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 = b2 g − 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 − =0 ∂x v2 ∂ t2 (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 Undisturbed gas longitudinal sound pulse in a Compressed gas gas-filled tube Compressed gas Compressed gas rods In fluids we replace τ with the bulk modulus B, and in rods we replace τ with Young’s modulus Y In both, we replace μ with the density ρ Then: v= ⎧√ ⎨ B/ρ elastic property = medium property ⎩ √Y /ρ (In fluids) (15.1) (In solid rods) Table 15.1 depicts the speed of sound in several different materials Table 15.1 The speed of sound in different materials Medium v(m/s) Gases Medium Solids Oxygen (0 ◦ C) 317 Rubber (0 ◦ C) Air v(m/s) 1,600 331 Lead 1,960 Air (20 ◦ C) 343 Lucite 2,680 Helium (0 ◦ C) 972 Gold 3,240 Hydrogen (0 ◦ C) 1,286 Brass 4,700 Liquids at (25 ◦ C) Copper 5,010 Kerosene 1,324 Pyrex 5,640 Mercury 1,450 Iron 5,950 Water 1,493 Granite 6,000 Sea water 1,533 Aluminum 6,420 For sound traveling through air, the relation between the speed and the temperature of the medium is given by the following relation: 506 15 Sound Waves Area A Oscillating piston with frequency f Δm Δx Fig 15.5 A piston oscillates with frequency f in an air-filled tube The piston transfers energy to an adjacent air element that has a mass m and length x, causing it to oscillate with an amplitude smax The piston transfers energy to this element and hence the energy is propagated away through the tube by the sound wave As the sound wave propagates away, the displacement of this element with respect to its equilibrium position will be given by Eq 15.3, i.e.: s(x, t) = smax cos(k x − ω t) (15.14) The speed of this element can be found by taking the partial time derivative of s as follows: v(x, t) = ∗ ∂ ∂t s(x, t) = ∂ ∂t The kinetic energy [smax cos(k x − ω t)] = + ω smax sin(k x − ω t) K associated with the air element of mass (15.15) m = ρA x, where ρ is the air density, will be given by: sin2 (k x − ω t) K = 21 m v = 21 ρ A x ω2 smax (15.16) When we allow x to approach zero, this relation becomes a differential relationship and will take the following form: sin2 (k x − ω t)dx dK = 21 ρ A ω2 smax (15.17) At a given instant, let us integrate this expression over all the elements in a complete sound wavelength, which will give us the total kinetic energy Kλ in one wavelength: λ Kλ = dK = 21 ρ A ω2 smax sin2 (k x − ω t)dx (15.18) 15.3 Energy, Power, and Intensity of Sound Waves 507 If we take a snapshot at time t = 0, then we can evaluate the above integral by performing the following steps: z = kλ = 2π x=λ sin2 (k x)dx = 1k sin2 z dz x=0 z=0 2π = 1k 1 [1 − cos 2z]dz = 2k z− sin 2z 2π 0 = 2k (2π − sin 4π ) − = λ λ 2π = 4π (15.19) where we have used z = k x, sin2 z = (1 − cos 2z)/2 and k = 2π/λ to arrive to the above result Of course, we get the same answer if we perform the above steps at any other time different from zero When we substitute the above result into Eq 15.18, we get: Kλ = 41 ρ Aω2 smax λ (15.20) A similar analysis to the total potential energy Uλ in one wavelength will give exactly the same result Thus: Uλ = 41 ρ A ω2 smax λ (15.21) The total energy in one wavelength of the sound wave (Eλ ) is the sum of the obtained kinetic and potential energies Thus: Eλ = Kλ + Uλ = 21 ρ A ω2 smax λ (15.22) As the sinusoidal sound wave travels along the tube, this amount of energy (Eλ ) will cross any given point in the tube during a time interval equal to one period of the oscillation Thus, the rate of energy (power) transferred by the sound wave through the air is: P= E Eλ = t T where we used the symbol P for the power in this section to avoid confusion with the symbol P for pressure Therefore: P = 21 ρ A ω2 smax λ T 508 15 Sound Waves Using the relation v = λ/T , given by Eq 14.38, we finally reach the following power form: P = 21 ρ A v ω2 smax (15.23) Thus, the power of a periodic sound wave is proportional to the square of the angular frequency and the square of the displacement amplitude (as in the case of periodic string waves) For a wave crossing a particular surface, we define its intensity I as the power per unit area, or the rate of energy transfer (power P) of the wave through a unit area perpendicular to the direction of the propagation of the wave, i.e I = P/A Therefore: I= P A ⇒ I = 21 ρvω2 smax (15.24) By using Eq 15.13, Pmax = ρvω smax , the last relation can be written in terms of the pressure amplitude Pmax as: I= Pmax 2ρv On the other hand, we can express the pressure amplitude measurable quantities ρ, v, and I as follows: Pmax = √ 2ρvI (15.25) Pmax in terms of the (15.26) In three dimensions, we consider a point source S emitting sound waves uniformly in all directions as spherical waves, see Fig 15.6 When we construct an imaginary sphere of radius r centered at the sound source, the power emitted by this source must be distributed uniformly over this spherical surface, which has an area 4π r From the definition of the intensity, I = P/A, given by Eq 15.24, the intensity I at any point on the spherical surface will be given by: I= P 4π r (15.27) This equation is known as the inverse square law and tells us that the intensity of sound waves emitted from an isotropic point source decreases with the square of the distance r from the source, i.e the intensity is inversely proportional to the square of the distance r 15.3 Energy, Power, and Intensity of Sound Waves 509 r r λ S r λ λ Point source S (a) Three dimensional view (b) Cross sectional view Fig 15.6 (a) A point source S emitting sound waves uniformly in all directions with the waves passing through an imaginary sphere of radius r (b) A cross-sectional view showing the wavelength λ between consecutive crests of the sound waves Example 15.3 At a frequency of 1,000 Hz, the human ear can detect the loudest and faintest sounds with intensities of about 1.0 W/m2 and 1.0 × 10−12 W/m2 , respectively For sound waves traveling with a speed of v = 343 m/s, find the pressure amplitude Pmax for the faintest and the loudest sound waves, assuming the air’s density is ρ = 1.21 kg/m3 Solution: From Eq 15.26, we can find the pressure amplitude loudest sound waves as follows: Pmax for the Pmax = 2ρvI = 2(1.21 kg/m3 )(343 m/s)(1 W/m2 ) = 28.8 N/m2 = 28.8 Pa (Loudest; threshold of pain) Also, from Eq 15.26, we find the pressure amplitude Pmax for the faintest sound waves as follows: Pmax = 2ρvI = 2(1.21 kg/m3 )(343 m/s)(1 × 10−12 W/m2 ) = 2.88 × 10−5 N/m2 = 2.88 × 10−5 Pa (Faintest; threshold of hearing) 510 15 Sound Waves Example 15.4 A point source emits sound waves with a power of 50 W (a) Find the intensity of the sound waves m away from the source ( b) Find the distance at which the intensity of the sound is 10−6 W/m2 Solution: (a) The point source S shown in Fig 15.7 emits energy in the form of spherical sound waves centered at the source Thus, when using Eq 15.27 we find that: I= P 4π r = 50 W = 0.995 W/m2 4π(2 m)2 which is close to the intensity of the threshold of pain, see Example 15.3 (b) Expressing r in Eq 15.27 in terms of P and I, we obtain: r= P 4π I = 50 W = 1,995 m 4π(10−6 W/m2 ) Fig 15.7 km I I r=2 m S(50W) I 15.4 I The Decibel Scale According to Example 15.2, the displacement amplitude smax for the human ear ranges from about 10−5 m for the loudest tolerable sound to about 10−11 m for faintest detectable sound, a ratio of 106 From Eq 15.24, we see that the intensity I varies as the square of smax , so the ratio of intensities at these two limits of the human audibility is 1012 This goes to show that the human ear can accommodate an enormous range of intensities We can better represent large ranges of I by using logarithms Now, consider the following logarithmic relation of the base 10: 15.4 The Decibel Scale 511 y = log x It is usual to suppress explicit references to the base 10 (such as log10 x) and instead write log x If x in this equation is multiplied by 10, then y increases by 1, i.e.: y = log10 x = log 10 + log x = + log x = + y Similarly, if we multiply x by 1012 , y increases only by 12 Consequently, instead of speaking of intensity I of a sound wave, it is much more convenient to speak of its sound level β (Greek beta), defined by: β = (10 dB)log I I◦ (15.28) Here dB is the abbreviation for decibel, the unit of sound level, a name chosen to recognize the work of Alexander Graham Bell The constant I◦ in Eq 15.28 is the reference intensity, taken to be near the threshold of hearing, i.e I◦ = 1.0 × 10−12 W/m2 The intensity I in the same equation is measured in watts per square meter On this scale, the threshold of hearing (I = 1.0 × 10−12 W/m2 ) corresponds to a sound level of: β = (10 dB)log 1.0 × 10−12 W/m2 I = (10 dB)log I◦ 1.0 × 10−12 W/m2 = (10 dB)log1 = dB (Threshold of hearing) So our threshold of hearing level corresponds to zero decibel Also, the threshold of pain (I = 1.0 W/m2 ) corresponds to a sound level of: β = (10 dB)log 1.0 W/m2 I = (10 dB)log I◦ 1.0 × 10−12 W/m2 = (10 dB)log1012 = (10 dB) × 12 = 120 dB (Threshold of pain) In general, β = 10 × n dB corresponds to an intensity that is 10n times the reference intensity, i.e corresponds to I = 10n I◦ = 10n−12 W/m2 Table 15.2 lists some soundlevel values for some environments 512 15 Sound Waves Table 15.2 Approximate sound levels (dB) for several sources Source of sound β(dB) I(W/m2 ) Threshold of hearing in human auditory system 10−12 Quiet rustling leaves, calm human breathing 10 10−11 Very calm room 20 10−10 Whispering 30 10−9 Mosquito buzzing 40 10−8 Normal talking (1 m distant) 50 10−7 TV set—typical home level, m distant 60 10−6 Vacuum cleaner 70 10−5 Traffic noise for a main road, 10 m distant 80 10−4 Machine gun 90 10−3 Jack hammer, m distant 100 10−2 Jet engine, 100 m distant 110 10−1 Threshold of pain in human auditory system 120 Example 15.5 (a) Find the sound level in decibels for a sound wave of intensity 1.59 × 10−5 W/m2 (b) Find the sound intensity of a source rated at a 35 dB sound level Solution: (a) From Eq 15.28, we find that: I I◦ 1.59 × 10−5 W/m2 = (10 dB)log 1.0 × 10−12 W/m2 β = (10 dB)log = (10 dB)log1.59 × 107 = 72 dB (b) Substituting in Eq 15.28 with β = 35 dB, dividing both sides by 10, and taking the antilog of both sides, we can find I by performing the following steps: β = (10 dB)log I I◦ 35 dB = (10 dB)log 3.5 = log I I◦ I I◦ 15.4 The Decibel Scale 513 antilog(3.5) = antilog log 103.5 = I I◦ I I◦ I = 103.5 × I◦ Thus, with the reference intensity I◦ = 1.0 × 10−12 W/m2 , we find that: I = 103.5 × 1.0 × 10−12 W/m2 = 3.16 × 10−9 W/m2 Example 15.6 Two identical point sources, S1 and S2 , have the same power and driven by one oscillator The positions of the two sources relative to an observer is depicted in Fig 15.8 The sound intensity at the observer’s location from S2 is found to be I2 = 3.0 × 10−6 W/m2 (a) Find the total intensity of the combined sound waves that is received by the observer from the two sources (b) Find the difference in sound level when the two sources operate simultaneously and when only the second source operates r2 = r1 r1 S1 S2 Fig 15.8 Solution: (a) If I1 and I2 are the intensities received by the observer from the point sources S1 and S2 , respectively, then their ratio will be: 514 15 Sound Waves I1 = I2 P P 4π r12 4π r22 = r22 r12 = (2r1 )2 =4 r12 ⇒ I1 = I2 This means that the intensity I1 from S1 is four times the intensity I2 from S2 Thus, the total intensity becomes: Itot = I1 + I2 = 4I2 + I2 = 5I2 = 5(3.0 × 10−6 W/m2 ) = 1.5 × 10−5 W/m2 (b) If β2 is the sound level when only the second source operates and βtot is the sound level when both sources operate together, then: β2 = (10 dB)log I2 Itot 5I2 and βtot = (10 dB)log = (10 dB)log I◦ I◦ I◦ Then: β = βtot − β2 = (10 dB)log 15.5 5I2 I2 5I2 − (10 dB)log = (10 dB)log = dB I◦ I◦ I2 Hearing Response to Intensity and Frequency The threshold of hearing in the human auditory system depends on the intensity of the sound (or the sound level in dB) and its frequency We learned in the previous section that the threshold of hearing at 1,000 Hz requires an intensity of 10−12 W/m2 and corresponds to a sound level of dB Conversely, at 100 Hz sound must have an intensity level of about 30 dB to be barely audible Figure 15.9 maps the sound regions that humans can respond to for a range of sound levels β (or intensity I) and sound frequencies f Tentatively, the figure also overlays some sample sources The lower blue curve of the white area shows the dependence of the threshold of hearing β on the frequency This curve indicates that humans are sensitive to frequencies ranging from about 20 to 20,000 Hz The upper bound to the white area is the threshold of pain, and does not depend much on frequency The lower left region of the white area shows that our ears are particularly insensitive to low frequencies and low intensity levels 15.6 The Doppler Effect We move to a different phenomenon that applies to all kinds of waves, not only sound waves You most probably have noticed that when a car moves toward you with a high speed and horns, you hear the horn with a higher frequency than when the car 15.6 The Doppler Effect 515 is at rest Contrary wise, when the car moves away, you hear the horn with a lower frequency This phenomenon is called the Doppler effect 10 10 10 Sonar 10 Sonic frequencies 10 Threshold of pain 10 10 -2 Threshold of hearing 10 -4 10 -6 Bats 10 -8 Whispering 10 -10 10 -12 100 1000 10000 100000 Large rocket engine 10 Ultrasonic frequencies 220 200 180 160 140 120 100 80 60 40 20 Ι (W /m ) Infrasonic frequencies Sound level β (dB) Frequency, f (Hz) Fig 15.9 The dependence of the sound level β on the frequency f for normal human hearing (the white area) and various sources Let us now examine this phenomenon quantitatively First, we consider a point source that emits sound waves radially in all directions in a uniform medium It is useful to represent the emitted waves using a series of concentric spheres with the source located at their centers Each sphere represents a wave crest, and it moves away from the source with the speed of sound We call such a sphere of constant phase a wave front Therefore, the distance between any two successive wave fronts equals the wavelength λ of the sound wave and has a frequency f and speed v In our analyses that follow, we restrict ourselves to the motion of a sound source S and observer O along the line joining them Moving Observer and Stationary Source Figure 15.10 shows an observer O (represented by an ear) moving with a speed vo toward a stationary source S that emits spherical sound waves of speed v(v > vo ), wavelength λ, and frequency f The frequency detected by the observer O is the rate at which O intercepts successive wave fronts (or wavelengths) 516 15 Sound Waves Fig 15.10 A stationary sound source S emits spherical wave fronts (each is one λ wavelength λ from the next) with a speed v An observer O Wave fronts (represented by an ear) moves with a speed vo towards the λ λ o o< S O source If the observer O were stationary, the interception rate of wave fronts would be f But if the observer O is moving toward the source S, then the interception rate f is greater than f When the observer O moves with a speed vo toward a stationary source S, the speed of the wave fronts relative to O is not v, but v = v + vo , while the wavelength λ is unchanged When we apply the general relation v = λ f to this case, i.e v = λ f , we find that the frequency f heard by the observer has the following relation: f = v + vo v + vo v = = λ λ v/f (15.29) This relation can be rewritten as: f = 1+ vo f v (O is moving towards S) (15.30) When the observer O moves with a speed vo away from a stationary source S, the speed of the wave fronts relative to O is not v, but v = v − vo , while the wavelength λ is unchanged Steps similar to those above lead to the frequency heard by the observer as: f = 1− vo f v (O is moving away from S) (15.31) Generally, for an observer O moving with a speed vo relative to a stationary source S, a positive sign is used when O moves toward S and a negative sign is used when O moves away from S Thus: f = 1± vo f v + when O is moving towards S − when O is moving away from S (15.32) 15.6 The Doppler Effect 517 Moving Source and Stationary Observer Figure 15.11 shows a source S moving with a speed vS toward an observer O while emitting spherical sound waves of speed v,wavelength λ, and frequency f The figure indicates that the wave fronts detected by the observer O are closer together than they would be if the source S was not moving Thus, the wavelength λ measured by the observer O is shorter than the wavelength λ of the source S Fig 15.11 A moving sound source S emits spherical wave fronts with a speed v and wavelength λ, while moving with a speed vS < v towards a s stationary observer O The W1 wave front W1 that arises from W2 λ' W3 S1 S2 S S the source when it was at point S1 , etc is shown O During a period T, the source S emits a wave front that moves a distance λ with a speed v, while the source itself moves a distance vS T before emitting the next wave front Thus, the wavelength λ is shortened by vS T Then, the observed wavelength λ will be: λ = λ − vS T = vS v − f f (15.33) Using v = λ f in this case, i.e v = λ f , we find that the frequency f that is heard by the observer O is related to f as follows: v v vS = (v − vS ) = − f f f f (15.34) This relation can be rewritten as: f = 1 − vS /v f (S is moving towards O) (15.35) When the source S is moving with a speed vS away from a stationary observer O, the wavelength λ is increased by vS T Therefore, the observed wavelength λ will be given by: 518 15 Sound Waves λ = λ + vS T = v vS + f f (15.36) With similar steps to those of Eqs 15.33–15.35, we get: f = 1 + vS /v (S is moving away from O) f (15.37) Generally, for a source S moving with a speed vS relative to a stationary observer O, a negative sign is used when S moves toward O and a positive sign is used when S moves away from O Thus: f = 1 ∓ vS /v f − when S is moving towards O + when S is moving away from O (15.38) One can find a generalized relation that includes all collinear motion of a source with speed vS and an observer with speed vo to be: f = ± vo /v ∓ vS /v f (General Doppler effect) (15.39) The upper signs in the numerator and denominator (+vo /v and −vS /v) refer to motion of one toward the other, while the lower signs (−vo /v and +vS /v) refer to motion of one away from the other Spotlight You can determine the signs in Eq 15.39 by remembering that: the word toward is associated with an increase in observed frequency, while the word away from is associated with a decrease in observed frequency Example 15.7 A car moves on a straight road with a speed of 20 m/s Its siren emits a sound with a frequency of 500 Hz Find the frequencies heard by a stationary person on the sidewalk when the car approaches him (Fig 15.12a) and then when it recedes from him (Fig 15.12b) Assume collinear motion of the source and observer Solution: When the car approaches the observer, we use the upper signs in the numerator and denominator of the general formula of Doppler effect given by Eq 15.39 In this formula, we take vo = for the stationary observer, vS = 20 m/s 15.6 The Doppler Effect 519 for the speed of the car, v = 343 m/s for the speed of sound in air, and f = 500 Hz for the siren frequency Thus, the frequency f heard by the observer will be: f = + vo /v − vS /v f= 1+0 − (20 m/s)/(343 m/s) × 500 Hz = 531 Hz s f f f (a) s f (b) Fig 15.12 When the car recedes from the observer, we use the lower signs in the numerator and denominator of Eq 15.39 Thus: f = − vo /v + vS /v f= 1−0 + (20 m/s)/(343 m/s) × 500 Hz = 473 Hz The change in frequency detected by the stationary observer is: f = f − f = 531 Hz − 473 Hz = 58 Hz This is about 8.6% of the actual frequency emitted from the siren Example 15.8 Submarines use sound propagation under water to navigate, communicate, or detect other objects; this technique is known as sonar (SOund NAvigation and Ranging) A submarine (sub 1) moves with a speed v1 = 10 m/s and emits a sonar wave of frequency f = 1,500 Hz A second submarine (sub 2) moves directly towards the first one with a speed v2 = m/s See Fig 15.13, and take the speed of sound in water to be 1,533 m/s (a) Find the frequency detected by an observer in sub (b) Find the reflected frequency detected by an observer in sub (c) Find the frequency detected by an observer in sub when the two submarines miss each other and pass 520 15 Sound Waves Reflected waves f f′ Emitted waves Fig 15.13 Solution: (a) When the submarines move toward each other, we use the upper signs in the numerator and denominator of Eq 15.39 Then we take: vo = v2 = m/s for observer (sub 2) vS = v1 = 10 m/s for the speed of the source (sub 1) v = 1,533 m/s for the speed of sound in water, f = 1,500 Hz for the emitted frequency from the source (sub 1) Thus, the frequency f received by an observer in sub will be: f = = + vo /v f − vS /v + (8 m/s)/(1,533 m/s) − (10 m/s)/(1,533 m/s) × 1,500 Hz = 1,518 Hz (b) The frequency calculated in part (a) will be reflected from sub (which acts as a moving source) and then be detected by sub (the moving observer) In this case we take: vo = v1 = 10 m/s for observer (sub 1) vS = v2 = m/s for the speed of the source (sub 2) v = 1,533 m/s for the speed of sound in water, f = 1,518 Hz for the emitted frequency from the source (sub 2) Thus, the frequency f received by an observer in sub will be: f = + vo /v − vS /v f = + (10 m/s)/(1,533 m/s) − (8 m/s)/(1,533 m/s) × 1,518 Hz = 1,536 Hz (c) When the submarines move away from each other, we use the lower signs in the numerator and denominator of Eq 15.39 All the parameters used in this [...]... human ear ranges from about 10−5 m for the loudest tolerable sound to about 10−11 m for faintest detectable sound, a ratio of 106 From Eq 15 .24 , we see that the intensity I varies as the square of smax , so the ratio of intensities at these two limits of the human audibility is 10 12 This goes to show that the human ear can accommodate an enormous range of intensities We can better represent large ranges... P for the power in this section to avoid confusion with the symbol P for pressure Therefore: 2 P = 21 ρ A 2 smax λ T 508 15 Sound Waves Using the relation v = λ/T , given by Eq 14.38, we finally reach the following power form: 2 P = 21 ρ A v 2 smax (15 .23 ) Thus, the power of a periodic sound wave is proportional to the square of the angular frequency and the square of the displacement amplitude (as... wave front Therefore, the distance between any two successive wave fronts equals the wavelength λ of the sound wave and has a frequency f and speed v In our analyses that follow, we restrict ourselves to the motion of a sound source S and observer O along the line joining them Moving Observer and Stationary Source Figure 15.10 shows an observer O (represented by an ear) moving with a speed vo toward... case of periodic string waves) For a wave crossing a particular surface, we define its intensity I as the power per unit area, or the rate of energy transfer (power P) of the wave through a unit area perpendicular to the direction of the propagation of the wave, i.e I = P /A Therefore: I= P A ⇒ 2 I = 21 ρv 2 smax (15 .24 ) By using Eq 15.13, Pmax = ρvω smax , the last relation can be written in terms of. .. Generally, for a source S moving with a speed vS relative to a stationary observer O, a negative sign is used when S moves toward O and a positive sign is used when S moves away from O Thus: f = 1 1 ∓ vS /v f − when S is moving towards O + when S is moving away from O (15.38) One can find a generalized relation that includes all collinear motion of a source with speed vS and an observer with speed vo to... word away from is associated with a decrease in observed frequency Example 15.7 A car moves on a straight road with a speed of 20 m/s Its siren emits a sound with a frequency of 500 Hz Find the frequencies heard by a stationary person on the sidewalk when the car approaches him (Fig 15.1 2a) and then when it recedes from him (Fig 15.12b) Assume collinear motion of the source and observer Solution: When... 100 m = 20 0 m Thus: t= 20 0 m x = = 0.138 s v 1,449 m/s 5 02 15 Sound Waves 15 .2 Periodic Sound Waves As a result of continuous push and pull of a piston in a gas tube, continuous regions of compressions and expansions (or called rarefactions) are generated, see Fig 15. 3a The darker-colored areas in the figure represent regions where the gas is compressed, and thus the pressure and density are above their... a speed v The wave consists of a moving pattern of compressions and expansions The wave is shown at an arbitrary time t (b) An element of thickness x is displaced at a distance s to the right from its equilibrium position Its maximum displacement, either right or left, is smax , where smax Consider a thin element of air of thickness λ x located at a position x along the tube As the wave passes through... indicates that humans are sensitive to frequencies ranging from about 20 to 20 ,000 Hz The upper bound to the white area is the threshold of pain, and does not depend much on frequency The lower left region of the white area shows that our ears are particularly insensitive to low frequencies and low intensity levels 15.6 The Doppler Effect We move to a different phenomenon that applies to all kinds of waves,... is a remarkably small number! This displacement amplitude is about onetenth the size of a typical atom (diameter ≈10−10 m) Indeed, the ear is an extremely sensitive detector for sound waves On the other hand, the ear can detect a sound-wave pulse whose total energy is about the same as the total energy required to remove an outer electron from a single atom 15.3 Energy, Power, and Intensity of Sound