16 - Superposition of waves

Question 1

Superposition

Answer 1

• When two or more waves of the same type meets at a point, the resultant oscillation is a sum of the individual oscillations
• Coherent waves = same type, frequency and wavelength
• two peaks meet to make constructive interference, increasing the sound, and a peak and a trough meet to make destructive interference, cancelling out the sound

Question 2

Interference Patterns

Answer 2

• two coherent sources can produce circular wave patterns that create interference patterns of Minima/Maxima where destructive/ constructive interference take place and create areas where the wave is smaller/larger
• if the path difference of the source to a point is equal to a whole multiple of the wavelength then the interference is constructive

Question 3

Light Slit experiment

Answer 3

• Light can be diffracted by a suitably small slit
• two slits cause the diffracted waves to overlap and interact in an interference pattern. If two separate waves interact they don’t do this as they aren’t coherent
• when the interference pattern shines on a surface, a series of dots are seen with gaps between them (these are the maxima)
• different colours of light diffract differently due to wavelengths
• wavelength = slit width*fringe separation/distance from slits
• order of maximawavelength = slit separationsin(theta)

Question 4

Standing waves in string

Answer 4

• made by vibrating a string and holding it either side so it reflects back in anti-phase and has a node at each side. A node is where the two waves are touching
• anti-nodes are where the two waves are furthest apart
• standing waves store energy
• frequency of standing wave in string = (1/2length)(sqroot of Tension/mass per unit length)
• fundamental frequency has one antinode, length is 1/2 wavelength
• first overtone = second harmonic = 1 node + 2 antinodes
• general rule is antinodes = harmonic, nodes = overtones
• speed of transverse wave in string = sqrt(Tension/mass per unit length)

Question 5

standing waves in tubes

Answer 5

• open ends have antinodes closed ends have nodes
• first frequency has length = 1/4 wavelength (closed) or 1/2 wavelength (open) - closed tube is the frequency of equivalent open pipe
• in wind instruments, holes are opened to create anti-nodes and decrease the wavelength

Question 6

Diffraction

Answer 6

• caused by the individual spherical wavelets of the wavefront being pushed through a gap and curving the edges of the wavefront
• diffraction is greater when slit width is close to wavelength
• diffraction grating experiment creates a series of maxima when laser light is shone through a grating with a tiny d, each of the maxima is a dot on the screen/wall the light is projected onto
• nwavelength = dsin(theta)
• n = order of maxima being used, d = diffraction spacing in slit experiment, theta = angle between central maxima and the diffracted maxima

Question 7

Electron Diffraction

Answer 7

• de broglie wavelength of a particle = plancks constant (6.63*10^-34)/momentum
• electrons are accelerated by an accelerated by an electric field towards a layer of graphite, which has a spacing similar to the wavelength of an electron at a specific voltage
• electrons are diffracted like waves, proving the theory, as they form concentric circle patterns (each circle is a maxima)

Question 8

Path difference, phase and wavefronts

Answer 8

• path difference is the distance from source to a point on the wave
• phase difference is the distance between the peaks of two identical waves, and it tells you how in -phase they are
• a wavefront is an imaginary line on the surface of a wave at which all points are in phase
• according to Huygens construction these wavefronts are made of spherical wavelets, and when the straight line of the wavefront is interrupted in diffraction, the curvature of the wavelets is revealed