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Transcript
Ch 14 – Waves and Sound 14.1 Types of Waves All mechanical waves require source of disturbance and a disturbable medium that can influence itself Point element (the black point) below shows an object in the water where a wave passes by. The object rises up then goes back down as the wave passes. There are two type of waves Transverse Waves: A traveling wave or pulse that causes the elements of the disturbed medium to moved perpendicular to the direction of propagation. As you can see below…the distance traveled by a wave from a point is Δx which equals vt. We can say the distance traveled by a wave in a given about of time is, vt. So the position of any wave is given by the starting position (x) and the distance traveled after time (vt). Longitudinal Waves: A traveling wave or pulse that causes the elements of the medium to move parallel Travels to the right: to the direction of the propagation. Travels to the left: y(x,t) = f(x – vt) y(x,t) = f(x + vt) Earthquakes are an example of both types of waves. Primary (P) waves are the longitude waves which travel faster than the transverse or Secondary (S) waves. The longitudinal waves travel about 7.5 km/s near the surface and the transverse waves travel near the surface through the earth at 4.5 km/s. Example: Ocean waves with a crest-to-crest distance of 10.0 m and v = 1.20 m/s can be described by the wave function y(x, t) = (0.800 m) sin[0.628(x – vt)] (a) Sketch y(x, t) at t = 0. (b) Sketch y(x, t) at t = 2.00 s Note how the entire wave form has shifted 2.40 m in the +x direction in this time interval. Note: sine or cosine can be used for harmonic wave motion, this example uses sine, most in your book use cosine. Longnitudnal & Transverse Waves: OW-B-LT 14.2 Waves on a String v2 = Ftension / μ; F=m a F = m v2/r The radial (centripetal) force is Tension(sinθ) Fr = FT sinθ (by one side) The rope is being pulled on two sides so total radial force is given by Fr = 2 FT sinθ If θ is small (<10°) then sinθ = θ Fr = 2 FT θ 2 FT θ 2 FT θ FT v2 = (m) v2/r = (2 μ R θ) v2/r = ( μ ) v2 = FT / μ Mass associated with Δs is μΔs (where μ is mass per unit length) m = μ (Δs) Δs = 2(R θ) m = μ (2 R θ) m = 2 μ R θ Example A piano string having a mass per unit length 10.0 10–3 kg/m is under a tension of 1000 N. Find the speed of a wave traveling on this string. 14.2 continues v2 v2 v = FT / μ = 1000 / 0.010 = 316 m/s When a rope fixed to an immoveable object reflects off the object, the wave reflects off in the opposite direction, with the negative amplitude. If the wave continues to another medium, some of the energy will be transmitted to the new different sized medium, and some of the energy will be reflected. Newton’s 3rd Law…for every action there is an opposite and equal reaction. The total energy will remain the same. (13.5: T = ½kA2) Why in the case of the high density rope to low density rope is the reflected wave not inverted? How would the above picture be different if instead of being attached to the wall, it was attached to a freely moving ring Similar explanation as the rings…the low density (thin) rope resists change much less than the high density rope. Thus the low density rope accelerates up to the new Amplitude, then the downward tension pulls the rope back down. (free to move up and down with the same amplitude as the wave)? Ans: For the thin rope to the thick rope…the thick rope resists change similarly to the rope fixed to the wall. Waves on a Rope: OW-B-WR Wave Machine with Rods: OW-B-WM 14.3 Harmonic Wave Functions A sinusoidal wave is created by plotting sin θ versus θ. Crest (+ max amplitude) In the form of y(x, t) = A cos (a x) Trough (- max amplitude) If the wave is moving to the right then y(x, t) = A cos [a (x – vt)] Wavelength (length of complete wave, until start of repetition) frequency = 1/T measured in hertz v=λ/T k = 2π / λ ; ω = 2π / T v = ω/k and the wave function is generally expressed as y(x, t) = A cos [kx – ωt + φ] Amplitude = A v = -ωAsin(kx – ωt); vmax = ωA Wave # is k a = -ω2Acos(kx – ωt); amax = ω2A Example A sinusoidal wave on a string is described by y(x, t) = A cos [ k x – ω t + φ] y(x, t) = 0.51 cos [310 x – 9.30 t] (3.1 rad/cm = 310 rad/m) y = (0.51 cm) sin(kx – t) where k = 3.10 rad/cm and = 9.30 rad/s. How far does a wave crest move in 10.0 s? Does it move in the (+) or (-) x direction? 14.4 Sound Waves v= ω /k v = 9.3/310 v = 0.030 m/s vave = ∆x / ∆t 0.030 = ∆x / 10.0 ∆x = 0.300 meters (+x-dir) Type of sound wave: Audible Infrasonic Ultrasonic The speed of sound is given below. The first is the speed of sound in air. The latter is the speed of sound in liquids and solids. 331 1 Tc v 273 Find vsound in mercury, with a bulk modulus of about 2.80 1010 N/m2 and a density of 13600 kg/m3. Find vsound in an Aluminum rod, which has a Young’s modulus of about 7.0 1010 N/m2 and a density of 2700 kg/m3. elastic _ prop inertial _ prop v v B stress / strain Y 2.80 1010 1.43 km s 13.6 103 7.0 1010 Pa m 5090 s 2.70 103 kg 3 m Y Flaming Standing Waves: OW-D-SF 14.5 Sound Intensity The piston transmits energy to the element of air in the tube. This energy is propagated away from the piston by the sound wave. We must use the velocity to determine the total kinetic energy in one wavelength. v(x,t) = -ω smaxsin(kx - ωt) How would we define intensity? Suppose you take a large magnifying glass, go outside on a nice sunny day ΔK = Δ½ mv2 ΔK = ½ Δm (ω smax)2sin2(kx - ωt) If t is constant (at a given instant in time) ΔK = ½ Δρ V (ω smax)2sin2(kx) dK = ½Δρ A dx (ω smax)2sin2(kx) K K K Kλ = ½ΔρA (ω smax) sin (kx)dx = ½ΔρA (ω smax)2 ½ x (see note below) = ½ΔρA (ω smax)2 ½λ 2 = ¼ ρA(ωsmax) λ 2 2 The total potential energy for one wavelength is the same as the kinetic so the total mechanical energy is E λ = K λ +U λ = ½ ρA(ωsmax)2λ And the rate of energy transfer is the power of the wave which is the energy that passes by a given point during one period of oscillation P = Work/time P = Eλ / T P = ½ ρA(ωsmax)2 (λ / T) P = ½ ρA(ωsmax)2 (v) P = ½ ρAv(ωsmax)2 NOTE explanation Double angle formulas cos(2x) = 1 – 2sin2(x) solve for sin sin2(x) = ½ - ½ cos(2x) near June 20th and play with ants. Would agree the intensity of the light rays that strike the ants is intense? The intensity I of a wave is defined as the power per unit area, I = Power / A = Power / 4πr2 This is the rate at which the energy being transported by the wave transfers through a spherical shell unit of radius, r, and area, A, perpendicular to the direction of the wave. The intensity decreases in proportion to the square of the distance from the source. I= Power /A I = ½ ρAv(ωsmax)2 / A I = ½ ρv(ωsmax)2 In terms of pressure amplitude (ΔPmax = ρvωsmax) P = F / A P = m a / A P = ρ(A*Δx) Δv/Δt / A P = ρ (Δx) Δv/Δt P =ρ (Δx)/Δt Δv Pmax = ρ -ω smaxsin(kx - ωt) Δv ΔPmax = ρ ω smax (1) Δv ΔPmax = ρvωsmax I = ΔP2 / 2ρv A logarithmic scale is best used to determine the intensity level, β = 10 log (I / Io) I0, 1.00 x 10-12 W/ m2, is the reference intensity also the threshold of hearing at 1000 Hz Threshold of pain Threshold of hearing Now we can integrate ∫sin2(x) dx = ∫½dx - ½∫cos(2x)dx ∫sin2(x) dx = ½x – ¼ ∫sin(2x) + const If x >> 1 then simplifies to ½ x Example What is the sound level that corresponds to an intensity of 10 x 10-7 W/m2 ? I = 1.00 W/m2 β = 120 dB I0 = 1.00 x 10-12 W/ m2 β = 0 dB β = 10 log ( I / Io ) β = 10 log (10 x 10-7 W/m2 / 1.0 x 10-12 W/m2) β = 10 log 106 Rule of thumb: A doubling in the loudness is β = 60 dB approximately equivalent to an increase of 3 dB Range of Hearing: OW-C-RH Example A loudspeaker is placed between two observers who are 30 m apart, along the line connecting them. If one observer records a sound level of 60.0 dB, and the other records a sound level of 66.0 dB, how far is the speaker from each observer? Derive the equation β2 – β1 = 20 log (r1 / r2) from β = 10 log (I / I0) β2 – β1 = 20 log (r1 / r2) 66 – 60 = 20 log (r1 / r2) 0.3 = log (r1 / r2) 2 r2 = r1 10 and 20 meters β1 = 10 log (I1 / I0); β2 = 10 log (I2 / I0) 14.6 The Doppler Effect The Doppler effect is the apparent change in frequency (or wavelength) that occurs (because of motion of the source or observer) of a wave. When the relative speed of the source and observer is higher than the speed of the wave, the frequency appears to increase (if lower, then freq appears to decrease) We also call an appearance of increased frequency, a blue shift or blue-shifted…and decreased frequency, a red shift. Why? low energy, freq high energy, high freq Radio micro IR R OYGB IV UV X γ β2 β2 β2 β2 - β1 = 10 β1 = 10 β1 = 10 β1 = 10 log(I2/I0) - 10 log(I1/I0) [log (I2/I0) - log(I1/I0)] log [ (I2 / I0) / (I1 / I0)] log ( I2 / I1 ) P2 = P1 β2 - β1 = 10 log (P2/4πr22 / P1/4πr12) β2 - β1 = 10 log ( r1 / r2) 2 β2 - β1 = 20 log ( r1 / r2) A convenient graphical representation is to use circular arcs concentric to the source which the fronts of which are called wave fronts When the observer moves toward the source, the speed of the waves relative to the observer is Actual wavelength doesn’t change so f’ / v’ = f / v f’ = f (v’/ v) Toward: Away: v ’ = v + vo vo is the observer speed The frequency heard by the observer appears higher when the observer approaches the source f’ = f ( v’ ) / v v ’ = v - vo vo is the observer speed The frequency heard by the observer appears lower when the observer moves away from the source f’ = f ( v’ ) / v f’ = f (v + vo)/ v f’ = f (v - vo)/ v Example At the Winter Olympics, an athlete rides her luge down the track while a bell just above the wall of the chute rings continuously. When her sled passes the bell, she hears the frequency of the bell fall by the musical interval called a minor third. That is, the frequency she hears drops to five sixths of its original value. (a) Find the speed of sound in air at the ambient temperature, 10.0C. (b) Find the speed (a) v = 331 + 0.6*TC v = 331 + 0.6(-10); (b) f’’ / f’ = 5/6 Approaching bell f’ = f (v + vo)/ v 5 / 6 = f (v - vo)/ v 5 / 6 = (v - vo) v = 325 m/s Leaving bell f’’ = f (v - vo)/ v / f (v + vo)/ v / (v + vo) of the athlete. 5(v + vo) = 6 (v - vo) vo = 29.5 m/s The below section is not covered during lecture. The speed of the source can exceed the speed of the wave. The envelope of these wave fronts is a cone whose apex half-angle is given by sin θ = vt / vsourcet θ = sin-1(v / vsource) (θ is the Mach angle) The ratio vsource / v is referred to as the Mach number The conical wave front produced when vsource > v is known as a supersonic “shock wave” The shock wave carries a great deal of energy concentrated on the surface of the cone There are correspondingly great pressure variations Doppler Ball - OW-B-DB 14.7 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 (c) For mechanical waves, linear waves have a medium, the resultant value of the wave function at amplitudes much smaller than their wavelengths any point is the algebraic sum of the values of the wave functions of the individual waves (d) Two traveling waves can pass through each other without being destroyed or altered, because of the (b) Waves that obey the superposition principle are superposition principle linear waves 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 To the right 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 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 S can reach R by two different paths. The upper path can be varied Constructive interference occurs @ Δr = |r2 – r1| Δr = n λ (n = 0, 1, …) Destructive interference occurs @ Δr = |r2 – r1| Δr = n ½λ (n is odd) φ occur unequal path lengths φ = Δr (2π λ) Constructive interference Δr = 2n λ/2 Δr = nλ Destructive Inteference Δr = (2n+1) λ/2 Δr = nλ + ½ 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. Use the trig identity sina + sinb = 2[cos(a-b)/2][sin(a+b)/2] 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 y = A sin (kx - ωt) + A sin (kx - ωt + φ) y = 2A cos (φ /2) sin (kx - ωt + φ /2) Destructive Interference Speakers: OW-B-Di 14.8 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 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 The amplitude of the individual waves, A The amplitude of the simple harmonic motion of the elements in the medium, 2A sin kx The amplitude of the standing wave, 2A Nodes occurs at a point of zero amplitude x = n λ /2 n = 0, 1, 2, … Antinode occurs at a point of maximum displacement, 2A x = n λ /2 + λ /4 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. k = 2 π / 0.4 = 2 π / =5π = 15.7 meters v= f v = 31.8 (15.7) v = 500 m/s ω = 2π f 200 = 2 π f f = 100/ π f = 31.8 Hz to apply trig identity sin(a±b) = sina cosb ± cosa sinb (a = kx and b = ωt) 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) Standing Waves, Cenco String: OW-B-Sc Standing Waves in a String Fixed at Both Ends Consider a string fixed at both ends The string has length L Standing waves are set up by a continuous superposition of waves incident on and reflected from the ends There is a boundary condition on the waves The wavelengths of the normal modes for a string of length L fixed at both ends are λ= 2L / n n = 1, 2, 3, … n is the nth normal mode of oscillation 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 Quantization is a common occurrence when waves are subject to boundary conditions 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 A musical note is defined by its fundamental frequency, (n = 1) Example: The frequency of the string can be changed by changing either its length or its tension (changing it’s boundary condition) 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. The above string is vibrating in its second harmonic A middle “C” on a piano has a fundamental frequency of 262 Hz. What are the next two harmonics of this string? v2 = FT / μ v2 = 20 / .009 v = 47.1 m/s fnat = n v/2L f1 = n v/2L f1 = 47.1/(2*30) f1 = 0.786 Hz Ans: ƒ1 = 262 Hz ƒ2 = 2ƒ1 = 524 Hz ƒ3 = 3ƒ1 = 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? Standing Waves in Air Columns ω = √(g/L) 2πf = √(g/L) 2πf = √(9.8/2) f = 0.352 The closed end of a pipe is a displacement node because the wall at this end doesn’t allow longitudinal motion Closed one end of the air molecules. Thus the reflected fnat = n v/4L wave is 180° out of phase with the v = f1 λ incident wave. Also the pressure wave is 90° out of phase with the displacement wave, 17.2, Open both ends we know the closed end of the tube fnat = n v/2L corresponds to a pressure ANTINODE. The open end is a displacement AND pressure node. 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. Extra … Standing Waves in Rods 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 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. 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. 14.9 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 v =f λ 5100 = 400(L) L = 1.16 meters Beats (Interference in Time) Beating is the periodic variation in amplitude at a y1 = A cos ω1t y1 = A cos 2πf1t y2 = A cos ω2t y2 = A cos 2πf2t given point due to the superposition of two waves having slightly different frequencies. where Anew = 2Acos((πt(f1 - f2)) 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) The above derivation isn’t explained in the book…only the result. 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 fbeat = |f1 – f2| 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 Differential Tuning Forks: OW-B-Df 14.11 Rate of Energy Transfer by Sinusoidal Waves on Strings (Optional) ΔK = ½ m v2 ΔK = ½(μΔx)v2 dK = ½ μ ω2A2 cos2(kx) dx ∫dK = ½ μ ω2A2 ∫cos2(kx) dx So our total energy is TE = ¼ μ ω2A2 λ + ¼ μ ω2A2 λ μ is mass per unit length v is the transverse speed integrate by parts K = ½ μ ω2A2 [x/2+sin(2kx)/4k] dK = ½ μ dx [ v2 ] dK = ½ μ dx [ω2A2 cos2(kx – ωt)] dK = ½ μ ω2A2 cos2(kx – ωt) dx @t=0 (ωt just shifts the phase, thus if Example A taut rope has a mass of 0.180 kg and a length of 3.60 m. What power must be supplied to the rope in order to generate sinusoidal waves having an amplitude of 0.100 m and a wavelength of 0.500 m and traveling with a speed of 30.0 m/s? The Linear Wave Equations ΣFy ΣFy m ay m (∂2y/∂t2) μΔx (∂2y/∂t2) = FT sinθB = FT (∂y/∂xB) = FT (∂y/∂xB) = FT (∂y/∂xB) = FT [∂y/∂x B – – – – – FT sinθA FT (∂y/∂x A) FT (∂y/∂x A) FT (∂y/∂x A) ∂y/∂xA] (μ/ FT) ∂2y/∂t2 = [∂y/∂xB – ∂y/∂xA] / Δx Work = ΔK TE = ΔK for one full period 2 Using similar derivation from above, we get our Potential energy function as U = ¼ μ ω2A2 λ we’re attempting to find Total Energy a phase shift is irrelevant) 14 from 0 to λ K=¼μω A λ 2 TE = ½ μ ω2A2 λ P = Work / time P = ΔK / T P = ½ μ ω2A2 λ / T v = f λ = λ/T P = ½ μ ω2A2 v v = fλ 30.0 = f (0.500) f =60.0 Hz ω = 2πf ω = 2π(60) ω = 120π rad/s v P =½(0.18/3.6) (120π)2 (0.100)2 30.0 P = 1070 Watts P=½ (optional) (eq 16.23) (eq 16.24) (eq 16.25) At this point we need to remember that at some point in time, tension at B was or will be the same as the tension at A at the current time, so that μ ω2 A2 Tsin(θB + ωt) - TsinθA = 0 at some time, t. Notes & explanations The partial derivative of any function is Associate f(x + Δx) with ∂y/∂xB (1st derivative of point B) and f(x) with ∂y/∂xA (1st derivative of point A) The derivative of a point is the slope (1st derivative), and the definition of a partial derivative gives us the 2nd derivative, thus ∂2y/∂x2 (μ/T) ∂2y/∂t2 = ∂2y/∂x2 This is the linear wave equation for a string. y(x, t) = A sin (kx – ωt) (from 16.2) ∂2y/∂t2 = (FT /μ) ∂2y/∂x2 ∂2y/∂t2 y(x, t) = A sin (kx – ωt) ∂y/∂t = -ω A cos (kx – ωt)] ∂2y/∂t2 = -ω2 A sin (kx – ωt)] ∂2y/∂t2 -ω2 A sin(kx – ωt)] (sinθ)max = (cosθ)max = 1 -ω2 A ω2 / k2 v2 T represents the tension which is a force, we should say FTension. ΣFy = TsinθB – TsinθA (TB is going up…thus positive) If θ is small, θ < 10° we know (always applicable at small amplitudes) sin θ = tan θ; & tan θ = y / x; And since our infinitesimal displacements (sin θ = dy/dx) are changing with time (& only concerned with components of “y” that are dependant on “x”) sin θ = ∂y/∂x = ∂2y/∂x2 y(x, t) = A sin (kx – ωt) ∂y / ∂x = k A cos (kx – ωt)] ∂2y/∂x2 = -k2 A sin (kx – ωt)] (FT /μ) ∂2y/∂x2 = (FT /μ) -k2A sin(kx – ωt)] = (FT /μ) -k2A = FT / μ = FT / μ ∂2y/∂t2 = (FT /μ) ∂2y/∂x2 ∂2y/∂x2 = (μ/ FT) ∂2y/∂t2 ∂2y/∂x2 = (1/v2) ∂2y/∂t2 Example Show that the wave function y = ln[b(x – vt)] is a solution to ∂2y/∂x2 = (1/v2) ∂2y/∂t2, where b is a constant. y = ln[b(x – vt)] ∂y/∂t = -bv / b(x-vt) ∂2y/∂t2 = -1(-bv)(bv) / b2(x-vt)2 ∂2y/∂t2 = v2 / (x-vt)2 (1/v2) ∂2y/∂t2 = Yes, it satisfies the equation y = ln[b(x – vt)] ∂y/∂x = -b / b(x-vt) ∂y/∂x = -1 / (x-vt) ∂2y/∂x2 = -1 / (x-vt)2 ∂2y/∂x2