Chapter 18 Superposition and Standing Waves 18.1 Superposition and Interferences = + y tx ),(' y1 tx ),( y2 tx ),( Overlapping waves algebraically add to produce a resultant wave (or net wave). Overlapping waves do not in any way alter the travel of each other.
Superposition of Sinusoidal Waves = −ω y1 tx ),( ym sin( kx t) = − ω + φ y2 tx ),( ym sin( kx t ) φ φ y tx ),(' = y + y = 2y sin( kx −ωt + )cos( ) 1 2 m 2 2 If two sinusoidal waves of the same amplitude and wavelength travel in the same direction along a stretched string, they interfere to produce a resultant sinusoidal wave traveling in that direction.
Phase Differences and Resulting Interference Types Amplitude of Resultant Type of Phase difference Wave Interference Degrees Radians Wavelengths
0 0 0 2ym Fully constructive
120 2π/3 0.33 ym Intermediate 180 π 0.50 0 Fully destructive
240 4/3 π 0.67 ym Intermediate
360 2π 1.00 2ym Fully constructive
865 15.1 2.40 0.60 ym Intermediate
Example:
1 Two identical sinusoidal waves, moving in the same direction along a stretched string, interfere with each other. The amplitude y m of each wave is 9.8 mm, and the phase difference φ between them is 100 o.
(a) What is the amplitude y’m of the resultant wave due to the interference of these two waves, and what type of interference occurs? (b) What phase difference, in radius and wavelength, will give the resultant wave an amplitude of 4.9 mm? φ φ y'= y sin( kx −ωt) + y sin( kx −ωt + φ) = 2y cos( )sin( kx −ωt + ) m m m 2 2 φ y' = 2y cos( ) = 2 ⋅ 8.9 ⋅ cos( 50 o ) = 6.12 mm m m 2 intermediate φ φ y' = 9.4 = 2y cos( ) = 2 ⋅ 8.9 ⋅ cos( ) , φ = 151 o = 636.2 rad m m 2 2 2π x = 636.2 , x = 42.0 ⋅ λ λ
Interference of Sound Waves: It is often useful to express path difference in terms of the phase angle φ between the two waves. ∆r φ = λ 2π λ ∆r = 2n for constructive interference 2 λ ∆r = ()2n +1 for destructive interference 2 Example: Two identical speakers placed 3.00 m apart are driven by the same oscillator. A listener is originally at point O, located 8.00 m from the center of the line connecting the two speakers. The listener then moves to point P, which is a perpendicular distance 0.350 m from O, and she experiences the first minimum in sound intensity. What is the frequency of the oscillator? λ ∆r = r − r = 85.1 2 + 82 − 15.1 2 + 82 = 13.0 ‰ = 13.0 2 1 2 ‰ f = v / λ = 343 26.0/ = 3.1 (kHz)
2 18.2 Standing Waves = − ω = + ω y1 ym sin( kx t) , y2 ym sin( kx t ) = + = ω y' y1 y2 2ym sin( kx )cos( t)
n 2y sin( kx ) = 0 , when kx = nπ , x = λ are the m 2 position of nodes
1 2y sin( kx ) = max imum , when kx = (n + )π , m 2 1 λ x = (n + ) are the position of antinodes 2 2
18.3 Standing Waves in Strings Fixed at Both Ends boundary condition: sin( kx = )0 = 0 , sin( kL ) = 0 2L kL = nπ , λ = n v v f = = n λ 2L
Example: A string, tied to a sinusoidal vibrator at P and running over a support at Q, is stretched by a block of mass m. The separation L between P and Q is 1.2 m, the linear density of the string is 1.6 g/m, and the frequency f of the vibrator is fixed at 120 Hz. The amplitude of the motion at P is small enough for that point to be considered a node. A node also exists at Q.
(a) What mass m allows the vibrator to set up the fourth harmonic on the string?
3 T mg 2L 2 ⋅ 2.1 v = = , λ = = = 6.0 , v = fλ = 120 ⋅ 6.0 = 72 µ .0 0016 n 4
m ⋅ 8.9 = 72 2 ⋅ .0 0016 , m = 846.0 kg
Example: Changing string vibration with water One end of a horizontal string is attached to a vibrating blade, and the other end passes over a pulley. A sphere of mass 2.00 kg hangs on the end of the string. The string is vibrating in its second harmonic. A container of water is raised under the sphere so that the sphere is completely submerged. After this is done, the string vibrates in its fifth harmonic. What is the radius of the sphere?
T v v T = mg = 2 ⋅ 8.9 = 6.19 N , v = , f = 1 = 2 , 1 µ λ λ 1 2
λ T L 5.2/ T 2 = 2 , = 2 λ 1 T1 L 1/ 6.19
4π 4π T = 136.3 = T − ρg R 3 = 6.19 −1000 ⋅ 8.9 R 3 2 1 3 3 R = .0 0737 m
Example: A middle C string on a piano has a fundamental frequency of 262 Hz, and the A note has a fundamental frequency of 440 Hz. (a) Calculate the frequency of the next two harmonics of the C string. (b) If the strings for A and C notes are assumed to have the same mass per unit length and the same length, determine the ratio of tensions in the two string. (c) In a real piano, the assumption we made in part (b) is only partial true. The string densities are equal, but the A string is 64% as long as the C string. What is the ratio of their tensions? = = ⋅ = = ⋅ = (a) C string: f1 262 Hz , f 2 2 262 524 Hz , f3 3 262 786 Hz
v 1 T f T µ 262 T T (b) f = = , C = C , = C , C = 355.0 µ µ 2L 2L f A TA 440 TA TA
f L T T T (c) C = A C = 64.0 C , C = 866.0 f A LC TA TA TA
4 18.4 Resonance
18.5 Standing waves in air columns
Example: Wind in a Culvert A section of drainage culvert 1.23 m in length makes a howling noise when the wind blows across its open ends. (a) Determine the frequencies of the first three harmonics of the culvert if it is cylindrical in shape and open at both ends. Take v=343 m/s as the speed of sound in air. λ = ( ) = λ 1 L 2/1/ ‰ f1 v / 1 λ = ( ) = λ = 2 L 1/ ‰ f2 v / 2 2 f1 λ = ( ) = λ = 1 L 2/3/ ‰ f3 v / 3 3 f1 (b) What are the three lowest natural frequencies of the culvert if it is blocked at one end? λ = ( ) 1 L 4/1/ λ = ( ) = 2 L 4/3/ ‰ f2 3 f1 18.6 Standing Waves in Rods and
5 Membranes 18.7 Beats: Interference in Time interference effect that results from the superposition of two waves with slightly different frequencies = −ω = −ω y1 Acos( kx 1t) , y2 Acos( kx 2t) = + = −ω + −ω y' y1 y2 Acos( kx 1t) Acos( kx 2t) ω + ω ω −ω ω −ω y'= 2Acos( kx − 1 2 t)cos( 1 2 t) ≈ 2Acos( kx −ω t)cos( 1 2 t) 2 2 1 2 ω −ω y = 2Acos( 1 2 t) 0 2 1 ω −ω the beat frequency = 1 2 = f − f π 2 1 2
18.8 Nonsinusoidal Wave Patterns Fourier’s theorem (xy ) is a wave on the string satisfy y(0) = y(L) = 0 ‰ ( ,txy ) = (a cos (kx )+ bsin (kx ))sin (ωt) ‰ (xy ) = a cos (kx )+ bsin (kx ) π ∞ π = π () = n () = n x ‰ kL n ‰ yn x bn sin x ‰ xy ∑bn sin L n=1 L f x)( = 10 , 0
6 π a π a π () = n x n x () = 2 n x xf ∑bn sin ‰ ∫sin 1 dx bn ∫sin dx a 0 a 0 a
a nπx a 2 sin ()1 dx = b ‰ b = 1[ − ()−1 n ] ∫ n n π 0 a 2 n
,0 otherwise 2 ∞ (1− (−1)n ) nπx ‰ ()xf = = sin ≤ ≤ π ∑ 0,1 x a n=1 n a
7