Superposition and Standing Waves. Solutions of Selected Problems

Transcription

1 Chapter 18 Superposition and Standing Waves. s of Selected Problems 18.1 Problem 18.8 (In the text book) Two loudspeakers are placed on a wall 2.00 m apart. A listener stands 3.00 m from the wall directly in front of one of the speakers. A single oscillator is driving the speakers at a frequency of 300 Hz. (a) What is the phase difference between the two waves when they reach the observer? (b) What If? What is the frequency closest to 300 Hz to which the oscillator may be adjusted such that the observer hears minimal sound? (a) The listener is 3 m away from one speaker and = 13 m away from the other speaker. The path difference x is then: and the wavelength is: and the phase difference φ is: x = 13 3 = m λ = v f = = 1.14 m φ = 2π x λ = 2π = 3.34 rad

2 2 CHAPTER 18. SUPERPOSITION AND STANDING WAVES. SOLUTIONS OF SELECTED PROBLEMS (b) Minimum sound means destructive interference, and for that we must have: x = 1 2 λ = 1 v 2 f or f = v 2 x = = 283 Hz

3 18.2. PROBLEM (IN THE TEXT BOOK) Problem (In the text book) Two sinusoidal waves combining in a medium are described by the wave functions y 1 = (3.0cm) sin π(x t) and y 2 = (3.0cm) sin π(x 0.60t) where x is in centimeters and t is in seconds. Determine the maximum transverse position of an element of the medium at (a) x = cm, (b) x = cm, and (c) x = 1.50 cm. (d) Find the three smallest values of x corresponding to antinodes. The combined wave function is: y = y 1 + y 2 = 3.00 sin[π(x t)] sin[π(x 0.600t)] Using: sin(a ± b) = sin(a) cos(b) ± cos(a) sin(b) we get for the combined wave: y = 3.00[sin(πx) cos(0.600t) + cos(πx) sin(0.600t)] [sin(πx) cos(0.600t) cos(πx) sin(0.600t)] = 6.00 sin(πx) cos(0.600t) (a) To find the maximum transverse position of an element of the medium; we start by setting cos(0.600t) = +1, we then get at x = 0.25 cm: y max = 6.00(cm) sin(0.250π) = 4.24 cm (b) At x = cm: y max = 6.00 sin(0.500π) = 6.00 cm

4 4 CHAPTER 18. SUPERPOSITION AND STANDING WAVES. SOLUTIONS OF SELECTED PROBLEMS (c) Now at x = 1.5 cm, we take cos(0.600t) = 1: (d) The antinodes occur when: y max = 6.00 sin(1.5π) = 6.00 cm x = nλ 4 n = 1, 3, 5, From the combined wave function above, we find k = π, we then get: k = 2π λ = π or λ = 2.00 cm The smallest values of x corresponding to antinodes are given by n = 1, 3, 5, so: x 1 = λ 4 x 2 = 3λ 4 x 3 = 5λ 4 = cm = 1.50 cm = 2.50 cm

5 18.3. PROBLEM (IN THE TEXT BOOK) Problem (In the text book) A standing-wave pattern is observed in a thin wire with a length of 3.00 m. The equation of the wave is y = (0.002m) sin(πx) cos(100πt) where x is in meters and t is in seconds. (a) How many loops does this pattern exhibit? (b) What is the fundamental frequency of vibration of the wire? (c) What If? If the original frequency is held constant and the tension in the wire is increased by a factor of 9, how many loops are present in the new pattern? The general wave function of a standing wave in a wire is: y = 2A sin(kx) cos(ωt) comparing this general function with the given function we get: k = π = 2π λ or λ = 2.00 m and ω = 100π = 2πf or f = 50.0 Hz (a) The distance between adjacent nodes d NN = λ/2 and the number of loops N on the wire with length L = 3.00 m is: N = and the speed of the wave in the wire is: L = L d NN 2λ = 2L λ = v = fλ = = 100 m/s = 3 loops (b) The standing wave that has the fundamental frequency fits one-half of a wavelength in the entire length of the wire, so its wavelength is: λ = 2L = 6.00 m This wave travels the wire with the same speed as the wave that fits 3 loops in the length of the wire, so fundamental frequency is then: f = v λ = 100 m/s 6.00 m = 16.7 Hz

6 6 CHAPTER 18. SUPERPOSITION AND STANDING WAVES. SOLUTIONS OF SELECTED PROBLEMS (c) Let the initial tension be T 1, the velocity v 1 = 100 m/s, the new tension T 2 = 9T 1 and the new velocity v 2, then T 2 v 2 = µ = 9T 1 µ = 3 T 1 µ = 3v 1 = = 300 m/s The new wavelength is: and the number of loops in this case: λ 2 = v 2 f = = 6.00 m d NN = 1 2 λ 2 = 3.00 m and N = L 1 2 λ = = 1 loop

7 18.4. PROBLEM (IN THE TEXT BOOK) Problem (In the text book) A student holds a tuning fork oscillating at 256 Hz. He walks toward a wall at a constant speed of 1.33 m/s. (a) What beat frequency does he observe between the tuning fork and its echo? (b) How fast must he walk away from the wall to observe a beat frequency of 5.00 Hz? The student and the tuning fork are moving toward the wall with a speed of v s = 1.33 m/s. The frequency at the wall f is: f = v f v v s Now waves with frequency f are reflected from the wall and heard by the student who is walking, with a speed of 1.33 m/s toward the wall which is the source of the the reflected waves. The frequency of the reflected waves as heard by the student f is: f = v + v s f = v + v s v f = v + v s f v v v v s v v s The student, actually, hears the wave produced by tunning fork and the wave reflected from the wall. The two waves combine and produce beats. The beats frequency f b = f f is then: f b = f f = v + v ( ) ( ) s v + vs v + vs v + v s f f = f 1 = f = f 2v s v v s v v s v v s v v s (a) The last formula is for the case when the student is walking toward the wall, so f b is then: f b = = 1.99 Hz (b) When student is walking away from the wall, v s changes sign and, In this case f b = f f, from that we find v s f = v v s v + v s f f b = f f = f v v s v + v s f = f ( ) v + vs v + v s v + v s f b v + f b v s = 2v s f or v s = f bv 2f f b = = f 2v s v + v s = 3.38 m/s

8 8 CHAPTER 18. SUPERPOSITION AND STANDING WAVES. SOLUTIONS OF SELECTED PROBLEMS 18.5 Problem (In the text book) Two wires are welded together end to end. The wires are made of the same material, but the diameter of one is twice that of the other. They are subjected to a tension of 4.60 N. The thin wire has a length of 40.0 cm and a linear mass density of 2.00 g/m. The combination is fixed at both ends and vibrated in such a way that two antinodes are present, with the node between them being right at the weld. (a) What is the frequency of vibration? (b) How long is the thick wire? (a) The frequency of the standing wave and the tension are the same along the entire length of the wire. Since there is a node at the weld, then there is a half-wave fitted on the length of the thin wire. So, the wavelength λ = 2L 1 = 80.0 cm, where the L 1 is the length of the thin wire. The frequency of the wave is then: f = 1 2L where µ 1 is the linear density of the this wire. T = = 59.9 Hz µ (b) Since the thick wire is twice the diameter of thin wire then its linear density µ 2 is four times that of the thin wire, i.e. µ 2 = 4µ 1 = kg/m. The length of the thick wire L 2 is then: L 2 = 1 T = 2f µ = 20.0 cm