Wave Phenomena

Related chapter: Waves


  • 9.1 Simple harmonic motion
    • State the defining equation for SHM and associate it with the graph of aa against xx.
    • Understand that xx refers to the displacement from the equilibrium position.
    • Associate ωω with the hardware of the system generating the SHM.
    • From ωω calculate TT and ff, understanding that they do not change with x0x0, the amplitude.
    • Identify (in the data booklet) and apply the equations relating to:
      • xx as a function of tt.
      • vv as a function of tt.
      • vv as a function of xx.
      • kinetic energy as a function of xx.
      • total energy as a function of amplitude.
    • Sketch the graphs for the relationships above.
    • Show graphically how the kinetic and potential energy changes with displacement and with time.
    • Discuss the parameters of a mass–spring system that affect its frequency and apply the correct equation.
    • Discuss the parameters of a pendulum that affect its frequency and apply the correct equation.


f=1Tv=fλa=ω2xTf=2πωEtotal=12mvmax2=12mA2ω2EK=12mv2=12mω2(A2x2)EP=12mω2x2θ=λbbsinθ=nλs=λDd2dn=(m+12)λθAλbR=λavgΔλ=mNλ=λ(1usv)f=f(vvus)ΔffvcΔλλvc \begin{aligned} f &= \frac{1}{T}\\[1.5em] v &= f \lambda \\[1.5em] a &= - \omega^2 x\\[1em] Tf &= \frac{2\pi}{\omega}\\[1em] E_{total} &= \frac{1}{2} mv_{max}^2 = \frac{1}{2}mA^2\omega^2\\[1em] E_{K} &= \frac{1}{2} mv^2 = \frac{1}{2}m\omega^2 (A^2 - x^2)\\[1em] E_{P} &=\frac{1}{2}m\omega^2x^2\\[1em] \theta &= \frac{\lambda}{b}\\[1em] b\sin\theta &= n\lambda\\[1em] s &= \frac{\lambda D}{d}\\[1em] 2dn &= (m + \frac{1}{2})\lambda\\[1em] \theta_A &\geq \frac{\lambda}{b}\\[1em] R &= \frac{\lambda_{avg}}{\Delta \lambda} = m N\\[1em] \lambda' &= \lambda(1- \frac{u_s}{v})\\[1.5em] f' &= f(\frac{v}{v - u_s})\\[1em] \frac{\Delta f}{f} &\approx \frac{v}{c}\\[1.5em] \frac{\Delta \lambda}{\lambda} &\approx \frac{v}{c} \end{aligned}


F=kxω2=kmT=2πmk \begin{aligned} F &= -kx\\[1em] \omega^2 &= \frac{k}{m}\\[1.5em] T&= 2\pi\sqrt{\frac{m}{k}}\\[1.5em] \end{aligned}


T=2πlg \begin{aligned} T&= 2\pi\sqrt{\frac{l}{g}} \end{aligned}

Symbol What it represents Unit/Value
FF restoring force NN
kk spring constant Nm1Nm^{-1}
xx displacement mm
ω\omega angular frequency rads1\text{rad} \cdot s^{-1}
ll length of pendulum mm
gg acceleration due to gravity 9.81ms29.81 ms^{-2}

Single slit diffraction

First location of destructive interference: bsinθ=λb \sin\theta= \lambda
For small angles of θ\theta: θ=λb\theta = \frac{\lambda}{b}

Double slit diffraction

Separation between maximums:
s=λDds = \lambda\frac{D}{d}

Multiple slit diffraction

With NN slits, there are N2N-2 maxima.
Intensity of the central maximum: N2×I0N^2 \times I_0, where I0I_0 is the intensity of the central maximum in single slit diffraction.


  • Simple harmonic motion can be modelled by the following equations:
    x=Asin(ωt+ϕ)v=ωAcos(ωt+ϕ)a=ω2Asin(ωt+ϕ)=ω2x=kmx \begin{aligned} x &= A\sin(\omega t + \phi)\\[1em] v &= \omega \cdot A\cos(\omega t + \phi)\\[1em] a &= -\omega^2 \cdot A\sin(\omega t + \phi) = -\omega^2 x=-\frac{k}{m} x \end{aligned}
    • AA: amplitude
    • ω\omega: angular frequency (how many radians on the graph correspond to 1 second)
    • ϕ\phi: the leftward phase shift
    • tt: time
    • xx: displacement from equilibrium position.
    • vv: velocity.
    • aa: acceleration.
  • In addition:
    vmax=ωAamax=ω2A \begin{aligned} v_{max} &= \omega A \\[1.5em] a_{max} &= -\omega^2 A \end{aligned}

Velocity and acceleration as a function of displacement

enter image description here


  • In an ideal system with no drag forces, energy is conserved:
    EK+EP=Etotal remains constant.E_K + E_P = E_{total}\text{ remains constant.}
  • The total energy of a system can be expressed in terms of the maximum kinetic energy of the system:
    Etotal=12mvmax2=12mA2ω2E_{total} = \frac{1}{2} mv_{max}^2 = \frac{1}{2}mA^2\omega^2
  • So kinetic and potential energy are:
    EK=12mv2=12mω2(A2cos2(ωt+ϕ))=12mω2(A2(1sin2(ωt+ϕ)))=12mω2(A2x2)EP=EtotalEK=12mA2ω212mω2(A2x2)=12mω2x2E_{K} = \frac{1}{2} mv^2 = \frac{1}{2}m\omega^2( A^2 \cos^2(\omega t + \phi))\\[1em] = \frac{1}{2}m\omega^2( A^2 (1 - \sin^2(\omega t + \phi)))\\[1em] = \frac{1}{2}m\omega^2 (A^2 - x^2)\\[2em] E_P = E_{total} - E_K= \frac{1}{2}mA^2\omega^2 -\frac{1}{2}m\omega^2 (A^2 - x^2)\\[1em] = \frac{1}{2}m\omega^2x^2
    • note that v2=A2cos2(ωt+ϕ)v^2 = A^2 \cos^2(\omega t + \phi) and x2=A2sin2(ωt+ϕ)x ^2 = A^2\sin^2(\omega t + \phi).

enter image description here

Single slit diffraction

  • For destructive interference, the path difference must be (n+12)λ(n + \frac{1}{2})\lambda, and the first minimum is obtained where path difference = 12λ\frac{1}{2}\lambda.

  • With the aid of the diagram, we know that AB is the path difference between the two waves, the slit width is bb, and λ\lambda is the wavelength:
    AB=path differencepath difference=12λ (for the first minimum)AB=12λAB=b2sinθ (using trignometry)12λ=b2sinθbsinθ=λsinθ=λbSince sinθθ for very small angles:θ=λb \begin{aligned} \text{AB} &= \text{path difference} \\ \text{path difference} &= \frac{1}{2}\lambda \text{ (for the first minimum)} \\ \text{AB} &= \frac{1}{2}\lambda \\ \text{AB} &= \frac{b}{2}\sin\theta \text{ (using trignometry)} \\ \frac{1}{2}\lambda &= \frac{b}{2}\sin\theta \\ b\sin\theta &= \lambda \\ \sin\theta &= \frac{\lambda}{b} \\ \text{Since } \sin\theta &\approx \theta \text{ for very small angles:} \\ \theta &= \frac{\lambda}{b} \end{aligned}
  • So, we can say that the first minimum for a single slit diffraction is observed at an angle θ\theta where θ=λb\theta = \frac{\lambda}{b}.
  • The formula to find the angle at which additional minima form becomes:
    bsinθ=nλb\sin\theta = n\lambda
  • Where:
    • bb : slit width.
    • nn: the nth minima.
    • λ\lambda: wavelength.

  • We can also conclude that:
    θλ\theta \propto \lambda
  • As wavelength \uparrow, θ\theta \uparrow, and the angular width of the central maxima \uparrow.
    θ1b\theta \propto \frac{1}{b}
  • As bb \uparrow, θ\theta \downarrow, and so the angular width of the central maxima \downarrow.

Young’s Double Slit Experiment

  • When light from two slits intefere in the following setup, a set of fringes are formed:
    • enter image description here
  • The following conditions are required:
    • The light from both slits must be coherent, ie, constant or zero phase difference.
    • The distance of between the slits and the slit width must be negligible compared to the distance between the screen and the slits.
    • The slit width should be comparable to the wavelength of the light for circular diffraction.

Screenshot 2020-03-22 at 4 42 10 PM

  • Since D>>dD >> d, it can be assumed that the two rays are parallel.
  • For constructive interference at point PP, Δx=nλ\Delta x = n\lambda, where n=0,1,2,3,4,n = 0, 1, 2, 3, 4, \dots.
    • In this case, n=1n = 1.
    • Δx=λ=dsinθ\Delta x = \lambda = d\sin\theta
    • tanθ=sD\tan\theta = \frac{s}{D}
    • We can assume tanθ=sinθ\tan\theta = \sin\theta because θ\theta is extremely small.
    • λ=d(tanθ)=dsD\lambda = d (\tan \theta ) = \frac{ds}{D}
  • Finally, we can conclude that:
    • s=λDds = \frac{\lambda D}{d}

The graph of light intensity vs angle from center for both single and double slit diffraction is as follows:
enter image description here

  • The intensity of the double slit pattern is modulated by the one-slit pattern.

Multi-slit diffraction

  • If the number of slits are increased (with the same length between each slit), the fringes are more distinctly pronounced:

enter image description here

  • For NN slits, there are N2N - 2 secondary maxima between two primary maxima.
  • With an increase in the number of slits to NN:
    • the primary maxima will become thinner and sharper
    • The N2N - 2 secondary maxima will become unimporant
    • The intensity of the central maximum is proportional to N2N^2.

Diffraction grating

  • used in spectroscopy to measure the wavelength of light.
  • Has rulings, which are slits, which help determine the slit seperation.
  • xx lines/rulings per millimetre corresponds to a 1x\frac{1}{x} slit seperation.
  • d=1xd = \frac{1}{x}
  • Since we know that the condition for constructive interference is dsinθ=nλd\sin\theta = n\lambda, we can use this to calculate the wavelength of light.

Thin-film interference

  • Upon reflection on a surface with a refractive index greater than the medium a ray is already in, the ray undergoes a phase change of π\pi.

  • If dd is the thickness of the film, the condition for constructive interference, when only one phase change occurs (light is reflected off a medium with a greater refractive index only once):
    2dn=(m+12)λ2dn = (m + \frac{1}{2})\lambda
  • Where:
    • dd : thickness of the film.
    • nn : refractive index.
    • mm : an integer.

Note: It is similar to the condition for destructive interference (n+12λn+\frac{1}{2}\lambda), but it is the condition for constructive interference in this case because of the phase shift of π\pi, resulting in crests becoming troughs and troughs becoming crests.

  • The condition for destructive interference, when only one phase change occurs(light is reflected off a medium with a greater refractive index only once):
    2dn=mλ2dn = m\lambda
  • The condition for constructive interference when there is no or two phase changes (light is reflected off a medium with a greater refractive index twice):
    2dn=mλ2dn = m\lambda
  • The condition for destructive interference when there is no or two phase changes (light is reflected off a medium with a greater refractive index twice):
    2dn=(m+12)λ2dn = (m + \frac{1}{2})\lambda


  • The angular separation of two objects is given by θA=sD\theta_A=\frac{s}{D} where ss is the distance between the objects, and DD is the distance between the observer and the objects.
  • According to the Rayleigh criterion, resolution is possible when the angular separation is greater than the angle fo the first diffraction minimum: θAλb\theta_A \geq \frac{\lambda}{b}
  • For a circular slit, the following criteria is used: θA1.22λb\theta_A \geq 1.22\frac{\lambda}b
    Screenshot 2020-06-02 at 5 35 55 PM

In a diffraction grating

R=λavgΔλ=mNR = \frac{\lambda_{avg}}{\Delta \lambda} = m N

  • RR: resolving power of the grating
  • λavg\lambda_{avg}: The average of the two wavelengths to be resolved
  • Δλ\Delta \lambda: difference in the wavelengths that are to be resolved
  • mm:

Doppler Effect

  • The doppler effect is the change in the observed frequency of a wave which happens whenever there is relative motion between the source and the observer.
  • If the wavelength of the light decreases, the source is moving towards the observer, which is called a blueshift.
  • If the wavelength of the light increases, the source is moving away from the observer, which is called a redshift.
  • If a source is moving towards an observer with velocity usu_s:
    λ=λ(1usv)f=f(vvus) \lambda' = \lambda(1- \frac{u_s}{v})\\[1.5em] f' = f(\frac{v}{v - u_s})
  • If an observer is moving towards a source with velocity uou_o:
    f=f(v+uov) f' = f(\frac{v+u_o}{v})

For light

  • If the speed of the observer is small compared to the speed of light, then: ΔffvcΔλλvc\frac{\Delta f}{f} \approx \frac{v}{c}\\[1.5em]\frac{\Delta \lambda}{\lambda} \approx \frac{v}{c}


  • K. A. Tsokos - Physics for the IB Diploma, Sixth Edition
  • Class Notes