Ch 21 – Superposition and Standing Waves

Superposition and Interference

Differences between waves and particles Particles have (1) zero size and (2) must exist at different location Waves have (1) characteristic size, λ, and can (2) combine at one point

(a) If two or more traveling waves are moving through a medium, the resultant value of the wave function at any point is the algebraic sum of the values of the wave functions of the individual waves (b) Waves that obey the superposition principle are linear waves

(c) For mechanical waves, linear waves have amplitudes much smaller than their wavelengths (d) Two traveling waves can pass through each other without being destroyed or altered, because of the superposition principle

Interference is the combination of separate waves in the same region of space which produces a resultant wave

The two pulses, y1 & y2, with elements of positive displacement are traveling in opposite directions with same speeds

Waves start to overlap, the resultant wave function is y1 + y2 When crest meets crest the resultant wave has a larger amplitude than either of the original waves

The two pulses separate They continue moving in their original directions The shapes of the pulses remain unchanged

Below two equal, symmetric pulses are traveling in opposite directions on a stretched spring, obeying the superposition principle.

Below (rope and spring) two pulses are traveling in opposite directions. Their displacements are inverted with respect to each other.

When they overlap, their displacements partially cancel each other

If two identical waves are traveling in the same direction, with the same frequency, wavelength and amplitude; BUT differ in phase the waves add together. y = y1 + y2 where y1 = A sin (kx - ωt) and y2 = A sin (kx - ωt + φ) y = A sin(kx - ωt) + A sin(kx - ωt + φ) Apply trig identity: sin a + sin b = 2 cos((a-b)/2) sin((a+b)/2) A sin ( a ) + A sin ( b ) = 2A cos((a-b)/2) sin((a+b)/2) y = 2A cos (φ /2) sin (kx - ωt + φ/2) The resultant sinusoidal wave has the same frequency and wavelength as the original waves, but the amplitude has changed: Amplitude equals 2A cos (φ /2) with a phase angle of φ/2

Constructive Interference When φ = 0 (crest to crest and trough to trough), then cos (φ /2) = 1. The amplitude of the

resultant wave is A1 + A2 = 2A. The waves are “in phase.”

Destructive Interference When φ = π (crest meets trough), then cos (φ/2) = 0 and the amplitude of the resultant wave is 0

The wave are “out of phase.”

When φ is other than 0 (or an even multiple of π), the amplitude of the resultant is between 0 and 2A The wave functions still add.

Sound from the speaker can reach the receiving ear, R, by two different paths. The upper path can be varied Constructive interference occurs @ Δr = |r2 – r1| = n λ (n = 0, 1, …) Destructive interference occurs @ Δr = |r2 – r1| = (n/2) λ (n is odd) φ occur à unequal path lengths φ = Δr (2π λ)

Constructive Interference Δr = 2n λ/2

Destructive Interference Δr = (2n+1) λ/2

Example Two waves are traveling in the same direction along a stretched string. The waves are 90.0° out of phase. Each wave has an amplitude of 4.00 cm. Find the amplitude of the resultant wave.

y = 4 sin(kx – ωt) + 4 sin(kx – ωt + 90°) Apply below trig identity y = 2(4) cos(90°/2) sin(kx-ωt + 90°/2) y = 8 cos(45°) sin (kx - ωt + 45°) A = 8 cos(45°) A = 5.66 cm

Use the trig identity à y = A sin (kx - ωt) + A sin (kx - ωt + φ) sina + sinb = 2[cos(a-b)/2][sin(a+b)/2] y = 2A cos (φ /2) sin (kx - ωt + φ /2)

Demo: Destructive Interference Speakers: OW-B-DI

Standing Waves

Given two identical waves, y1 = A sin (kx – ωt) and y2 = A sin (kx + ωt), they interfere according to the superposition principle & with the trig identity: sin(a±b) = sina cosb ± cosa sinb (Application Exercise: Apply the above identity) we know y = y1 + y2 is y = (2A sin kx) cos ωt ß This is the wave function of a standing wave There is no kx – ωt term; therefore it is not a traveling wave, thus it’s called a standing wave (appears to stands still) There are three types of amplitudes used in describing waves v The amplitude of the individual waves, A v The amplitude of the simple harmonic motion of the elements in the medium, § 2A sin kx v The amplitude of the standing wave, 2A

Nodes occurs at a point of zero amplitude x = n λ /2 à n = 0, 1, 2, …

Antinode occurs at a point of maximum displacement, 2A x = n λ /2 + λ /4 à n = 0, 1, 2, …

Recall: Resonance in Air Column from lab

Example Two sinusoidal waves traveling in opposite directions interfere to produce a standing wave with the wave function, y = (1.50 m) sin(0.400x) cos(200t), where x is in meters and t is in seconds. Determine the wavelength, frequency, and speed of the interfering waves.

apply trig identity sin(a±b) = sina cosb ± cosa sinb (a = kx and b = ωt) to y = A sin (kx - ωt) + A sin (kx + ωt) sin(a±b) = sina cosb ± cosa sinb sin (kx - ωt) = sin kx cos ωt - cos kx sin ωt sin (kx + ωt) = sin kx cos ωt + cos kx sin ωt sin (kx - ωt) + sin (kx + ωt) = 2 sin kx cos ωt y = (1.50 m) sin(0.400x) cos(200t)

k = 2 π /l 0.4 = 2 π /l l = 5 π l = 15.7 meters

ω = 2π f 200 = 2 π f f = 100/ π f = 31.8 Hz

v = f l v = 31.8 (15.7) v = 500 m/s

Demo: Standing Waves, Cenco String: OW-B-SC

Standing Waves in a String Fixed at Both Ends

Consider a string fixed at both ends of length L v Standing waves are set up by a continuous superposition of waves incident on and reflected from the ends v There is a boundary condition on the waves

v The wavelengths of the normal modes for a string of length L fixed at both ends are λ= 2L / n n = 1, 2, 3, … v n is the nth normal mode of oscillation v These are the possible modes for the string

The natural frequencies are fnat = n v / 2L fnat = n √(FT/μ)/ 2L

The above situation in which only certain frequencies of oscillation are allowed is called quantization The fundamental frequency occurs at n = 1, and is the lowest frequency, ƒ1 Harmonics are integral multiples or the fundamental frequency and together they form the harmonic series

Quantization is a common occurrence when waves are subject to boundary conditions The above string is vibrating in its second harmonic

A musical note is defined by its fundamental frequency, (n = 1) The frequency of the string can be changed by changing either its length or its tension (changing it’s boundary condition)

Example: A middle “C” on a piano has a fundamental frequency of 262 Hz. What are the next two harmonics of this string?

Ans: ƒ1 = 262 Hz ƒ2 = 2ƒ1 = 524 Hz ƒ3 = 3ƒ1 = 786 Hz

Example Find the fundamental frequency and the next three frequencies that could cause standing-wave patterns on a string that is 30.0 m long, has a mass per length of 9.00 x 10^–3 kg/m, and is stretched to a tension of 20.0 N.

v^2 = FT / μ v^2 = 20 / .009 v = 47.1 m/s fnat = n v/2L f1 = n v/2L f1 = 47.1/(2*30) f1 = 0.786 Hz

f2 = 2(47.1)/(2*30) f2 = 1.57 Hz f3 = 3(47.1)/(2*30) f3 = 2.36 Hz f4 = 4(47.1)/(2*30) f4 = 3.14 Hz

Resonance

Resonance can occur in systems that are capable of oscillating in one or more normal modes

If a periodic force is applied to such a system, the amplitude of the resulting motion is greatest when the frequency of the applied force is equal to one of the natural frequencies of the system Because an oscillating system exhibits a large amplitude when driven at any of its natural frequencies, these frequencies are referred to as resonance frequencies, ƒo = largest amplitude The maximum amplitude is limited by friction in the system (of the medium the wave is traveling)

Example The chains suspending a child's swing are 2.00 m long. At what frequency should a big brother push to make the child swing with largest amplitude?

ω = √(g/L) 2πf = √(g/L) 2πf = √(9.8/2) f = 0.352

Standing Waves in Air Columns

The closed end of a pipe is a displacement node because the wall at this end doesn’t allow longitudinal motion of the air molecules. Thus the reflected wave is 180° out of phase with the incident wave. Also the pressure wave is 90° out of phase with the displacement wave, 17.2, we know the closed end of the tube corresponds to a pressure ANTINODE. The open end is a displacement AND pressure node.

Closed one end fnat = n v/4L v = f1 λ Open both ends fnat = n v/2L

Example The overall length of a piccolo is 32.0 cm. The resonating air column vibrates as in a pipe open at both ends. (a) Find the frequency of the lowest note that a piccolo can play, assuming that the speed of sound in air is 340 m/s. (b) Opening holes in the side effectively shortens the length of the resonant column. If the highest note a piccolo can sound is 4000 Hz, find the distance between adjacent antinodes for this mode of vibration.

(a) dAtoA = 0.320 m L = λ /2 λ = 0.32 (2) λ = 0.64 meters v = f λ 340 = f 0.64 f = 531 Hz

(b) v = f λ 340 = 4000 λ λ = 0.0850 m dAtoA = 0.0425 m

Standing Waves in Rods and Membranes

If a rod is clamped at a node, then a standing wave may be set up in a rod. Just add energy by rubbing rosin pad or similar material along the length of the unclamped portion of the rod. Demo: Singing Rods: OW-D-SR

An aluminum rod is clamped one quarter of the way along its length and set into longitudinal vibration by a variable-frequency driving source. The lowest frequency that produces resonance is 4400 Hz. The speed of sound in an aluminum rod is 5100 m/s. Determine the length of the rod.

When the rod is clamped at one-quarter of its length, the vibration pattern reads ANANA and the rod length is L = 2dAtoA = λ. v = f L à v = f λ 5100 = 400(L) L = 1.16 meters

Beats: Interference in Time

Beating is the periodic variation in amplitude at a given point due to the superposition of two waves having slightly different frequencies.

y1 = A cos ω1t y1 = A cos 2πf1t

y2 = A cos ω2t y2 = A cos 2πf2t

cos a + cos b = 2cos((a-b)/2)cos((a+b)/2) y = y1 + y2 y = A cos 2πf1t + A cos 2πf2t apply the above trig identity y = 2Acos((2πf1t - 2πf2t)/2)cos((2πf1t + 2πf2t)/2) y = 2Acos((2πt(f1 - f2)/2) cos((2πt(f1 + f2)/2) y = Anew cos((2πt(f1 + f2)/2)

where Anew = 2Acos((πt(f1 - f2)) ß

And cos((2πt(f1 - f2)/2) = ± 1 Since amplitude varies with ½(f1 - f2) we get two max amplitudes during one period So we get a beat à fbeat = |f1 – f2|

Example In certain ranges of a piano keyboard, more than one string is tuned to the same note to provide extra loudness. For example, the note at 110 Hz has two strings at this frequency. If one string slips from its normal tension of 600 N to 540 N, what beat frequency is heard when the hammer strikes the two strings simultaneously?

fλ = v = (FT/μ)^1/2 fnew / f = (FT-new/ FT)^1/2 fnew / 110 = (540/ 600)^1/2 fnew = 104.4 Hz Δf = 110 – 104.4 Δf = 5.64 beats/sec

Demo: Differential Tuning Forks: OW-B-DF

Non-sinusoidal Wave Patterns

Since the wave pattern of non-sinusoidal wave patterns are periodic, we can represent this as a combination of sinusoidal waves that form a harmonic series by implementing Fourier’s theorem (by Jean Baptiste Joseph Fourier, 1786-1830). y(t) = Sn (An sin 2pfnt + Bn cos 2pfnt) where f1 = 1/T and fn = nf1 and An, Bn are amplitudes of the waves corresponding to each n as n = 1 to n

Example Suppose that a flutist plays a 523-Hz C note with first harmonic displacement amplitude A1 = 100 nm. Read, by proportion, the displacement amplitudes of the above flute for harmonics 2 through 7. Take these as the values A2 through A7 in the Fourier analysis of the sound, and assume that B1 = … = B7 = 0. Construct a graph of the waveform of the sound. Your waveform will not look exactly like the flute waveform because you simplify by ignoring cosine terms; nevertheless, it produces the same sensation to human hearing.

We evaluate y(t) = (Ansin2πfnt + Bncos2πfnt)

s = 100sinθ + 157sin2θ + 62.9sin3θ + 105sin4θ + 51.9 sin5θ + 29.5 sin6θ + 25.3 sin7θ

where s represents particle displacement in nanometers and θ represents the phase of the wave in radians. As θ advances by 2π, time advances by (1/523) s.