_{x+dx}-(\partial y/\partial x)_x] = T\dfrac{\partial^2 y}{\partial x^2}dx$. (1)
• Newton: $\mu\,dx\,\dfrac{\partial^2 y}{\partial t^2}=T\dfrac{\partial^2 y}{\partial x^2}dx$. (1)
• ⇒ $\dfrac{\partial^2 y}{\partial t^2}=\dfrac{T}{\mu}\dfrac{\partial^2 y}{\partial x^2}$. (1)

(b) [2 marks]

• $v=\sqrt{T/\mu}=\sqrt{80/0.005}=\sqrt{16000}=126.5\ \text{m s}^{-1}$. (2)

(c) [2 marks]

• $f_1 = v/(2L)=126.5/2.4 = 52.7\ \text{Hz}$. (2)

Question 5

(a) [4 marks]

• $y=A\cos(2\pi f_1 t)+A\cos(2\pi f_2 t)$. (1)
• Sum-to-product: $=2A\cos\!\big(2\pi\tfrac{f_1-f_2}{2}t\big)\cos\!\big(2\pi\tfrac{f_1+f_2}{2}t\big)$. (1)
• Envelope amplitude $2A\cos(\pi(f_1-f_2)t)$ modulates carrier. (1)
• Intensity maxima occur twice per envelope period ⇒ $f_{beat}=|f_1-f_2|$. (1)

(b) [3 marks]

• $f' = f\dfrac{v\pm v_o}{v\mp v_s}$. (1)
• Observer motion changes the number of wavefronts encountered per second (relative speed of waves changes). (1)
• Source motion compresses/stretches the wavelength itself (source chases its own waves), so it enters as a denominator; these are physically distinct mechanisms. (1)

(c) [3 marks]

• Source approaching, observer stationary: $f'=f\dfrac{v}{v-v_s}=600\dfrac{343}{343-30}$. (1)
• $=600\times343/313 = 600\times1.0958$. (1)
• $=657.5\ \text{Hz}$. (1)

Question 6

(a) [4 marks]

• Mach number $M = v_{source}/v_{sound}$. (1)
• In time $t$: source travels $v_s t$, wavefront radius $= v\,t$. (1)
• Cone half-angle: $\sin\theta = \dfrac{v t}{v_s t} = \dfrac{v}{v_s}$. (1)
• $=1/M$. (1)

(b) [2 marks]

• $\sin\theta=1/2.5=0.4$ ⇒ $\theta=\arcsin 0.4 = 23.6°$. (2)

(c) [2 marks]

• $\Delta L = 10\log_{10}(I_2/I_1)=10\log_{10}(10^{-3}/10^{-6})=10\log_{10}(1000)=30\ \text{dB}$. (2)

[
  {"claim":"Q1d max speed 0.8 and energy 0.08 J","code":"import sympy as sp\nomega=sp.sqrt(100/sp.Rational(1,4))\nvmax=omega*sp.Rational(4,100)\nE=sp.Rational(1,2)*100*(sp.Rational(4,100))**2\nresult=(vmax==sp.Rational(8,10)) and (E==sp.Rational(8,100))"},
  {"claim":"Q3c Q factor equals 25","code":"Q=50*sp.Rational(2,10)/sp.Rational(4,10)\nresult=(Q==25)"},
  {"claim":"Q4 wave speed 126.5 and fundamental 52.7 Hz","code":"v=sp.sqrt(80/sp.Rational(5,1000))\nf1=v/sp.Rational(24,10)\nresult=(abs(float(v)-126.49)<0.1) and (abs(float(f1)-52.7)<0.1)"},
  {"claim":"Q5c Doppler observed freq approx 657.5 Hz","code":"fobs=600*343/(343-30)\nresult=abs(float(fobs)-657.5)<0.5"},
  {"claim":"Q6c decibel increase is 30 dB","code":"dL=10*sp.log(sp.Rational(10)**-3/sp.Rational(10)**-6,10)\nresult=(sp.simplify(dL)==30)"}
]