Physics 17 Part N Dr. Alward

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1 Physics 17 Part N Dr. Alward String Waves L = length of string m = mass μ = linear mass density = m/l T = tension v = pulse speed = (T/μ) Example: T = 4.9 N μ = 0.10 kg/m v = (4.9/0.10) 1/2 = 7.0 m/s Shake repeatedly: a train of pulses, all traveling at the same speed, represents a traveling wave. The wave speed is the same as the speed of any pulse: v. Peaks are maxima. Valleys are minima. They re also called maximums and minimums. f =frequency = number of oscillations per second Frequency units: per second, or s -1 1 hertz (Hz) = 1 s -1 λ = wavelength = distance between consecutive peaks (maxima) = distance beween consecutive valleys (minima) = distance between alternate zeros ( nodes ) 1

2 The Wave Equation: Three Different Forms λf = v λ = v/f f = v/λ Example A: The speed of a wave on a string is 12.0 m/s. At what frequency will the wavelength of a wave be 4.0 m? f = 12.0/4.0 = 3.0 Hz Example B: T = 100 N L = 1.6 m λ =? µ = 0.4/1.6 = 0.25 v = (T/µ) 1/2 = (100/0.25) 1/2 = 20 m/s λ = v/f = 20/10 = 2 m m = 0.4 kg f = 10 Hz Resonances At certain frequencies, a resonance, called a standing wave, occurs in which comparatively small-amplitude vibrations of one end of the string leads to comparatively large amplitudes at certain points along the string, and zero amplitude at other points. Recall: λ = distance between alternating nodes. The standing wave consists of loops that have a width of λ/2. In the figure below, there are two loops. N: node A: anti-node 2

3 Each loop has a width equal to half a wave-length: λ/2. The sum of all of the loop widths equals the length of the string. Note that the distance between a node and an anti-node is λ/4. Example A: The speed of pulses on a string of length 1.40 m is 2.60 m/s. What frequency of oscillation will create a standing wave with four anti-nodes? Example B: Resonance on 1.60 m string is occurs with six anti-nodes when the string is oscillated at 6.0 Hz. The tension in the string is 14.0 N. What is the linear mass density of the string? Solution: Four loops. Each loop has a width of λ/2. The sum of these widths equals the length of the string. 4(λ/2) = 1.40 λ = 0.70 m f = v/λ = 2.60/0.70 = 3.71 Hz First, determine the wavelength, then the speed, then the density. 6(λ/2) = 1.60 λ = 0.53 m v = λf = 0.53 (6.0) = 3.18 m/s (T/µ) 1/2 = v (14.0/µ) 1/2 = 3.18 Solve for µ: µ = 1.38 kg/m If one end of a rope is free, then resonances on the rope have an anti-node at the free end. 3

4 Example: A rope 0.90 meters long is hanging vertically; the bottom of the rope is free. An oscillator attached to the top of the rope is vibrating at a frequency of 4.0 Hz, which causes a resonance with four anti-nodes. (See figure at the right.) What is the speed of waves on this rope? Solution: Note that there are 3.5 loops. 3.5 (λ/2) = 0.90 Another way to find λ: Use the fact that the distance between an N and an A is λ/4. These distances are called half-loop widths. There are seven halfloop widths. 7 (λ/4) = 0.90 λ = 0.51 m λ = 0.51 m v = λf = 0.51 (4.0) = 2.1 m/s Sound Waves ΔP = P - Po N = normal pressure (atmospheric pressure, 1010 millibars (mb) H = 1015 mb, so ΔP = 5 mb L = 1005 mb, so ΔP = - 5 mb Speed of sound in air: 340 m/s 4

5 Most sound occurs at frequencies between Hz. Range of human hearing: 20-20,000 Hz Range of wavelengths: 20 Hz: λ = 340/20 = 17 m = 56 ft 20,000 Hz: λ = 340/2 x 10 4 = m = 1.7 cm Ultrasound: greater than 20,000 Hz Interaction of Ultra-Sound Waves with Matter Example: Lithotripsy is a medical procedure in which ultrasound is used to break up kidney stones. Wavelength long compared to object: gentle rise and fall. If the object above is an automobile traveling over mall parking lot speed bumps, the bumps would hardly be noticed. diameter = 1.0 cm λ = 1.0 cm = 0.01 m In kidney tissue, the speed of sound is v = 1500 m/s (not 340 m/s) f = v/λ = 1500/0.01 = 150,000 Hz Wavelength comparable to object s size: maximum disturbance. Rapid rise and fall shakes stone apart. If the object above is an automobile traveling over those speed bumps, sever damage would be done to the car. Rule: When the object size and sound wavelength are comparable, maximum damage is done to the object. 5

6 The Doppler Effect for Sound If either one is moving toward the other, the observed frequency will be higher: choose the sign that makes the ratio larger. If either one is moving away from the other, choose the sign that makes the ratio smaller. fo = fs (340 ± vo) / (340 ± vs) vo = observer speed vs = source speed For example, if the listener and the source are headed toward each other, the top/bottom signs are +/-. If both are moving away from the other, the top/bottom signs are -/+. If the observer is chasing the source, the signs are +/+. If the source is chasing the observer, the signs are -/-, as shown in Example B, below. Example A: In the diagram above, suppose the observer is at rest, and the police car is moving away from the observer at a speed vs = 30 m/s. If the siren s frequency fs is 5,000 Hz, what does the observer hear? Source is moving away, so the heard frequency will be lower, so we need to choose the sign for the source speed that will decrease the ratio. This will happen if we add 30 to the 340 in the denominator, rather than subtract. fo = 5000 ( ) / ( ) = 4595 Hz Example B: A police car emitting 4000 Hz sound is chasing a speeder. What frequency does the driver of the speeding car hear? fo = 4000 (340-50) / (340-30) = 3742 Hz Example C: A fire truck emitting 2500 Hz and traveling east at 40 m/s is racing toward an automobile traveling to the west. The automobile driver hears 3000 Hz. What is the automobile s speed? 3000 = 2500 (340 + vo) / (340 40) vo = 20 m/s 6

7 Sound Wave Resonances in Tubes The closed end of a tube in which a standing wave exists is always a node; open ends are always anti-nodes. The harmonics of a wind instrument are those frequencies that are integer-multiples of the lowest frequency. Example: What are the four lowest frequencies of sound that will resonate in an open-closed tube 0.85 meter long? 1(λ/4) = (λ/4) = (λ/4) = (λ/4) = 0.85 λ = 3.40 m λ = 3.40/3 m λ = 3.40/5 m λ = 3.40/7 m = 100 Hz = 300 Hz = 500 Hz = 700 Hz Note: All of the resonant frequencies ( harmonics ) are odd-integer multiples of the lowest frequency, 100 Hz. Other harmonics include 900, 1100, 1300, 1500, and 1700 Hz, and so on. The even-integer multiples of 100 Hz cannot resonate in this tube. 7

8 Example: What are the four lowest frequencies of sound that will resonate in the 0.85 meter tube in the previous example if the closed end is opened? 2(λ/4) = (λ/4) = (λ/4) = (λ/4) = 0.85 λ = 1.70 m λ = 3.40/4 m λ = 3.40/6 m λ = 3.40/8 m = 200 Hz = 400 Hz = 600 Hz = 800 Hz Note: All of the resonant frequencies ( harmonics ) are even-integer multiples of the lowest frequency, 200 Hz. Other harmonics include 1000, 1200, 1400, 1600, and 1800 Hz, and so on. The odd-integer multiples of 200 Hz cannot resonate in this tube. 8