Physics 11 Chapter 15/16 HW Solutions

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1 Physics Chapter 5/6 HW Solutions Chapter 5 Conceptual Question: 5, 7 Problems:,,, 45, 50 Chapter 6 Conceptual Question:, 6 Problems:, 7,, 0, 59 Q5.5. Reason: Equation 5., string T / s, gies the wae speed on a stretched string with tension T s and linear mass density mlwe /. will inestigate how Ts and are changed in each case below and how that affects the wae speed. Use a subscript for the original string, and a subscript for the altered string. We are gien that ( ) 00cm/s. string (a) ( T ) ( T ) s s ( string) ( Ts) / ( Ts) / ( Ts) / ( ) ( T ) / ( T) / ( T ) / string s s s (b) ( T ) ( T ) m 4m 4 s s ( ) ( ) (00cm/s) 80cm/s string string ( string) ( Ts) / ( Ts) / ( Ts) /4 ( ) ( T ) / ( T) / ( T) / string s s s ( string) ( string) (00cm/s) 00cm/s (c) ( T ) ( T ) L 4L s s 4 ( string) ( Ts) / ( Ts) / ( Ts) / 4 ( ) ( T ) / ( T) / ( T ) / string s s s ( ) ( ) (00 cm/s) 400 cm/s string string (d) ( Ts) ( Ts) m 4m and L 4L ( string) ( Ts) / ( Ts) / ( Ts) / ( ) ( T ) / ( T) / ( T) / string s s s ( ) ( ) 00cm/s string string Assess: Notice as in part (d) that if both the mass and length of the string are increased by the same factor, then is not changed, so the speed is the same (with no change in tension). Q5.7. Reason: The speed of sound in air depends on the temperature of the air. The distance across the stadium can be measured to gie the path length of the sound. Then, you can take the path length and diide by the time between the

2 emission and detection of the pulse to get the speed of the sound. Finally, once the speed of sound is known, you can find the corresponding temperature either by consulting a chart like Table 5. or by using Equation 5.. Assess: The adantage of measuring temperature this way is that it gies you an idea of the aerage temperature for the whole stadium. It also determines the temperature quickly the time of measurement is simply the time it takes for sound to trael across the stadium. P5.. Prepare: The wae is a traeling wae on a stretched string. We will use Equation 5. to find the tension that corresponds to a wae speed of 80 m/s. To be able to obtain the tension from 5., we will first obtain from the first part of the problem. Sole: The wae speed on a stretched string with linear density is string For a wae speed of 80 m/s, the required tension will be TS 75.0 N 50 m/s. 0 kg/m TS string (. 0 kg/m)(80 m/s) 0 N Assess: Increased wae speed must lead to increased tension, as is preiously obtained. P5.. Prepare: Since the lab temperature is 0 C, we know the speed of sound is 4 m/s, as gien in Table 5.. Sole: (a) f 40 khz 4 m/s λ 8.6 mm f 40 0 Hz (b) For the round trip the distance is 5.0 m. Δx 5.0 m Δ t 0.05 s 5 ms 4 m/s x Assess: 5 ms is quick by human reaction time standards, but it is easy to hae electronics in the detector record time interals such as this. P5.. Prepare: From Table 5., the speed of sound in room temperature air is 4 m/s and in water is 480 m/s. On the other hand, the speed of an electromagnetic wae is c. We will use Equation 5.0 for the wae speed. Sole: (a) The frequency is to two significant figures. (b) The frequency is air 4 m/s f 700 Hz λ 0.0 m f 8 c.0 0 m/s Hz.5 GHz λ 0.0 m (c) The speed of a sound wae in water is water 480 m/s. The waelength of the sound wae would be water 480 m/s 7 λ m 990 nm 9 f.5 0 Hz Assess: Because speed frequency waelength, we expected a much higher frequency for the electromagnetic wae. P5.45. Prepare: First compute f 0.0/0 s 0.0 Hz. When the detector (you in the boat) is moing, the measured frequency is f.0/0 s 0.0 Hz. The problem gies o.5 m/s. Use Equation 5.7, f ( o/ ) f0, and sole for, the speed of the waes.

3 Sole: f o f0 o f f0 o f f0 f 0 o f0 f 0.0 Hz (.5 m/s) 4.5 m/s 0.0 Hz 0.0 Hz Assess: These are pretty fast water waes, but within reason. The speed of surface water waes depends on the depth of the water, but is typically in the range of m/s. Howeer, tsunamis can trael much faster than this. Carefully note the subscripts: The o stands for obserer while the 0 stands for the initial or original (frequency). They look similar but are different. P5.50. Prepare: The wae pulse is a traeling wae on a stretched string. We will use Equations 5. and 5. to find the string mass. Sole: The wae speed on a stretched string with linear density is T T LT.0 m (.0 m)(0 N) m 0.05 kg 5 g ml / m 50 0 s m S S S string Assess: A mass of 5 g for the.0 m long string is realistic. Q6.. Reason: See Figure 6.0 in the text. (a) In Chapter 5 we saw that the speed of a wae on a stretched string is T/. s Since the left side of the string has a lower speed, the linear density must be greater there. (b) You would start a pulse from the left side, in the part with the greater linear density, in order to hae the reflection not inerted. Assess: This would be a fairly easy experiment to set up at home with two different strings tied together. Q6.6. Reason: When the frequency is doubled the waelength is haled. This haling of the waelength will increase the number of antinodes to six. Assess: Calculating the number of antinodes in this situation is easier than calculating the number of nodes because there are nodes on each end of the string, so the number is not doubled: It goes from four to seen. P6.. Prepare: The principle of superposition comes into play wheneer the waes oerlap. The waes are approaching each other at a speed of m/s, that is, each part of each wae is moing m eery second. Sole: The graph at t s differs from the graph at t 0 s in that the left wae has moed to the right by m and the right wae has moed to the left by m. This is because the distance coered by the wae pulse in s is m. The snapshot graphs at t 4 s and t 6 s are a superposition of the left- and the right-moing waes. The oerlapping parts of the two are shown by the dotted lines.

4 Assess: This is an excellent problem because it allows you to see the progress of each wae and the superposition (addition) of the waes. As time progresses, you know exactly what has happened to each wae and to the superposition of these waes. P6.7. Prepare: Reflections at both ends of the string cause the formation of a standing wae. Figure P6.7 indicates that there are three full waelengths on the.0-m-long string and that the wae speed is 40 m/s. We will use Equation 5.0 to find the frequency of the standing wae. Sole: The waelength of the standing wae is λ (.0 m) m. The frequency of the standing wae is thus 40 m/s f 60 Hz λ m Assess: The units are correct and this is a reasonable frequency for a ibrating string. P6.. Prepare: A string fixed at both ends forms standing waes. Three antinodes means the string are ibrating as the m standing wae. The waelengths of standing wae modes of a string of length L are gien by Equation 6.. Sole: (a) The frequency is f f, so the fundamental frequency is f (40 Hz) 40 Hz. The fifth harmonic will hae the frequency f 5 5f 700 Hz. (b) The waelength of the fundamental mode is λ L.0 m. The wae speed on the string is λ f (.0 m)(40 Hz) 68 m/s. Alternatiely, the waelength of the n mode is λ (L) 0.40 m, from which λ f (0.40 m)(40 Hz) 68 m/s. The wae speed on the string, gien by Equation 5., is

5 T S TS (0.000 kg/m)(68 m/s) 56 N Assess: You must remember to use the linear density in SI units of kg/m. Also, the speed is the same for all modes, but you must use a matching λ and f to calculate the speed. P6.0. Prepare: We assume that the speakers are identical and that they are emitting in phase. Since you don t hear anything, the separation between the two speakers corresponds to the condition of destructie interference. Sole: Equation 6.9 for destructie interference is Since the waelength is λ λ 5λ Δ d m+ λ Δ d,, λ 40 m/s.0 m f 70 Hz three possible alues for d are.0 m,.0 m, and 5.0 m. Assess: The units worked out and these are reasonable distances. P6.59. Prepare: The waes constructiely interfere when speaker is located at 0.75 m and.00 m, but not in between. Assume the two speakers are in phase (helpful for isualization, but the result will be generally true as long as the two frequencies are the same). For constructie interference the path length difference must be an integer number of waelengths, 0.75 m nλ, and.00 m ( n + ) λ. Subtracting the two equations gies λ 0.5 m. Sole: 40 m/s f 60 Hz 400 Hz λ 0.5 m Assess: 400 Hz is near the middle of the range of human hearing, so it is probably right.