Physics Solution Manual for 1100 and 2101

# 0 hz so the beat frequency increases since the

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Unformatted text preview: structive interference. First, we will determine the wavelength being produced by the speakers. Then, we will determine the difference in path lengths between the speakers and the observer and compare the differences to the wavelength in order to decide which type of interference occurs. SOLUTION According to Equation 16.1, the wavelength λ is related to the speed v and frequency f of the sound as follows: λ= v 343 m/s = = 0.800 m f 429 Hz Since ABC in Figure 17.7 is a right triangle, the Pythagorean theorem applies and the difference ∆d in the path lengths is given by 2 2 ∆d = d AC − d BC = d AB + d BC − d BC We will now apply this expression for parts (a) and (b). a. When d BC = 1.15 m , we have 2 2 ∆d = d AB + d BC − d BC = ( 2.50 m )2 + (1.15 m )2 − 1.15 m = 1.60 m Since 1.60 m = 2 ( 0.800 m ) = 2λ , it follows that the interference is destructive (the speakers vibrate out of phase). b. When d BC = 2.00 m , we have 2 2 ∆d = d AB + d BC − d BC = ( 2.50 m )2 + ( 2.00 m )2 − 2.00 m = 1.20 m ( ) λ , it follows that the interference is Since 1.20 m = 1.5 ( 0.800 m ) = 1 speakers vibrate out of phase). 1 2 constructive (the 908 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA 9. REASONING When the difference l1 − l 2 in path lengths traveled by the two sound ( 1 , 1 1 , 2 1 , K) 2 2 2 waves is a half-integer number of wavelengths, destructive interference occurs at the listener. When the difference in path lengths is zero or an integer number (1, 2, 3, K) of wavelengths, constructive interference occurs. Therefore, we will divide the distance l1 − l 2 by the wavelength of the sound to determine if constructive or destructive interference occurs. The wavelength is, according to Equation 16.1, the speed v of sound divided by the frequency f ; λ = v/f . SOLUTION a. The distances l1 and l 2 can be determined by applying the Pythagorean theorem to the two right triangles in the drawing: l1 = ( 2.200 m ) l2 = ( 2.200 m )2 + (1.187 m )2 Therefore, 2 1.813 m 1.187 m 2.200 m l2 l1 + (1.813 m ) = 2.851 m 2 = 2.500 m P l1 − l 2 = 0.351 m. The v 343 m/s = = 0.234 m . Dividing the distance l1 − l 2 by f 1466 Hz the wavelength λ gives the number of wavelengths in this distance: wavelength of the sound is λ = Number of wavelengths = l1 − l 2 λ = 0.351 m = 1.5 0.233 m () Since the number of wavelengths is a half-integer number 1 1 , destructive interference 2 occurs at the listener. v 343 m/s = = 0.351 m . Dividing the distance f 977 Hz l1 − l 2 by the wavelength λ gives the number of wavelengths in that distance: b. The wavelength of the sound is now λ = Number of wavelengths = l1 − l 2 λ = 0.351 m =1 0.351 m Since the number of wavelengths is an integer number (1) , constructive interference occurs at the listener. Chapter 17 Problems 909 10. REASONING When the listener is standing midway between the speakers, both sound waves travel the same distance from the speakers to the listener. Since the speakers are vibrating out of phase, when the diaphragm of one speaker is moving outward (creating a condensation), the diaphragm of the other speaker is moving inward (creating a rarefaction). Whenever a condensation from one speaker reaches the listener, it is met by a rarefaction from the other, and vice versa. Therefore, the two sound waves produce destructive interference, and the listener hears no sound. When the listener begins to move sideways, the distance between the listener and each speaker is no longer the same. Consequently, the sound waves no longer produce destructive interference, and the sound intensity begins to increase. When the difference in path lengths l1 − l 2 traveled by the two sounds is one-half a wavelength, or l1 − l 2 = 1 λ , constructive 2 interference occurs, and a loud sound will be heard. SOLUTION The two speakers are vibrating out of phase. Therefore, when the difference in path lengths l1 − l 2 traveled by the two sounds is one-half a wavelength, or l1 − l 2 = 1 λ , constructive interference occurs. Note that this 2 condition is different than that for two speakers vibrating in phase. The frequency f of the sound is equal to the speed v of sound divided by the wavelength λ; f = v/λ (Equation 16.1). Thus, we have that l1 − l 2 = 1 λ = 2 v 2f or f= Midpoint 4.00 m v 2 ( l1 − l 2 ) 0.92 m l1 1.50 m l2 1.50 m The distances l1 and l 2 can be determined by applying the Pythagorean theorem to the right triangles in the drawing: l1 = l2 = ( 4.00m )2 + (1.50 m + 0.92 m )2 = 4.68 m ( 4.00 m )2 + (1.50 m − 0.92 m ) = 4.04 m 2 The frequency of the sound is f= v 343 m/s = = 270 Hz 2 ( l1 − l 2 ) 2 ( 4.68 m − 4.04 m ) 910 THE PRINCIPLE OF LINEAR SUPERPOSITION AND INTERFERENCE PHENOMENA 11. REASONING AND SOLUTION Since v = λ f, the wavelength of the tone is λ= 343 m/s v = = 4.70 m f 73.0 Hz The figure below shows the line between the two speakers and the distances in question. L L-x x B A P Constructive interference will occur...
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## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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