Oscillations & Waves

Question 1

(a) Equilibrium compression: $kx_0 = mg \Rightarrow x_0 = \dfrac{mg}{k} = \dfrac{0.40\times9.81}{250} = 0.0157\ \text{m}$. (1)

The block is pushed to $d = 0.15\ \text{m}$ below the natural length, i.e. $0.15 - 0.0157 = 0.1343\ \text{m}$ below the SHM equilibrium. So amplitude $A = 0.1343\ \text{m}$. (2)

(b) $\omega = \sqrt{k/m} = \sqrt{250/0.40} = 25.0\ \text{rad s}^{-1}$. (1) At equilibrium $v_{\max} = \omega A = 25.0 \times 0.1343 = 3.36\ \text{m s}^{-1}$. (2) (Why: at equilibrium all energy is kinetic; $v=\omega\sqrt{A^2-x^2}$ with $x=0$.)

(c) The block leaves the spring when the spring force can no longer act, i.e. when spring reaches its natural length — that is at displacement $x = +x_0 = 0.0157\ \text{m}$ above SHM equilibrium (spring uncompressed). (1) $v = \omega\sqrt{A^2 - x_0^2} = 25.0\sqrt{0.1343^2 - 0.0157^2} = 25.0\times0.1334 = 3.34\ \text{m s}^{-1}$. (2) (Why: contact ends when normal/spring force → 0, which is at natural length, not at SHM equilibrium.)

(d) After leaving the spring the block is a free projectile: only gravity acts, and it must still decelerate over the remaining rise. The SHM energy $\tfrac12 kA^2$ is partly converted to gravitational PE while in contact and partly to KE at separation. The naive expression ignores that (i) the block leaves before all spring PE is delivered as height, and (ii) it must gain height against gravity during the contact phase too. Correct method: total launch KE at separation + subsequent projectile rise, using energy conservation from release point. (3 for coherent energy accounting)

Question 2

(a) Amplitude falls by $1/e$ over 10 cycles: $10T_0 = \tau \Rightarrow \tau = 8.0\ \text{s}$. (2) Since amplitude $\propto e^{-\gamma t}$ with decay time $\tau = 1/\gamma$: $\gamma = 1/\tau = 0.125\ \text{s}^{-1}$. (1)

(b) $Q = \dfrac{\omega_0}{2\gamma}$ with $\omega_0 = 2\pi/T_0 = 7.854\ \text{rad s}^{-1}$. (1) $Q = \dfrac{7.854}{2\times0.125} = 31.4$. (2) Approximation: light damping, $\omega_d \approx \omega_0$. (equivalently $Q \approx \pi \times 10 = 31.4$)

(c) $\omega_d = \sqrt{\omega_0^2 - \gamma^2}$. Fractional difference $\approx \dfrac12(\gamma/\omega_0)^2 = \dfrac12(0.125/7.854)^2 = 1.27\times10^{-4}$, i.e. about $0.013\%$. (2) Negligible, so treating $\omega_d \approx \omega_0$ in (a) (period $\approx T_0$) is fully justified. (1)

(d) Critical damping requires $\gamma = \omega_0 = 7.854\ \text{s}^{-1}$ (i.e. increase damping enormously, factor ~63). (1) At critical damping the system returns to equilibrium in the shortest time without overshoot/oscillation. (1) Desirable for an instrument because it settles fastest and does not ring, giving a clean, prompt response ready for the next measurement. (1)

Question 3

(a) Path difference $\Delta = 31.5 - 24.0 = 7.5\ \text{cm} = 2.5\lambda$. (1) This is $(m+\tfrac12)\lambda$ with $m=2$, an odd multiple of half-wavelengths → destructive interference. (2)

(b) Adding a $\pi$ phase lag is equivalent to an extra $\lambda/2$ path. The former destructive condition $2.5\lambda$ becomes effectively $3\lambda$ → now constructive. (2)

(c) On the line between sources, path difference ranges from $-40$ to $+40\ \text{cm}$. Destructive when $|\Delta| = (m+\tfrac12)\lambda = 1.5, 4.5, 7.5, ...\ \text{cm}$. (1) Half-integer multiples of $3.0\ \text{cm}$ with $|\Delta| < 40$: values $1.5, 4.5, ..., 39.0\ \text{cm}$ → $(m+\tfrac12)\times3 < 40 \Rightarrow m+\tfrac12 < 13.33 \Rightarrow m = 0..12$, giving 13 values on each side. (2) Total nodal points strictly between: $13 + 13 = 26$. (1)

(d) Constructive: amplitudes add, $I_{\max} = (\sqrt{I_0}+\sqrt{I_0})^2 = 4I_0$. Destructive: $I_{\min}=0$. (2) Energy conserved: the average over the pattern is $2I_0$ (= simple sum of the two source intensities); energy is redistributed, not created or destroyed. (1)

Question 4

(a) $\sin\mu = 1/M \Rightarrow M = 1/\sin42^\circ = 1/0.6691 = 1.494$. (2) $v = M \times 343 = 512.5\ \text{m s}^{-1}$. (1)

(b) Subsonic, source approaching stationary observer: $f = f_0\dfrac{v_s}{v_s - v_{\text{src}}}$ with $v_{\text{src}} = 0.85\times343$. $f = 1200 \times \dfrac{343}{343 - 291.55} = 1200\times\dfrac{343}{51.45} = 8000\ \text{Hz}$. (3) (Why: denominator $1 - 0.85 = 0.15$; $f = 1200/0.15 = 8000$ Hz.)

(c) When $v_{\text{src}} \to v_s$ the denominator $\to 0$ and $f\to\infty$; beyond $v_s$ the formula gives a negative/nonphysical result because the source outruns its own wavefronts. (1) Physically, no sound reaches the observer ahead of the jet; the wavefronts pile up into a shock (Mach cone). (1) The observer hears silence as the jet approaches, then a sudden sonic boom as the cone sweeps past, followed by the receding-source sound. (1)

(d) The boom reaches the observer when the Mach cone (trailing the jet at half-angle $\mu$) sweeps over. Geometry: after the jet is overhead, it must travel a horizontal distance $x = h/\tan\mu$ for the cone edge to reach the observer. $x = 1500/\tan42^\circ = 1500/0.9004 = 1666\ \text{m}$. (1) Time = $x/v = 1666/512.5 = 3.25\ \text{s}$. (2)

Question 5

(a) Fundamental: $f_1 = \dfrac{1}{2L}\sqrt{T/\mu}$. (1) $v = 2Lf_1 = 2\times0.650\times196 = 254.8\ \text{m s}^{-1}$. (1) $T = \mu v^2 = 5.0\times10^{-3}\times254.8^2 = 324.7\ \text{N}$. (1)

(b) Touching the midpoint forces a node there, suppressing harmonics that have an antinode at the middle — i.e. odd harmonics (1st, 3rd). Surviving: even harmonics (2nd, 4th). (2) Lowest surviving = 2nd harmonic = $392\ \text{Hz}$. (1)

(c) Beats: $|f_2 - 196| = 3 \Rightarrow f_2 = 199\ \text{Hz}$ or $193\ \text{Hz}$. (2) Since $f \propto \sqrt{T}$, $\dfrac{\Delta f}{f} = \dfrac12\dfrac{\Delta T}{T}\Rightarrow \dfrac{\Delta T}{T} = 2\times\dfrac{3}{196} = 0.0306$, i.e. about $\pm3.1\%$. (1)

(d) Open–open pipe fundamental $f_1 = v_{\text{sound}}/(2L_p) \Rightarrow L_p = 343/(2\times196) = 0.875\ \text{m}$. (1) Open–open pipe supports all harmonics $n=1,2,3,...$ (1) — same as the string (both have $f_n = nf_1$), so their harmonic series is identical, differing only in relative amplitudes (timbre). (1)

[
  {"claim":"Q1 vmax = omega*A ≈ 3.36 m/s","code":"m=0.40;k=250;g=9.81;x0=m*g/k;A=0.15-x0;omega=(k/m)**0.5;vmax=omega*A;result=abs(float(vmax)-3.357)<0.02"},
  {"claim":"Q1 speed at separation ≈ 3.34 m/s","code":"m=0.40;k=250;g=9.81;x0=m*g/k;A=0.15-x0;omega=(k/m)**0.5;v=omega*(A**2-x0**2)**0.5;result=abs(float(v)-3.335)<0.02"},
  {"claim":"Q2 Q factor ≈ 31.4","code":"import sympy as sp;T0=0.80;tau=10*T0;gamma=1/tau;w0=2*sp.pi/T0;Q=w0/(2*gamma);result=abs(float(Q)-31.416)<0.05"},
  {"claim":"Q3 constructive=4 I0, destructive=0","code":"import sympy as sp;I0=sp.Symbol('I0');Imax=(sp.sqrt(I0)+sp.sqrt(I0))**2;result=sp.simplify(Imax-4*I0)==0"},
  {"claim":"Q4 Mach number = 1/sin42 ≈ 1.494","code":"import sympy as sp;M=1/sp.sin(sp.rad(42));result=abs(float(M)-1.4944)<0.005"},
  {"claim":"Q4b observed freq = 8000 Hz","code":"f=1200*343/(343-0.85*343);result=abs(float(f)-8000)<1"},
  {"claim":"Q4d boom delay ≈ 3.25