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IB Physics – Theme C: Wave Behaviour

Simple harmonic motion, wave phenomena, standing waves, Doppler effect.

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12 questions  ·  68 marks  ·  answers included  ·  IB Physics SL/HL (2023 syllabus)
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This worksheet covers:

  • C.1 simple harmonic motion
  • C.2 wave model
  • C.3 wave phenomena (refraction, diffraction, interference, polarisation)
  • C.4 standing waves
  • Resonance
  • C.5 Doppler effect

Take $g = 9.81\,\text{m s}^{-2}$, speed of sound $340\,\text{m s}^{-1}$ and $c = 3.00 \times 10^8\,\text{m s}^{-1}$. Calculator allowed; allow about 75 minutes. Questions marked HL use HL-only content.

Questions

Section A – Skills

Short questions.
1

A $0.25$ kg mass hangs from a spring of spring constant $40\,\text{N m}^{-1}$ and oscillates vertically with amplitude $5.0$ cm.(a) Show that the period of oscillation is about $0.50$ s.(b) Calculate the maximum speed of the mass.(c) Calculate the maximum acceleration of the mass.

Show answer
(a) $T = 2\pi\sqrt{\frac{0.25}{40}} = 0.497 \approx 0.50$ s(b) $\omega = 12.6\,\text{rad s}^{-1}$; $v_{max} = \omega A = 0.63\,\text{m s}^{-1}$(c) $a_{max} = \omega^2 A = 8.0\,\text{m s}^{-2}$
[6]
2

A simple pendulum has length $0.80$ m.(a) Calculate its period on Earth.(b) State two conditions required for the motion to be simple harmonic.

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(a) $T = 2\pi\sqrt{\frac{0.80}{9.81}} = 1.79$ s(b) Acceleration proportional to displacement from equilibrium and directed towards it; small amplitude (small angle); negligible friction/air resistance
[4]
3

A sound wave of frequency $250$ Hz travels through air at $340\,\text{m s}^{-1}$.(a) Calculate its wavelength.(b) State the difference between a transverse and a longitudinal wave, giving sound as an example of one of them.

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(a) $\lambda = \frac{340}{250} = 1.36$ m(b) Transverse: oscillations perpendicular to direction of energy transfer; longitudinal (sound): oscillations parallel to direction of energy transfer
[4]
4

A ray of light in air is incident on a glass block (refractive index $1.50$) at an angle of incidence of $40^\circ$.(a) Calculate the angle of refraction.(b) Calculate the critical angle for the glass-air boundary.(c) Calculate the speed of light in the glass.

Show answer
(a) $\sin r = \frac{\sin 40^\circ}{1.50} \Rightarrow r = 25.4^\circ$(b) $\sin c = \frac{1}{1.50} \Rightarrow c = 41.8^\circ$(c) $v = \frac{3.00 \times 10^8}{1.50} = 2.00 \times 10^8\,\text{m s}^{-1}$
[6]
5

Unpolarised light of intensity $I_0$ passes through a polariser and then an analyser whose transmission axis is at $60^\circ$ to that of the polariser.(a) State the intensity of light leaving the polariser.(b) Calculate the intensity of light leaving the analyser in terms of $I_0$.

Show answer
(a) $\frac{I_0}{2}$(b) $I = \frac{I_0}{2}\cos^2 60^\circ = 0.125\,I_0$
[3]

Section B – Exam style

6

Laser light of wavelength $633$ nm illuminates two narrow slits separated by $0.25$ mm. A screen is placed $2.0$ m from the slits.(a) Explain how bright fringes are formed on the screen.(b) Calculate the separation of adjacent bright fringes.(c) State and explain what happens to the fringe separation if the slit separation is increased.

Show answer
(a) Light from the two slits overlaps (coherent sources); where path difference is a whole number of wavelengths the waves arrive in phase and interfere constructively, giving a bright fringe(b) $s = \frac{\lambda D}{d} = \frac{633 \times 10^{-9} \times 2.0}{0.25 \times 10^{-3}} = 5.1 \times 10^{-3}$ m(c) Fringe separation decreases since $s \propto \frac{1}{d}$
[7]
7

A diffraction grating has $600$ lines per mm. Light of wavelength $500$ nm is incident normally on the grating.(a) Calculate the angle of the second-order maximum.(b) Determine the highest order maximum that can be observed.

Show answer
(a) $d = 1.67 \times 10^{-6}$ m; $\sin\theta = \frac{2 \times 500 \times 10^{-9}}{1.67 \times 10^{-6}} \Rightarrow \theta = 36.9^\circ$(b) $n_{max} = \frac{d}{\lambda} = 3.33$, so the third order is the highest
[5]
8

A pipe of length $0.60$ m is closed at one end and open at the other. The speed of sound is $340\,\text{m s}^{-1}$.(a) Describe the positions of the displacement nodes and antinodes for the fundamental (first harmonic).(b) Calculate the frequency of the first harmonic.(c) Calculate the frequency of the next harmonic that can be produced.(d) Explain why a standing wave is formed in the pipe.

Show answer
(a) Node at the closed end, antinode at the open end; pipe length is $\frac{\lambda}{4}$(b) $\lambda = 2.4$ m; $f_1 = \frac{340}{2.4} = 142$ Hz(c) Only odd harmonics: $f_3 = 3f_1 = 425$ Hz(d) The wave reflects at the closed end and superposes with the incident wave; at resonant frequencies the two waves of equal frequency and amplitude travelling in opposite directions produce fixed nodes and antinodes
[8]
9

A police siren emits sound at $500$ Hz. The car moves at $30\,\text{m s}^{-1}$ towards a stationary observer.(a) Calculate the frequency heard by the observer as the car approaches.(b) Calculate the frequency heard as the car moves away.(c) The observer instead runs at $20\,\text{m s}^{-1}$ towards a stationary siren. Calculate the frequency heard.

Show answer
(a) $f’ = 500 \times \frac{340}{340 – 30} = 548$ Hz(b) $f’ = 500 \times \frac{340}{340 + 30} = 459$ Hz(c) $f’ = 500 \times \frac{340 + 20}{340} = 529$ Hz
[6]

Section C – Extended

10

(HL) A particle performs SHM with amplitude $5.0$ cm and angular frequency $12.6\,\text{rad s}^{-1}$. At $t = 0$ it is at the equilibrium position moving in the positive direction.(a) Write an equation for the displacement $x$ as a function of time.(b) Calculate the speed of the particle when its displacement is $3.0$ cm.(c) The mass of the particle is $0.25$ kg. Calculate the total energy of the oscillation.(d) Sketch, by description, how the kinetic energy varies with displacement.

Show answer
(a) $x = 0.050\sin(12.6t)$ (metres)(b) $v = \omega\sqrt{A^2 – x^2} = 12.6\sqrt{0.05^2 – 0.03^2} = 0.51\,\text{m s}^{-1}$(c) $E = \frac{1}{2}m\omega^2A^2 = 0.050$ J(d) Inverted parabola: maximum ($0.050$ J) at $x = 0$, zero at $x = \pm A$
[8]
11

The hydrogen-alpha line has a wavelength of $656.3$ nm in the laboratory. In the spectrum of a distant galaxy the same line is observed at $656.4$ nm.(a) State whether the galaxy is moving towards or away from Earth.(b) Estimate the speed of the galaxy relative to Earth.(c) State one assumption made in your calculation.

Show answer
(a) Away (redshift, wavelength increased)(b) $v = c\frac{\Delta\lambda}{\lambda} = 3.00 \times 10^8 \times \frac{0.10}{656.3} = 4.6 \times 10^4\,\text{m s}^{-1}$(c) $v \ll c$ (non-relativistic approximation)
[5]
12

Light of wavelength $600$ nm passes through a single slit of width $0.10$ mm.(a) Calculate the angular width of the central maximum.(b) Describe the effect on the diffraction pattern of using blue light instead.(c) The slit is now replaced by two slits. Explain why the intensity of the double-slit fringes is modulated by the single-slit pattern.

Show answer
(a) $\theta = \frac{\lambda}{b} = 6.0 \times 10^{-3}$ rad, so angular width $= 1.2 \times 10^{-2}$ rad ($0.69^\circ$)(b) Shorter wavelength: central maximum narrower, minima closer together(c) Each slit diffracts independently, so the interference fringes have intensities limited by the single-slit envelope; fringes near single-slit minima are missing
[6]

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