PHYS Summer Professor Caillault Homework Solutions. Chapter 14

Transcription

1 PHYS Summer Professor Caillault Homework Solutions Chapter 14

2 5. Picture the Problem: A wave of known amplitude, frequency, and wavelength travels along a string. We wish to calculate the distance traveled horizontally by the wave in 0.5 s and the distance traveled by a point on the string in the same time period. Strategy: Multiply the time by the wave speed, where the wave speed is given by equation 14-1, to calculate the horizontal distance traveled by the wave. A point on the string travels up and down a distance four times the amplitude during each period. Calculate the fraction of a period by dividing the time by the time of a full period. Set the period equal to the inverse of the period and multiply by four times the amplitude to calculate the distance traveled by a point on the string. Solution: 1. (a) Calculate the d w vt ( λ f )t ( m) ( 4.5 Hz) ( 0.50 s) 0.61 m horizontal distance: 2. (b) Calculate the vertical distance: 4.5 Hz t d k ( 4A) T 4Aft m ( 0.50 s) 1.1 m 3. (c) The distance traveled by a wave peak is independent of the amplitude, so the answer in part (a) is unchanged. The distance traveled by the knot varies directly with the amplitude, so the answer in part (b) is halved. Insight: A point on the string travels four times the wave amplitude in the same time that the crest travels one wavelength. 11. Picture the Problem: Sound takes 0.94 seconds to travel across a wire of known length and density. We want to calculate the tension in the wire. Strategy: Solve equation 14-2 for the tension in the wire. The velocity is given by the length of the wire divided by the time for the sound to travel across it. The linear mass density is the mass divided by the length. Solution: 1. (a) Solve equation 14-2 for the tension: v F µ F µv 2 m L L t 2. Insert the given mass, length and time: ( kg) 7.3 m F 0.94 s 2 ml t N 2 3. (b) The mass is proportional to the tension (if L and t remain constant). So increased mass means increased tension. 4. (c) Solve with a mass of kg. ( kg) ( 7.3 m) F 0.94 s N Insight: A heavier string requires greater tension for a wave to travel across it in the same time.

3 3. Sum the two times to equal the total time: t t 1 + t 2 2d g + d v s 23. Picture the Problem: The figure represents you dropping a rock down a well and listening for the splash. From the time lapse between dropping the rock and hearing the splash we want to calculate the depth of the well. Strategy: The time to hear the splash, t 1.5 s, is the sum of the time for the rock to fall to the water, t 1, and the time for the sound of the splash to reach you, t 2. Solve the free-fall equation (equation 2-13) for the time to fall and displacement at constant velocity to calculate the time for the sound to return. Set the sum of these times equal to the time to hear the splash and solve for the distance. Solution: 1. (a) Solve equation 2-13 for the falling time: t 1 2d g 2. Solve for the time for the sound to travel up the well: t 2 d v s 4. Rewrite as a quadratic equation in terms of the variable d : 0 1 d v s g d t m/s ( d ) m/s 2 d 1.2s 5. Solve for d using the quadratic formula and square the result: d m d m 11 m 6. (b) The time to hear the sound would be less then 3.0 seconds because, although the sound travel time would double, the fall time would less than double. Insight: The time to hear the sound for a 21-meter-deep well is 2.1 s, which is indeed less than 3.0 s. 31. Picture the Problem: Twenty identical violins play simultaneously. Strategy: Write equation 14-8 for the intensity level of 20 violins and, using the rules of logarithms solve for the intensity level of one violin. Solution: 1. (a) Write equation 14-8 for 20 violins: β log 20I I 0 I 10 log β log 20 I log( 20) 2. Solve for the intensity level of one violin: β 1 β log( 20) db 3. (b) Doubling the number of violins will increase the intensity level by 10 log( 2) 3.0 db, therefore the intensity of 40 violins will be 82.5 db db 85.5 db, which is less than 165 db. Insight: The intensity is proportional to the number of instruments. However, the intensity level is related to the logarithm of the intensity, so it is not proportional to the number of instruments.

4 37. Picture the Problem: A bat, flying toward a stationary moth at 3.25 m/s, emits a sound at 34.0 khz, as shown in the figure. We wish to calculate the frequency heard by the moth. Strategy: This is a moving source problem, so use equation Since the bat is approaching the moth, you will use the minus sign in the equation. Solution: 1. (a) Insert speed and emitted frequency into equation 14-10: 1 f 1 u v f 1 ( 35.0 kh 1 ( 3.25 m/s 343 m/s) 35.3 khz 2. (b) Since the bat is approaching the moth, the Doppler shift is to higher frequencies. Increasing the speed will increase the Doppler shift so the moth will hear a higher frequency.. 3. (c) Insert the higher speed into equation 14-10: 1 f 35.0 khz 1 ( 4.25 m/s 343 m/s) 35.4 kh Insight: If the bat were moving away from the moth, it would hear a frequency lower than 35.0 khz. This frequency would become even lower as the speed of the bat increased.

5 55. Picture the Problem: The image shows two out-of-phase speakers separated by 3.5 meters and an observer standing by a wall 5.0 meters away. When the observer moves 0.84 meters along the wall he goes from the central destructive interference to the first constructive interference. We want to calculate the frequency of sound emitted by the speaker. Strategy: Use the Pythagorean Theorem to calculate the distance of the observer from each speaker. Calculate the difference in distances to each speaker. Since the speakers are out of phase, constructive interference occurs when the difference in distances is equal to a half wavelength. Set the difference in distances equal to a half wavelength and use equation 14-1 to calculate the resulting frequency. Solution: 1. Calculate d 1 : d 1 ( 5.0 m) m m 2. Calculate d 2 : d m 3. Set λ 2 d 1 d 2 d 1 d 2 1 λ : 2 4. Use equation 14-1 to calculate the frequency: m ( 3.5 m ) 0.84 m 2 2( m m) m f v 343 m/s 0.31 khz λ m) m Insight: Constructive interference will also occur at the observer s position for other frequencies for which d 1 d 2 is an integer multiple of half the wavelength, corresponding to f 0.94 khz, 1.6 khz, etc. 59. Picture the Problem: The image shows an ear canal of length 2.4 cm. Strategy: Treating the ear canal as a pipe closed at one end, we wish to calculate the fundamental and third harmonic frequencies and wavelengths. Use equations to calculate the frequencies and wavelengths. For the fundamental use n 1, and for the third harmonic use n 3. Solution: 1. (a) Calculate the fundamental frequency and wavelength using equation with n 1: 2. (b) Calculate the third harmonic frequency and wavelength using equation with n 3: f 1 nv 4L m/s m λ 1 4L n cm f 3 nv 4L 3.6 khz cm m/s m λ 3 4L n cm 11 khz cm 3. (c) The fundamental frequency is inversely proportional to the length of the ear canal. Therefore, if the ear canal is shorter, the fundamental frequency will be greater. Insight: For an ear canal of length 2.2 cm the fundamental frequency will be 3.9 khz.

6 71. Picture the Problem: We are given the length and frequency of one cello string. A second string, which is identical to the first, is shortened. When sounding together, the two strings produce a beat frequency of 4.33 Hz. Strategy: We want to calculate the length of the shortened string. Since the two strings are identical, the speed of the waves on each string is the same. Use equation to write a relation between the frequencies and lengths, given the same velocities. Solve this relation for the length of the second string. Since the second string has a shorter length, it has the higher frequency. Write the frequency of the second string as the sum of the first frequency and the beat frequency. Solution: 1. Solve equation for the velocity in terms of frequency and length: f v 2L v 2 f 1 L 1 2 f 2 L 2 2. Solve for the length of the second string: L 2 f 1 f 2 L 1 3. Set the second frequency equal to the sum of the first frequency and the beat frequency: L 2 f 1 f 1 + f beat L Hz Hz Hz 1.25 m Insight: The string has been shortened by a distance of 1.25 m 1.21 m 0.04 m or 4.0 cm m