Lecture15 - Superposi0on of sound waves Superposi0on...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Superposi0on of sound waves Superposi0on of waves In many situa0ons different waves interact with each other. For example, if two different rocks hit a lake, the ripples they make will meet and interact. When two waves meet, they create a new wave, the resultant wave, which has a displacement that is simply the sum of the displacements of the individual waves: y ( x, t ) = y1 ( x, t ) + y2 ( x, t ) Chapter 18 This is the principle of superposi5on for waves. As a result of superposi0on, if two pulses move in opposite direc0on, they do not alter each others travel. Waves interfere at the moment they pass each other. The example of ripples created by two rocks we men0oned above is also an example of interference of waves. We study interference of waves using the principle of superposi0on of waves. Interference of sinusoidal waves Let’s assume we have two waves of the same amplitude A and same angular frequency ω and wave number k. The only difference is that one of the waves started at a different 0me than the other and therefore has a phase shiJ f with respect to the other wave. y1 ( x, t ) = A sin ( kx − ω t ) y2 ( x, t ) = A sin ( kx − ω t + ϕ ) The resultant wave: y( x, t ) = A sin( kx − ω t ) + A sin( kx − ω t + ϕ ) Now use: 1⎞ ⎛ 1⎞ ⎛ y( x, t ) = ⎜ 2 A cos ϕ ⎟ sin ⎜ kx − ω t + ϕ ⎟ ⎝ 2⎠ ⎝ 2⎠ Amplitude Oscilla0ng term Interference of sinusoidal waves (contd…) ϕ⎞ ⎛ 1⎞ ⎛ y( x, t ) = ⎜ 2 A cos ⎟ sin ⎜ kx − ω t + ϕ ⎟ ⎝ 2⎠ ⎝ 2⎠ ϕ = 0 (In phase) ϕ = π (totally out of phase) ϕ = 2π/3 Construc0ve Interference Destruc0ve Interference Intermediate case Standing waves What happens if two con0nuous traveling waves travel in opposite direc0ons and interfere with each other ? The resultant wave will develop nodes, i.e. points at which the there is no mo0on due to con0nuous destruc0ve interference. The points in ­between where there is a maximum amplitude are called an5nodes. Since the wave does not seem to move, but just oscillate, such a wave is called standing wave. Standing Waves (contd…) Assume two waves are traveling in opposite direc0on and interfere with each other: y1 ( x, t ) = A sin ( kx − ω t ) y2 ( x, t ) = A sin( kx + ω t ) y( x, t ) = y1 ( x, t ) + y2 ( x, t ) ⇒ y( x, t ) = 2 A sin kx cos ω t Thus for all 0mes, the amplitude of the resultant wave will be zero where: λ 3λ 5λ , λ, , 2 λ, .... 2 2 2 π 3π 1 1λ Similarly, the an0nodes are at: sin kx = 1 ⇒ kx = , , ... = (n + )π ⇒ x = (n + ) 22 2 22 λ 3λ 5 λ i.e., at: x = , , .... 444 i.e., the posi0ons of the nodes are: x= Note that: two adjacent nodes are separated by λ/2 two adjacent an0nodes are separated by λ/2 the distance between a node and adjacent an0node is λ/4 Wave reflec0on Standing waves occur most oJen in cases where waves are reflected at a boundary. There are two types of reflec0on: Reflec0on off a fixed end and reflec0on off a loose end. At a fixed end the incoming pulse will exert a force on the support. The support by Newton’s third law will exert an opposite force on the string, causing a pulse with the opposite deflec0on (see (a)). The fixed end is a node in the standing wave. At a loose end, the end is moved up by the pulse and causes a pulse with the same sign to be reflected. A loose end acts as an an0node. Standing waves & resonance When a string is clamped on both sides, it ‘wants’ to oscillate with wavelengths that correspond to having a node at each clamped end. Thus there are certain natural wavelengths (and frequencies) associated with a certain length L of a clamped string (as in a string instrument): λ 2L n = 1, 2, 3, .... L = n ⇒ λn = 2 n We already know that the speed of the wave v is determined by the thickness (strictly, linear mass density) of the string and the tension in the string, but not by the length of the string: v = T / µ The corresponding natural frequencies are given by: f= v v nT ⇒ fn = n = λ 2L 2L µ ⇒ fn = nf1 Thus the frequency is quan0zed. The most basic mode of vibra0on is that with frequency = f1. This is called the fundamental mode or the first harmonic. The second harmonic has n=2 and so on. All of them together make up the harmonic series. Standing waves in air columns We can have standing waves of sound in air columns in a pipe. There are two cases: A.  Open ended pipes The open end of an air column is approximately a displacement an0node and a pressure node. Here we see that Since v = fλ, v fn = n 2L nλn 2L = L ⇒ λn = 2 n Recall that v = f1 = B ρ Thus the harmonics are v 2L f 2 = 2 f1 , f3 = 3 f1 , f 4 = 4 f1 , ..... In open ended pipes, all integer mul5ples of the fundamental frequency are excitable. In a flute the effec0ve length changes as the musician alters the placement of her fingers, thus crea0ng higher or lower frequencies. Pipe with 1 closed end In single end ­closed pipes: only odd ­integer mul0ples of the fundamental frequency are allowed From the figure we have f n = n v , where n = 1, 3, 5, … . 4L But note that n can have only odd integral value. Even mul0ples of the fundamental frequency are absent. What is the effect of temperature? It comes from the speed of sound v. Recall that v = T , where T is the absolute temperature. Example In the setup shown in the picture, a 1.2 m long string is vibrated at 120 Hz. The string has linear density of 1.6 g/m. What mass m do we have to hang on the string to set up the fourth harmonic of the string ? Fourth harmonic, n = 4: Solve for m: f =n ν 1 T 2 mg =4 = 2L 2L µ L µ Example A ver5cal pipe with both ends open is submerged in water, and a tuning fork is vibra5ng at an unknown frequency is placed near the top. The length L of the pipe is ver5cally adjustable. The Amplitude of the sound is magnified when the length of the pipe corresponds to a resonant wavelengths of the tuning fork. For a certain pipe, the smallest value of L which a peak occurs is = 9.0cm. (a) what is the frequency of the tuning fork? (b) what are the L2 and L3 values (a) Note that the water acts like a closed end to the pipe so we need to use the 1 end closed frequency equa0on. fn = n v , v = 343m / s 4L f1 = v 343m / s = = 953Hz 4 L 4 ⋅ 0.09 m (b) Recall that closed ended pipes resonate only odd ­ integer mul0ples of frequency so the next 2 are f3 and f5. Beats When two oscillators are simultaneously oscilla0ng with different but rela0vely close frequencies, a periodic “beat” is heard. This bea0ng is the result of interference in 0me: meaning that periodically the superposi0on of the two waves are in phase (construc0vely interfering) and periodically out of phase (destruc0ve). It is clear in graph(a) that the two frequencies are just out of 0me with one another, crea0ng a paeern of construc0ve and destruc0ve interference like 2 tuning forks of similar frequency. Beats (contd …) Previously we’ve discussed the superposi0on of waves and their types of interference. Beats arise from alterna0ng construc0ve and destruc0ve interference between two frequencies. y1 = A cos( 2π f1t ) , y2 = A cos( 2π f 2t ) y = y1 + y2 = A(cos( 2π f1t ) + cos( 2π f 2t )) ⎡ ⎛ f1 − f 2 ⎞ ⎤ ⎛ f1 + f 2 ⎞ y = ⎢ 2 A cos 2π ⎜ t ⎥ cos 2π ⎜ t ⎝ 2 ⎟⎦ ⎠⎥ ⎝2⎟ ⎠ ⎢ ⎣ ⎛ f + f2 ⎞ y = ⎡ yenvelope ⎤ cos 2π ⎜ 1 t ⎣ ⎦ 2⎟ ⎝ ⎠ ⎡ ⎛ f − f2 ⎞ ⎤ yenvelope = ⎢ 2 A cos 2π ⎜ 1 t⎥ 2 ⎟⎥ ⎝ ⎠⎦ ⎢ ⎣ ⎛f−f⎞ Maximum amplitude is found when: cos 2π ⎜ 1 2 ⎟ t = ±1 ⎝2⎠ Thus there are two maxima in each beat period (period of the envelope). Thus the beat frequency is given by: ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online