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Home , Longitudinal wave, Standing wave

149: Lecture 24

• Chapter 11: – 11.8 and Refraction – 11.10 Standing Waves

• Chapter 12: – 12.1 Sound Waves – 12. 4 Standing Sound Waves

Lecture 24 Purdue University, Physics 149 1 ILQ 1

A thick string and a thin string made of the same material hliWhihhhldfhave equal tensions. Which has the larger speed of waves on these strings?

A) thin string B) thi c k s tr ing C) waves have the same speed

Lecture 24 Purdue University, Physics 149 2 ILQ 2

Of these properties of a , the one that is independent ofhf the ot hers is its

A) B) C) D) speed

Lecture 24 Purdue University, Physics 149 3 Waves y(x,t) = A sin(ωt – kx) A = Amplitude= Maximum displacement of a point on the wave λ = Wavelength: Distance between identical points on the wave T = Period: Time for a point on the wave to undergo one complete oscillation. f = frequency=1/T ω = angular frequency= 2π/T k = wave number = 2π/λ λ ω v = = λ f = T k Lecture 24 Purdue University, Physics 149 4 Wave Speed

y(x,t) = Acos(ωt − kx) ωtkx−=constant so ωΔ−tkx Δ = 0 gives Δx ω ==v Δtk

A wave y = A cos(ωt - kx) travels in +x direction A wave y = A cos(ωt + kx) travels in -x direction

Lecture 24 Purdue University, Physics 149 5 Suppperposition Princip le

• When two or more waves pass through the same region the actual displacement is the sum of the separate displacements. y′(x,t) = y (x,t) + y (x,t) 1 2 • If two waves pass through the same region they continue to move independently.

Lecture 24 Purdue University, Physics 149 6 Interference

• Interference describes what happens when two waves are travelling in t he same direct ion t hroug h t he same reg ion o f space. • Su perposition principle: the res ulting displacement is the algebraic sum of their separate displacement

yy(x’(x, t) = y1(x, t) + y2(x, t) – It the two waves arrive with the same amplitude (both waves have a crest): constructive interference and the resultant pulse is larger then either pulses. – It the two waves arrive with opposite amplitude (one has a crest, the other a valley): destructive interference the resulting pulse could be zero.

Lecture 24 Purdue University, Physics 149 7 Interference and Suppperposition

Lecture 24 Purdue University, Physics 149 8 Reflection

• When a wave travels from one boundary to another, reflection occurs. Some of the wave travels backwards from the boundary – Traveling from fast to slow inverted – Traveling slow to fast upright

Lecture 24 Purdue University, Physics 149 9 Reflection and Transmission z When a wave strikes an obstacle or comes at the end of the medium it is travelling in, it is reflected (at least in part). z If the end of the rope is fixed the reflected pulse is inverted z If the end of the rope is free Antinode the reflected pulse is not inv erted

Lecture 24 Purdue University, Physics 149 10 Reflected Wave: Inverted or Not?

• The reflected wave will be inverted if it reflects from a medium with a higher mass density (that is, with a lower wave speed; recall ). • The reflected wave will not be inverted if it reflects from a medium with a lower mass density (that is, with a higher wave speed).

Å This phenomenon can be explained by

(1) The principle of superposition at the fixed point at the end, or

(2) Newton’s third law fthffor the forces b btetween th thtie string and dthll the wall.

Lecture 24 Purdue University, Physics 149 11 ILQ

• A on a string is described by y(x,t) = A cos(ωt + kx). It arrives at the point x = 0 where the string is fixed in place. Which function describes the reflected wave?

a) A cos(ωt + kx) b) A cos(ωt – kx) c) –A cos(ωt – kx) d) –A sin(ωt + kx)

Lecture 24 Purdue University, Physics 149 12 Example

• You send a wave pulse (upright) down a lightweight rope, which is tied to a very heavy rope. At the boundary,

a) the re fl ected wa ve will be upri gh t an d th e tr an smi tted wave will be inverted. b) the reflected wave will be inverted and the transmitted wave will be upright. c) the reflected wave will be inverted and the transmitted wave will be inverted. d) the reflected wave will be upright and the transmitted wave will be upr ig ht.

Lecture 24 Purdue University, Physics 149 13 ILQ

• You send a wave pulse (upright) down a very heavy rope, which is tied to a lightweight rope. At the boundary,

a) the re fl ected wa ve will be upri gh t an d th e tr an smi tted wave will be inverted. b) the reflected wave will be inverted and the transmitted wave will be upright. c) the reflected wave will be inverted and the transmitted wave will be inverted. d) the reflected wave will be upright and the transmitted wave will be upr ig ht.

Lecture 24 Purdue University, Physics 149 14 Diffraction Diffraction is the spreading of a wave around an obstacle in its path .

Diffraction: Bending of waves around obstacles

Lecture 24 Purdue University, Physics 149 15 Speed of Waves Depends on Medium

When a wave reaches a boundary between two different media the speed and the wavelength change, but the frequency remains the same.

v v f = 1 = 2 λ1 λ2

Lecture 24 Purdue University, Physics 149 16 Law of Refraction

θ1

θ2 sin(θ ) v 1 = 1 sin(θ2) v2 Angle of incidence and angle of refraction Lecture 24 Purdue University, Physics 149 17 Standing Waves • Standing waves occur when a wave is reflected at a boundary in such a way that the wave appears to stand still. • In a on a string, every point moves (as a whole) in simple harmonic motion (SHM) with the same frequency. • Every point reaches its maximum amplitude simultaneously, and every point also reaches its minimum amplitude (namely, zero) simultaneously as well. • Nodes are points of zero amplitude (that is, points that never move); antinodes are points of maximum amplitude. The distance between two adjacent nodes is ½ λ. • Fixed end of a string is always a node, because it never moves.

Lecture 24 Purdue University, Physics 149 18 Standing Waves

• Nodes: points of destructive interference where the cord remains st ill at a ll t imes. We have a lways no des at fixe d ends. • Antinodes: points of constructi ve interference where the cords oscillates with maximum amplitude. • The nodes and antinodes remain in a fixed position for a given frequency. Standing wave can occur at resonant . • The lowest frequency that produces the standing wave yields a wave with one antinode. If you double the frequency there will be two anti-nodes etc.

Lecture 24 Purdue University, Physics 149 19 How Do We Get Standing Waves? • Graphical Understanding

Standing wave: superposition of incident and reflected waves • Mathematical Understanding – IidtIncident wave: A cos(ωt + kx) Å tthlftto the left – Reflected wave: –A cos(ωt – kx) Å to the right, inverted – Accordinggpppp, to the principle of superposition, y(x,t) = A cos(ωt + kx) – A cos(ωt – kx) = 2A cos(ωt) sin(kx) Every point moves in SHM!

Lecture 24 Purdue University, Physics 149 20 Possible and Frequencies

= 2L/1

= 2L/2

= 2L/3

• f1 is called the fundamental

= 2L/4 frequency.

• f1, 2f1, 3f1,…,nf1,… (tha t i s fn ) are called the natural frequencies or resonant frequencies. Don’t forget that • occurs when a system fixed ends of the string (like a ) is driven at one of its are always nodes. natural frequencies.

Lecture 24 Purdue University, Physics 149 21 Standing Waves z Shake the end of a string y(,)xt=+ Asin (()ω t kx ) z If the other end is fixed, a wave travels down and it is reflected back ⇒ interference y(,)xt=− A sin(ω t − kx ) nodes z The two waves i n terf ere => st andi ng wave yxt(,)= 2 A cos()sin()ω t kx z Nodes when sin (nπ)=0: nnπ λ Anti-nodes x == n = 0120,1,2 k 2 z Anti-nodes sin (nπ)=+/- 1, half way between nodes

Lecture 24 Purdue University, Physics 149 22 Standing Waves

• The natural frequency is related to the length of the string L. The lowest frequency (first harmonic) has one antinode 1 L = λ1 λ1 = 2L 2 • The second harmonic has two antinodes λ = L 2 • The n-th harmonic 2 λn = L n Lecture 24 Purdue University, Physics 149 23 Standing Waves

z Once you know the wavelength you also know the fddhdiRllfrequency needed to have a standing wave. Recall:

v v nv f = fn = = = nf1 λ λn 2L

z Note that energy is not transmitted by a standing wave. Since the string is at rest at the nodes no energy flows.

Lecture 24 Purdue University, Physics 149 24 ILQ

• A violin string of length L is fixed at both ends. Which one of these is not a wavelength of a standing wave on the string?

a) L b) 2L c) L/2 d) L/3 e) 3L/2

Lecture 24 Purdue University, Physics 149 25 ILQ

• A violinist discovers while tuning her violin that her string is flat (has a lower frequency). She should

a) tighten the string b) loosen the string c) plftlay faster

Lecture 24 Purdue University, Physics 149 26 Standing Waves L = λ / 2

f1 = (lowest possible)

A guitar’s E-stringgg has a length of 65 cm and is stretched to a tension of 82N. If it vibrates with a fundamental frequency of 329.63 Hz, what is the mass of the string? F v = f = v / λ tells us v if we know f and λ μ v2 = F / μ v=v = λ f μ = F/vF / v2 = 2 (0.65 m) (329.63 s-1) m= F L / v2 = 428.5 m/s = 82 (0.65) / (428.5)2 = 2.9 x 10-4 kg Lecture 24 Purdue University, Physics 149 27 Sound Waves

• Sound is a , with compressidftifiions and rarefactions of air.

• The wave can be described by the gauge pressure p (a measure of the compression and rarefaction of the air) as a function of time.

• The wave can also be described by the displacement s (= Δx) of an element of the air, which will oscillate around an equilibrium position (in the direction of energy transport). The restoring force is caused by the air pressure. When p is zero (node), s is maximum or minimum (antinode)!

Lecture 24 Purdue University, Physics 149 28

Lecture 22 Purdue University, Physics 220 29 Frequencies of Sound Waves

• Humans with excellent hearing can hear frequencies from 20 Hz to 20 kHz (audible range).

• The terms and are used to describe soundithfibl20Hdb20d waves with frequencies below 20 Hz and above 20 kHz, respectively.

• Some animals (such as dogs and dolphins) can here ultrasound, while some others (such as elephants) can here infrasound.

Lecture 24 Purdue University, Physics 149 30 Velocity ILQ

A sound wave having frequency f0, speed v0 and wavelength λ0, is traveling through air when it encounters a large helium- filled balloon. Inside the balloon the frequency of the wave is

f1, its speed is v1, and its wavelength is λ1. CthdfthdCompare the speed of the sound wave inside and outside the balloon

A) v1 < v0

B) v1 = v0

C) v1 > v0

V =965m/s V0=343m/s 1

Lecture 22 Purdue University, Physics 220 31 Freqqyuency ILQ

A sound wave having frequency f0, speed v0 and wavelength λ0, is traveling through air when it encounters a large helium- filled balloon. Inside the balloon the frequency of the wave is f1, its speed is v1, and its wavelength is λ1. Compare the frequency of the sound wave inside and outside the balloon

A) f1 < f0

B) f1 = f0 f0 f1 C) f1 > f0

Time between wave peaks does not change! Lecture 22 Purdue University, Physics 220 32 Wavelength ILQ

A sound wave having frequency f0, speed v0 and wavelength λ0, is traveling through air when it encounters a large helium- filled balloon. Inside the balloon the frequency of the wave is f1,,p its speed is v1,,g and its wavelength is λ1. Compare the wavelength of the sound wave inside and outside the balloon A) λ < λ 1 0 λ0 λ1

B) λ1 = λ0

C) λ1 > λ0

λ = v / f

Lecture 22 Purdue University, Physics 220 33