490 14 Oscillations and Wave Motion the phase angle φ, the maximum speed vmax , and the maximum acceleration amax (c) Write down the position, velocity, and acceleration in terms of time t, then substitute with t = π/8 s and find their values Equilibrium position x x x=0 x=0 (a) xi (b) Fig 14.33 See Exercise (12) (14) A bullet of mass m = 10 g is fired horizontally with a speed v into a stationary wooden block of mass M = kg The block is resting on a horizontal smooth surface and attached to a massless spring with spring constant kH = 150 N/m, where the other end of the spring is fixed through a wall, as shown in Fig 14.34a In a very short time, the bullet penetrates the block and remains embedded before compressing the spring, as shown in Fig 14.34b The maximum distance that the block compresses the spring is cm, as shown in Fig 14.34c (a) What is the speed of the bullet? (b) Find the period T and frequency f of the oscillating system Before collision (a) M Just after collision M+m V 8cm (c) At maximum compression Fig 14.34 See Exercise (14) M+m Stage (b) v Stage m 14.8 Exercises 491 Section 14.2 Damped Simple Harmonic Motion (15) An object of mass m = 0.25kg oscillates in a fluid at the end of a vertical spring of spring constant kH = 85 N/m, see Fig 14.35 The effect of the fluid resistance is governed by the damping constant b = 0.07kg/s (a) Find the period of the damped oscillation (b) By what percentage does the amplitude of the oscillation decrease in each cycle? (c) How long does it take for the amplitude of the damped oscillation to drop to half of its initial value? Fig 14.35 See Exercise (15) kH m (16) A simple pendulum has a length L and a mass m Let the arc length s and the angle θ measure the position of m at any time t, see Fig 14.36 (a) When a damped force Fd = −bvs exists, show that the equation of motion of the pendulum is given for small angles by: m d2θ mg dθ + θ =0 +b dt dt L (b) By comparison with Eq 14.25, show that the above differential equation has a solution given by: θ = θ◦ e−bt/2m cos(ωd t), ωd = g b2 − L 4m2 where θ◦ is initial angular amplitude at t = and ωd = 2π fd is damped angular frequency, see Fig 14.9b (c) When the pendulum has L = m, m = 0.1 kg, and the angular amplitude θ becomes 0.5 θ◦ after minute, find the damping constant b and the ratio (f − fd )/f , where f is the undamped frequency 492 14 Oscillations and Wave Motion Fig 14.36 See Exercise (16) θ L m s s O −b s θ mg −m g sinθ Section 14.3 Sinusoidal Waves (17) Given a sinusoidal wave represented by y = (0.2 m) sin(k x − ω t), where k = rad/m, and ω = rad/s, determine the amplitude, wavelength, frequency, and speed of this wave (18) A harmonic wave traveling along a string has the form y = (0.25 m) sin(3 x − 40 t), where x is in meters and t is in seconds (a) Find the amplitude, wave number, angular frequency, and speed of this wave (b) Find the wavelength, period, and frequency of this wave? Section 14.4 The Speed of Waves on Strings (19) A uniform string has a mass per unit length of × 10−3 kg/m The string passes over a massless, frictionless pulley to a block of mass m = 135 kg, see Fig 14.37, and take g = 10 m/s2 Find the speed of a pulse that is sent from one end of the string toward the pulley Does the value of the speed change when the pulse is replaced by a sinusoidal wave? Fig 14.37 See Exercise (19) At time t τ mg 14.8 Exercises 493 (20) Assume a transverse wave traveling on a uniform taut string of mass per unit length μ = × 10−3 kg/m The wave has an amplitude of cm, frequency of 50 Hz, and speed of 20 m/s (a) Write an equation in SI units of the form y = A sin(kx − ω t) for this wave (b) Find the magnitude of the tension in the string Section 14.5 Energy Transfer by Sinusoidal Waves on Strings (21) A sinusoidal wave of amplitude 0.05 m is transmitted along a string that has a linear density of 40 g/m and is under 100 N of tension If the wave source has a maximum power of 300 W, what is the highest frequency at which the source can operate? (22) A long string has a mass per unit length μ of 125 g/m and is taut under tension τ of 32 N A wave is supplied by a generator as shown in Fig 14.38 This wave travels along the string with a frequency f of 100 Hz and amplitude A of cm (a) Find the speed and the angular frequency of the wave (b) What is the rate of energy that must be supplied by a generator to produce this wave in the string? (c) If the string is to transfer energy at a rate of 100 W, what must be the required wave amplitude when all other parameters remain the same? Fig 14.38 See Exercise (22) y Vibrator A V x (23) A sinusoidal wave is traveling along a string of linear mass density μ = 75 g/m and is described by the equation: y = (0.25 m) sin(2 x − 40 t) where x is in meters and t in seconds (a) Find the speed, wavelength, and frequency of the wave (b) Find the power transmitted by the wave 494 14 Oscillations and Wave Motion Section 14.6 The Linear Wave Equation (24) A one-dimensional wave traveling with velocity v is found to satisfy the partial differential equation [see Eq 14.58]: ∂ 2y ∂ 2y − =0 ∂x v2 ∂ t2 Show that the following functions are the solutions to this linear wave equation: (a) y = A sin(k x − ω t) (b) y = A cos(k x − ω t) (c) y = exp[b(x − v t)], where b is a constant (d) y = ln[b(x − v t)], where b is a constant (e) Any function y having the form y = f (x − v t) (25) If the linear wave functions y1 = f1 (x, t) and y2 = f2 (x, t) satisfy the wave Eq 14.58, then show that the combination y = C1 f1 (x, t)+C2 f2 (x, t) also satisfies the same equation, where C1 and C2 are constants Section 14.7 Standing Waves (26) A standing wave having a frequency of 20 Hz is established on a rope 1.5 m long that has fixed ends Its wavelength is observed to be twice the rope’s length Determine the wave’s speed (27) A stretched string of length 0.6 m and mass 30 g is observed to vibrate with a fundamental frequency of 30 Hz The amplitude of any antinodes in the standing wave is 0.04 m (a) What is the amplitude of a transverse wave in the string? (b) What is the speed of a transverse wave in the string? (c) Find the magnitude of the tension in the string (28) A student wants to establish a standing wave with a speed 200 m/s on a string that is fixed at both ends and is 2.5 m long (a) What is the minimum frequency that should be applied? (b) Find the next three frequencies that cause standing wave patterns on the string (29) Two identical waves traveling in opposite directions in a string interfere to produce a standing wave of the form: y = [(2 m) sin(2 x)] cos(20 t) where x is in centimeters, t is in seconds, and the arguments of the sine and cosine are in radians Find the amplitude, wavelength, frequency, and speed of the interfering waves 14.8 Exercises 495 (30) A standing wave is produced by two identical sinusoidal waves traveling in opposite directions in a taut string The two waves are given by: y1 = (2 cm) sin(2.3 x − t) and y2 = (2 cm) sin(2.3 x + t) where x and y are in centimeters, t is in seconds, and the argument of the sine is in radians (a) Find the amplitude of the simple harmonic motion of an element on the string located at x = cm (b) Find the position of the nodes and antinodes on the string (c) Find the maximum and minimum y values of the simple harmonic motion of a string element located at any antinode (31) A guitar string has a length L = 64 cm and fundamental frequency f1 = 330 Hz, see part (a) of Fig 14.39 By pressing down with your finger on the string, it is found that the string is shortened in a way so that it plays an F note with a fundamental frequency f1 = 350 Hz, see part (b) of Fig 14.39 [Assume the speed of the wave remains constant before and after pressing] How far is your finger from the near end of the string? L = λ /2 L′= λ ′1 /2 f1 n =1 (a) f1′ n =1 (b) Fig 14.39 See Exercise (31) (32) A violin string oscillates at a fundamental frequency of 262 Hz when unfingered At what frequency will it vibrate if it is fingered two-fifths of the length from its end? (33) A string that has a length L = m, mass per unit length μ = 0.1 kg/m, and tension τ = 250 N is vibrating at its fundamental frequency What effect on the fundamental frequency occurs when only: (a) The length of the spring is doubled (b) The mass per unit length of the spring is doubled (c) The tension of the spring is doubled (34) Show that the resonance frequency fn of standing waves on a string of length L and linear density μ, which is under a tensional force of magnitude τ , is given √ by fn = n τ/μ/2L, where n is an integer 496 14 Oscillations and Wave Motion (35) Show by direct substitution that the standing wave given by Eq 14.62, y = (2 A sin k x) cos ω t is a solution of the general linear wave Eq 14.58: ∂ 2y ∂ 2y − 2 =0 ∂x v ∂t (36) End A of a string is attached to a vibrator that vibrates with a constant frequency f, while the other end B passes over a pulley to a block of mass m, see Fig 14.40 The separation L between points A and B is 2.5 m and the linear mass density of the string is 0.1 kg/m When the mass m of the block is either 16 or 25 kg, standing waves are observed; however, standing waves are not observed for masses between these two values Take g = 10 m/s2 (Hint: The greater the tension in the string, the smaller the number of nodes in the standing wave) (a) What is the frequency of the vibrator? (b) Find the largest m at which a standing wave could be observed L Vibrator A B m Fig 14.40 See Exercise (36) (37) Two identical sinusoidal waves traveling in opposite directions on a string of length L = m interfere to produce a standing wave pattern of the form: y = [(0.2 m) sin(2π x)] cos(20π t) where x is in meters, t in seconds, and the arguments of the sine and cosine are in radians (a) How many loops are there in this pattern? (b) What is the fundamental frequency of vibration of the string? (38) Two strings and 2, each of length L = 0.5 m, but different mass densities μ1 and μ2 , are joined together with a knot and then stretched between two 14.8 Exercises 497 fixed walls as shown in Fig 14.41 For a particular frequency, a standing wave is established with a node at the knot, as shown in the figure (a) What is the relation between the two mass densities? (b) Answer part (a) when the frequency is changed so that the next harmonic in each string is established Knot μ1 Srting μ2 Srting Fig 14.41 See Exercise (38) (39) The strings and of exercise 38 have L1 = 0.64 m, μ1 = 1.8 g/m, L2 = 0.8 m, and μ2 = 7.2 g/m, respectively, and both are held at a uniform tension τ = 115.2 N Find the smallest number of loops in each string and the corresponding standing wave frequency (40) In the case of the smallest number of loops in exercise 39, determine the total number of nodes and the position of the nodes measured from the left end of string Sound Waves 15 Sound waves are the most common examples of longitudinal waves The speed of sound waves in a particular medium depends on the properties of that medium and the temperature As discussed in Chap 14, sound waves travel through air when air elements vibrate to produce changes in density and pressure along the direction of the wave’s motion Sound waves can be classified into three frequency ranges: (1) Audible waves: within the range of human ear sensitivity and can be generated by a variety of ways such as human vocal cords, etc (2) Infrasonic waves: below the audible range but perhaps within the range of elephant-ear sensitivity (3) Ultrasonic waves: above the audible range and lie partly within the range of dog-ear sensitivity 15.1 Speed of Sound Waves The motion of a one-dimensional, longitudinal pulse through a long tube containing undisturbed gas is shown in Fig 15.1 When the piston is suddenly pushed to the right, the compressed gas (or the change in pressure) travels as a pulse from one region to another toward the right along the pipe with a speed v The speed of sound waves depends on the compressibility and density of the √ medium We can apply equation v = τ/μ, which gives the speed of a transverse wave along a stretched string, to the speed of longitudinal sound waves in fluids or H A Radi and J O Rasmussen, Principles of Physics, Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-23026-4_15, © Springer-Verlag Berlin Heidelberg 2013 499 500 15 Sound Waves Fig 15.1 Motion of a longitudinal sound pulse in a gas-filled tube Undisturbed gas Compressed gas Compressed gas Compressed gas 536 16 Superposition of Sound Waves originally located at point O, which is a distance R = m from the center of the line connecting the two speakers The listener walks to point P, which is a distance y = 0.5 m above O, and thus reaches the first minimum in sound intensity Find the wavelength λ of the sound wave S1 L1 d/2-y P d/2 R d/2 y O d/2+y L2 S2 Fig 16.4 Solution: The first minimum in sound intensity occurs when the two waves reaching the listener at point P are 180◦ out of phase In other words, when their path difference equals λ/2 As per Fig 16.4, we first calculate the path lengths L1 and L2 as follows: L1 = R2 + (d/2 − y)2 = (5 m)2 + [(2 m)/2 − 0.5 m]2 = 5.0249 m R2 + (d/2 + y)2 = (5 m)2 + [(2 m)/2 + 0.5 m]2 = 5.2202 m and L2 = Thus, from Eq 16.6, the first destructive interference at point P leads to the following: |L2 − L1 | = λ ⇒ |5.2202 m − 5.0249 m| = λ/2 Therefore: λ = 0.3906 m = 39.06 cm ⇒ 0.1953 m = λ/2 16.3 Standing Sound Waves 16.3 537 Standing Sound Waves Assume we have two identical sound sources that face each other as shown in Fig 16.5 and driven by the same oscillator In this case, they produce two identical traveling waves each with a speed v These waves would be moving in opposite directions in the same medium Of course, these two waves combine according to the superposition principle Fig 16.5 Two identical sound sources emitting traveling waves towards each other, each with a speed v When the two waves overlap, they produce standing waves (not shown in the figure) To analyze this situation, we assume that the two sound sources generate sound waves that have the same frequency f, wavelength λ, and amplitude A but differ by traveling in opposite directions Therefore, we can write these two waves in the following form: y1 = A sin(kx − ω t), y2 = A sin(kx + ω t) (16.8) where y1 represents a wave traveling in the positive x-direction and y2 represents a wave traveling in the negative x-direction The superposition of y1 and y2 gives the following resultant: y = y1 + y2 = A [sin(kx − ω t) + sin(kx + ω t)] (16.9) To simplify this expression, we use the trigonometric identity: sin(a ± b) = sin a cos b ± cos a sin b (16.10) If we substitute in this identity with a = k x and b = ω t, then the resultant wave y reduces to: y = (2 A sin kx) cos ω t (16.11) 538 16 Superposition of Sound Waves The resultant y represented by Eq 16.11 gives a special kind of simple harmonic motion in which every element of the medium oscillates in simple harmonic motion with the same angular frequency ω (through the factor cos ω t) and an amplitude (given by the factor A sin kx) that varies with position x This wave is called a standing wave because there is no motion of the disturbance along the x-direction A standing wave is distinguished by stationary positions with zero amplitudes called nodes (see Fig 16.6) This happens when x satisfies the condition sin kx = 0, that is, when: kx = 0, π, 2π, 3π, When using k = 2π/λ, these values give x = 0, x = 0, λ 3λ , λ, , , that is: 2 3λ λ λ , λ, , = n , (n = 0, 1, 2, ) 2 (Nodes) (16.12) In addition, a standing wave is distinguished by elements with greatest possible displacements called antinodes (see Fig 16.6) This happens when x satisfies the condition sin kx = ±1, that is, when: kx = π 3π 5π , , , 2 λ 3λ 5λ Also, using k = 2π/λ, these values give x = , , , , that is: 4 x= λ 3λ 5λ λ , , , = (n + 21 ) , (n = 0, 1, 2, ) 4 (Antinodes) (16.13) Equations 16.12 and 16.13 indicate the following general features of nodes and antinodes (see Fig 16.6): Soptlight (1) The distance between adjacent nodes is λ/2 (2) The distance between adjacent antinodes is λ/2 (3) The distance between a node and an adjacent antinode is λ/4 16.3 Standing Sound Waves 539 y A When t= When t = p /2 When t= p o A sin k x A A A x N N N Antinode=A Node=N N N λ Fig 16.6 The time dependence of the vertical displacement (from equilibrium) of any individual element in a standing wave y is governed by cos ω t Each element vibrates within the confines of the envelope A sin kx The nodes (N) are points of zero displacement, and the antinodes (A) are points of maximum displacement In Fig 16.7a, at t = (ω t = 0), the two oppositely traveling waves are in phase, producing a wave pattern in which each element of the medium is experiencing its maximum displacement from equilibrium In Fig 16.7b, at t = T /4 (ω t = π/2), the traveling waves have moved one quarter of a wavelength (one to the right and the other to the left) At this time, each element of the medium is passing through the equilibrium position in its simple harmonic motion The result is zero displacement for each element at all values of x In Fig 16.7c, at t = T /2 (ω t = π ), the traveling waves are again in phase, producing a wave pattern that is inverted relative to the t = pattern The patterns at t = 3T /4 and t = T are similar to t = T /4 and t = 0, respectively y y y y y A A (a) t = y x N N N A N A N y N N N N (b) t = T / N x A y y A N N A N N A N x (c) t = T / Fig 16.7 Standing-wave patterns resulting from two oppositely traveling identical waves y1 and y2 at different phases The displacement is zero for each node (N), and is maximum for each antinode (A) 540 16 Superposition of Sound Waves Example 16.3 Two opposing speakers are shown in Fig 16.8 A standing wave is produced from two sound waves traveling in opposite directions; each can be described as follows: y1 = (5 cm) sin(4x − t), y2 = (5 cm) sin(4x + t) where x and y, are in centimeters and t is in seconds (a) What is the amplitude of the simple harmonic motion of a medium element lying between the two speakers at x = 2.5 cm? (b) Find the amplitude of the nodes and antinodes (c) What is the maximum amplitude of an element at an antinode? Fig 16.8 Standing wave Speaker Speaker x Solution: (a) Using the general form of a standing wave given by Eq 16.11, we find A = cm, k = rad/cm, and ω = rad/s Thus: y = (2 A sin kx) cos ω t = [(10 cm) sin(4x)] cos(2t) The amplitude of the simple harmonic motion of an element lying between the two speakers at x = 2.5 cm is the absolute value of the coefficient of cos(2t) evaluated at this point Thus: Amplitude = |(10 cm) sin(4x)|x = 2.5 | = |(10 cm) sin(10 rad)| = |−5.4 cm| = 5.4 cm (b) With k = 2π/λ = rad/cm, we have λ = π/2 cm Then, from Eq 16.12 we find that the nodes are located at: x=n λ π = n cm, (n = 0, 1, 2, ) From Eq 16.13, we find that the antinodes are located at: x = (n + 21 ) λ π = (n + 21 ) , (n = 0, 1, 2, ) (c) The maximum amplitude of antinodes will be A = 10 cm 16.3 Standing Sound Waves 541 Example 16.4 Two sinusoidal sound waves, equal in amplitude and traveling in opposite directions along the x-axis, are superimposed on each other The resultant wave is of the form: y = (2 m) sin π π x cos t L T where x is in meters and t in seconds and the arguments of the sine and cosine functions are in radians (a) What are the mathematical formulas of the two sinusoidal sound waves that are superimposed to give this resultant? (b) Find the values of the wavelength and the frequency of the two sinusoidal waves when L = m and T = s (c) What are the velocities of the two sinusoidal waves? Solution: (a) Using the general form of the standing waves given by Eq 16.11, we find A = m, k = π/L rad/m, and ω = π/T rad/s Using Eq 16.8, we find the two sinusoidal waves as follows: π x− L π y2 = (1 m) sin x+ L y1 = (1 m) sin π t , T π t T (b) Using k = 2π/λ, and ω = 2π f when L = m and T = s, we have: k= π π 2π = = λ L 2m ω = 2π f = π π = T 1s ⇒ ⇒ λ=4m f = 0.5 s−1 = 0.5 Hz (c) Using v = ω/k, we find the speed of each of the sinusoidal waves as follows: v= 2π f ω = = λf = (4 m)(0.5 s−1 ) = m/s k 2π/λ The velocity of y1 is v1 = +2 m/s (in the direction of increasing x) and the velocity of y2 is v2 = −2 m/s (in the direction of decreasing x) 16.4 Standing Sound Waves in Air Columns In Chap 14, we saw how a standing wave can be generated either on a stretched string with fixed ends or when one end is fixed and the other is left free to move We learned 542 16 Superposition of Sound Waves that this happens when the wavelengths of the waves suitably match the length of the string, in which case the superposition of the traveling and reflecting waves produce a standing wave pattern For such a match, the wavelength corresponds to the resonant frequency of the string We can set up standing sound waves in air-filled pipes in a way similar to that for strings Here is how we can compare the two: The closed end of a pipe is similar to the fixed end of a string in that it must be a displacement node This is because the pipe’s wall at this end does not allow longitudinal motion of the air and acts like a pressure antinode (point of maximum pressure variation) The open end of a pipe acts like the end of a string that is free to move, so there must be a displacement antinode there1 This is because the pipe’s open end allows longitudinal motion of the air and acts like a pressure node (point of no pressure variation, since the end must remain at atmospheric pressure) It is interesting to know how sound waves reflect from the open end of a pipe To get insight into this, we start with the fact that sound waves are in fact pressure waves Next, we know that any compression region must be contained inside the pipe (between its two ends) Furthermore, any compression region that exists at an open end is free to expand into the atmosphere This change in behavior of the air inside and outside the pipe is sufficient to allow some reflection With the boundary conditions of nodes and antinodes at the ends of air columns, we must set the normal modes of oscillations as we did in the case of stretched strings Air Columns of Two Open Ends First, we consider a pipe of length L that is open at both ends By representing the horizontal displacement of air elements on the vertical axis and applying the boundary condition that meets the case of two open ends, see Fig 16.9, the normal modes of oscillations can be explained by considering the following first three patterns: (1) The first normal mode (the first harmonic, or the fundamental): The simplest pattern is shown in Fig 16.9a There are two imposed antinodes The antinode of an open end of a pipe is located slightly beyond the end because sound compression reaching an open end does not reflect until it passes the end Therefore, the effective length of the air column is little greater than the true length L of the pipe 16.4 Standing Sound Waves in Air Columns 543 at the two ends and only one node in the middle of the pipe Also, there is only half a wavelength in the length L Thus, this standing wave pattern has: λ1 = 2L and f1 = v v = λ1 2L (2) The second normal mode (the second harmonic): The second pattern is shown in Fig 16.9b This pattern has three antinodes and two nodes This standing wave pattern has: v v = = f1 λ2 L λ2 = L and f2 = (3) The third normal mode (the third harmonic): The third pattern is shown in Fig 16.9c This pattern has four antinodes and three nodes This standing wave pattern has: v 3v = = f1 λ3 2L λ3 = 2L/3 and f3 = L λ1 = L (a) A A N λ1 = f1= n=1 First harmonic n=2 Second harmonic n=3 Third harmonic 2L λ2= L (b) A A N A N f2= λ2 = L= f λ3 = L (c) A N A N A N A f3= λ3 =3 2L= f Fig 16.9 The first three standing wave patterns (a), (b), and (c) of a longitudinal sound wave established in an organ pipe that is open to the atmosphere at both ends The horizontal motion of air elements in the pipe is displayed vertically by using a red color The difference between successive harmonics is the fundamental frequency f1 , and each harmonic is an integer multiple of the fundamental frequency f1 Generally, the relation between the wavelength λn of the various normal modes and the length L of a pipe of two open ends is: λn = 2L , (n = 1, 2, 3, ) n (Pipe, two open ends) (16.14) 544 16 Superposition of Sound Waves Also, according to the relation f = v/λ, where the speed v of the sound wave is the same for all frequencies, the resonance frequencies fn associated with these modes are (see Fig 16.9): fn = v v = n , (n = 1, 2, 3, ) λn 2L (Pipe, two open ends) (16.15) The expressions of λn and fn are the same as for the string, except that v is the speed of waves on the strings as in Eq 14.66, whereas v in Eq 16.15 is the speed of sound in air The relation between the resonance frequencies and the fundamental frequency is: fn = n f1 , (n = 1, 2, 3, ) (Pipe, two open ends) (16.16) Air Columns of One Closed End Second, we consider a pipe of length L that is open at one end and closed at the other By applying the boundary condition that meets this case, the normal modes of oscillations can be explained by considering the following first three patterns: (1) The first normal mode (the first harmonic, or the fundamental): Fig 16.10a shows the simplest pattern The standing wave extends from an antinode at the open end to the adjacent node at the closed end The fundamental standing wave pattern has: λ1 = L and f1 = v v = λ1 4L (2) The third normal mode (the third harmonic): The next pattern is shown in Fig 16.10b This pattern has two antinodes and two nodes Thus, this standing wave pattern has: λ3 = 4L/3 and f3 = v 3v = = f1 λ3 4L (3) The fifth normal mode (the fifth harmonic): The next pattern is shown in Fig 16.10c This pattern has four antinodes and four nodes Thus: λ5 = 4L/5 and f5 = v 5v = = f1 λ5 4L 16.4 Standing Sound Waves in Air Columns L 545 λ1 = L (a) A f1= N λ1 = 4L n=1 First harmonic n=3 Third harmonic n=5 Fifth harmonic λ 3= 4L/ (b) A N N A f3= λ3 = L= f λ5= 4L (c) A N A N A N f5= λ5 = L= f Fig 16.10 The first three standing wave patterns (a), (b), and (c) of a longitudinal sound wave established in an organ pipe that is open to the atmosphere at only one end The horizontal motion of air elements in the pipe is displayed vertically by using a red color The harmonic frequencies are the odd-integer multiples of f1 , and the successive difference is f1 Generally, λn and fn of the various normal modes for a pipe of length L with only one end open are given as (see Fig 16.10): λn = fn = 4L , (n = 1, 3, 5, ) n (Pipe, one open end) v v = n , (n = 1, 3, 5, ) λn 4L fn = nf1 , (n = 1, 3, 5, ) (Pipe, one open end) (Pipe, one open end) (16.17) (16.18) (16.19) Figure 16.11 shows a simple apparatus for demonstrating the resonance of sound waves in air columns A tube that is open from both ends is immersed into a container filled with water, and a tuning fork of unknown frequency f and wavelength λ is placed at its top The sound waves generated by the fork are reinforced when the length L corresponds to one of the resonance frequencies of the tube Thus: λ= v v 4Ln , f = =n , (n = 1, 3, 5, ) n λ 4Ln (16.20) 546 16 Superposition of Sound Waves f =? 3l l4 L n=5 n=3 n=1 First harmonic 5l Third harmonic Fifth harmonic Fig 16.11 An apparatus used to demonstrate the resonance of sound waves in a tube closed at one end At resonance, L and λ are related Example 16.5 When wind blows through a cylindrical drainage culvert of 2.5 m length, see Fig 16.12, a howling noise is established Take v = 343 m/s as the speed of sound in air (a) Find the frequencies of the first three harmonics if the pipe is open at both ends (b) How many of the harmonics fall within the normal human hearing range (from about 20 Hz → 20,000 Hz) (c) Answer part (a) if the pipe is blocked at the other end Fig 16.12 Solution: (a) When the pipe is open at both ends, we use Eq 16.15 with n = to find the fundamental frequency as follows: f1 = × 343 m/s v = = 68.6 Hz 2L × 2.5 m Also, all harmonics are available for a pipe open at both ends; thus: f2 = 2f1 = 137.2 Hz and f3 = f1 = 205.8 Hz 16.4 Standing Sound Waves in Air Columns 547 (b) We can express the frequency of the highest harmonic heard as fn = n f1 , where fn = 20,000 Hz and n is the number of harmonics that can be heard Therefore: n= fn 20,000 Hz = 292 = f1 68.6 Hz Although we get n = 292, practically, only the first few harmonics have amplitudes that are sufficient to be heard (c) Using Eq 16.18 and substituting with n = 1, the fundamental frequency of a pipe closed at one end will be given by: f1 = × v 343 m/s = = 34.3 Hz 4L × 2.5 m In this case, only the odd harmonics can exist Thus: f3 = f1 = 102.9 Hz and f5 = f1 = 171.5 Hz Example 16.6 A background noise in a hall sets up a fundamental standing wave frequency in a tube of length L = 0.7 m What is the value of this fundamental frequency if your ear blocks one end of the tube (see Fig 16.13a) and when your ear is far from the tube (see Fig 16.13b)? Take v = 343 m/s as the speed of sound in air Noise L (a) Noise Ear L (b) Ear Fig 16.13 Solution: When the tube is blocked by your ear (see Fig 16.13a) the fundamental frequency is given by Eq 16.18 with n = 1: f1 = × 343 m/s v = = 122.5 Hz 4L × 0.7 m In addition, you can hear frequencies that are odd integer multiples of 122.5 Hz provided that the standing waves are formed with sufficient amplitudes 548 16 Superposition of Sound Waves When you move your head away enough (see Fig 16.13b) the pipe becomes open at both ends and the fundamental frequency will be given by Eq 16.15 with n = 1: f1 = × 343 m/s v = = 245 Hz 2L × 0.7 m In addition, you can hear frequencies that are multiples of 245 Hz if the standing waves are formed with sufficient amplitudes Example 16.7 Resonance can occur in Fig 16.14 when the smallest length of the air column is L = 9.8 cm Take v = 343 m/s as the speed of sound in air (a) What is the frequency f of the tuning fork? (b) What is the value of L for the next two resonances? Fig 16.14 L = 9.8 cm n=1 First resonance Solution: (a) When the tube is blocked by the water’s surface, it acts as if the tube is closed at one end Thus, for the smallest air column L1 , the fundamental frequency is given by Eq 16.20 with n = 1: f =1× 343 m/s v = = 875 Hz 4L1 × (0.098 m) First resonance First harmonic This frequency must be equal to the frequency f of the tuning fork (b) We know from Fig 16.14 and Eq 16.20 that the wavelength of the fundamental mode is four times the length of the air column Thus: λ= 4L1 = 4(0.098 m) = 0.392 m 16.4 Standing Sound Waves in Air Columns 549 Because the frequency of the tuning fork is constant, then according to Fig 16.11, the values of L for the next two normal modes are: L3 = × (0.392 m) 3λ = = 0.294 m = 29.4 cm 4 L5 = 16.5 5λ × (0.392 m) = = 0.49 m = 49 cm 4 Second resonance Third harmonic Third resonance Fifth harmonic Temporal Interference of Sound Waves: Beats Previously, we discussed the spatial interference of waves of same frequencies, where at fixed time the amplitude of the oscillating elements varies with the position in space The standing waves in strings and air columns are good examples of this kind of interference Now, we consider another type of interference of waves having a slight difference in their frequencies, where at fixed position, the amplitude of the oscillating elements varies periodically with time The standing wave produced by two tuning forks having a slight difference in their frequencies is a good example of this kind of interference We refer to this interference in time by temporal interference, and this phenomenon is called beating: Beating Beating is defined as the periodic variation in amplitude at a fixed position due to the variation in the constructive and destructive interference between waves having slightly different frequencies Consider the time-dependent variations of the displacements of two sound waves of equal amplitude and slightly different frequencies f1 and f2 (angular frequencies ω1 = 2π f1 and ω2 = 2π f2 ) such that: y1 = A cos(k1 x − ω1 t), y2 = A cos(k2 x − ω2 t) (16.21) 550 16 Superposition of Sound Waves At the fixed point x = (chosen for convenience), the two wave functions become (see Fig 16.15a): y1 = A cos ω1 t, (16.22) y2 = A cos ω2 t y = A cos π f t , f = 11 Hz y = A cos π f t , f = Hz y (a) t y y y = y + y = [2 A cos π ( f − f ) t ] cos π ( f + f ) t Oscillates with an average frequency f av = ( f 1+f ) /2 (b) t [± A cos π ( f − f ) t ] Tbeat = 1/ f − f Fig 16.15 (a) Formation of beats by combining two waves of slightly different frequencies f1 and f2 (f1 = 11 Hz and f1 = Hz) (b) The slowly varying amplitude envelope ±2 A cos π(f1 − f2 ) t limits the amplitude of the rapid sinusoidal function cos π(f1 + f2 ) t, which proceeds with an average frequency fav = (f1 + f2 )/2 The superposition of y1 and y2 gives the following resultant: y = y1 + y2 = A [cos ω1 t + cos ω2 t] (16.23) To simplify this expression, we use the trigonometric identity: cos a + cos b = cos 21 (a − b) cos 21 (a + b) (16.24)