Oscillations & Waves

Question 1

(a) [4 marks] Newton's law vertical (down positive), total spring stretch $= y_0 + x$ where $ky_0 = mg$ at equilibrium: $m\ddot{x} = mg - k(y_0 + x) - b\dot{x} + F_0\cos\omega_d t.$ Since $ky_0 = mg$, the constant terms cancel: (2 for setup, 2 for cancellation) $m\ddot{x} + b\dot{x} + kx = F_0\cos\omega_d t.$ Gravity only sets the equilibrium point, not the dynamics.

(b) [5 marks]

• $\omega_0 = \sqrt{k/m} = \sqrt{200/0.5} = \sqrt{400} = 20\ \text{rad/s}$. (1)
• $\zeta = b/(2\sqrt{mk}) = 2.0/(2\sqrt{0.5\cdot200}) = 2.0/(2\cdot10) = 0.10$. (1)
• $\zeta < 1 \Rightarrow$ underdamped. (1)
• Damped frequency: $\omega_1 = \omega_0\sqrt{1-\zeta^2} = 20\sqrt{1-0.01} = 20\sqrt{0.99} \approx 19.90\ \text{rad/s}$. (2)

(c) [3 marks] $Q = \omega_0 m/b = 20\cdot0.5/2.0 = 5.0$. (1) Fractional energy lost per radian $\approx 1/Q = 0.20$ (20%), because $E \propto e^{-\omega_0 t/Q}$ over a phase $\omega_0 t$, so per radian the exponent decreases by $1/Q$. (2)

(d) [6 marks] Substitute trial $x = A\cos(\omega_d t - \delta)$ into $m\ddot x + b\dot x + kx = F_0\cos\omega_d t$. Balancing gives (standard phasor/complex method): (3 for derivation) $A(\omega_d) = \frac{F_0/m}{\sqrt{(\omega_0^2-\omega_d^2)^2 + (b\omega_d/m)^2}}.$ Maximize $A$ ⟺ minimize the radicand $g(\omega_d^2)=(\omega_0^2-\omega_d^2)^2+(b/m)^2\omega_d^2$. Setting $dg/d(\omega_d^2)=0$: (2) $\omega_{\text{res}} = \sqrt{\omega_0^2 - \tfrac{b^2}{2m^2}} = \sqrt{400 - \tfrac{4}{2\cdot0.25}} = \sqrt{400-8}=\sqrt{392}\approx 19.80\ \text{rad/s}.$ This is slightly below $\omega_0=20$ because damping shifts the amplitude peak down. (1)

(e) [4 marks] Pseudo-code (velocity-Verlet style), any correct scheme accepted: (3)

x=0; v=0; dt=1e-4
for n in range(N):
    a = (F0*cos(wd*n*dt) - b*v - k*x)/m
    v += a*dt
    x += v*dt      # or half-step Verlet
record x, v

Check: run past transient (many $1/\zeta\omega_0$ decay times), fit steady $x(t)$ to $A\cos(\omega_d t-\delta)$ and confirm $A$ matches the analytic $A(\omega_d)$ of part (d) to within tolerance. (1)

Question 2

(a) [5 marks] Wave equation from a string element: net vertical force $= T\,\partial^2 y/\partial x^2\,dx$ (small slope $\sin\theta\approx\tan\theta=\partial y/\partial x$), mass $=\mu\,dx$: (3) $\mu\,\partial_{tt}y = T\,\partial_{xx}y \Rightarrow v=\sqrt{T/\mu}.$ $v=\sqrt{320/(4.0\times10^{-3})}=\sqrt{80000}\approx 283\ \text{m/s}$ ($=200\sqrt2$). (2)

(b) [5 marks] Fixed–fixed: $L=n\lambda_n/2 \Rightarrow \lambda_n=2L/n$, $f_n=v/\lambda_n=nv/(2L)$. (2) $f_1 = 283/(2\cdot0.8)=283/1.6\approx176.8\ \text{Hz}$. (1) $f_3=3f_1\approx530.3\ \text{Hz}$. (1) Third harmonic: 4 nodes (including 2 ends) and 3 antinodes; pattern N–A–N–A–N–A–N. (1)

(c) [5 marks] Superpose $y_1=a\cos(2\pi f_1 t)$, $y_2=a\cos(2\pi f_2 t)$: $y=2a\cos\!\big(2\pi\tfrac{f_1-f_2}{2}t\big)\cos\!\big(2\pi\tfrac{f_1+f_2}{2}t\big).$ Envelope amplitude modulation gives $f_{\text{beat}}=|f_1-f_2|$. (3) So $f_2 = 176.8 \pm 3.0 = 179.8$ or $173.8\ \text{Hz}$. Since $f\propto\sqrt{T}$, $T_2=T(f_2/f_1)^2$: (1)

• $T_2^{(+)}=320(179.8/176.8)^2\approx331\ \text{N}$
• $T_2^{(-)}=320(173.8/176.8)^2\approx309\ \text{N}$. (1)

(d) [7 marks] Source moving toward stationary observer: $f'=f_1\frac{c}{c-u_s}=176.8\cdot\frac{340}{340-30}=176.8\cdot\frac{340}{310}\approx193.9\ \text{Hz}.$ (4) Lower-tension detuned string still sounds its fundamental $f_2=173.8\ \text{Hz}$ (stationary, unshifted). (1) Beat between $f'$ and $f_2$: $|193.9-173.8|\approx 20.1\ \text{Hz}$. (2)

Question 3

(a) [5 marks] Mach number $M = v/c$. Wavefronts form a cone; in time $t$ source moves $vt$, sound travels $ct$, so $\sin\theta = ct/(vt)=1/M$. (3) $M=900/300=3.0$; $\theta=\arcsin(1/3)\approx19.47^\circ$. (2)

(b) [4 marks] $L=10\log_{10}(I/I_0)\Rightarrow I=I_0\,10^{L/10}=10^{-12}\cdot10^{13}=10\ \text{W/m}^2$. (4)

(c) [4 marks] $I\propto1/r^2$: $\Delta L = 10\log_{10}(I_2/I_1)=10\log_{10}(r_1^2/r_2^2)=20\log_{10}(r_1/r_2)$. (2) $=20\log_{10}(500/2000)=20\log_{10}(0.25)\approx-12.04\ \text{dB}$. $L_2=130-12.04\approx118.0\ \text{dB}$. (2)

(d) [3 marks] Acceptable answers, e.g.: Rocket structures must be designed so their natural resonant frequencies avoid the dominant driving frequencies of engine thrust oscillations (POGO) / aerodynamic buffeting, otherwise resonance amplifies vibration to destructive amplitudes. Shock/sonic-boom overpressure imposes transient loads on the airframe and launch structures that must be damped or structurally accommodated. (3 for one coherent point)

[
  {"claim":"omega0=20, zeta=0.1 underdamped, omega1~19.8997","code":"w0=sqrt(200/Rational(1,2)); zeta=2/(2*sqrt(Rational(1,2)*200)); w1=w0*sqrt(1-zeta**2); result=(w0==20) and (zeta==Rational(1,10)) and abs(float(w1)-19.8997)<1e-3"},
  {"claim":"Amplitude resonance omega_res=sqrt(392)~19.7990","code":"wres=sqrt(400-4/(2*Rational(1,4))); result=abs(float(wres)-19.79899)<1e-3"},
  {"claim":"String speed 200*sqrt2 and f1~176.78","code":"v=sqrt(320/Rational(4,1000)); f1=v/(2*Rational(8,10)); result=(v==200*sqrt(2)) and abs(float(f1)-176.777)<1e-2"},
  {"claim":"Doppler f'~193.85 Hz, beat ~20.1 Hz","code":"f1=200*sqrt(2)/(2*Rational(8,10)); fp=f1*340/310; f2=float(f1)-3; beat=abs(float(fp)-f2); result=abs(float(fp)-193.85)<0.3 and abs(beat-20.1)<0.3"},
  {"claim":"Mach cone: M=3, theta=asin(1/3), dB drop 20log10(0.25)","code":"import math; M=900/300; theta=math.degrees(math.asin(1/M)); dropdb=20*math.log10(500/2000); result=(M==3.0) and abs(theta-19.4712)<1e-3 and abs(dropdb+12.041)<1e-2"}
]