Week 12 homework IMPORTANT NOTE ABOUT WEBASSIGN: In the WebAssign versions of these problems, various details have been changed, so that the answers will come out differently. The method to find the solution is the same, but you will need to repeat part of the calculation to find out what your answer should have been. WebAssign Problem 1: A longitudinal wave with a frequency of 3.0 Hz takes 1.7 s to travel the length of a 2.5-m Slinky (see Figure 16.3). Determine the wavelength of the wave. REASONING AND SOLUTION From Equation 16.1, we have λ = v/f. But v = x/t, so we find v x 2.5 m λ = = = = 0.49 m f t f ( 1.7 s ) ( 3.0 Hz ) WebAssign Problem 2: Tsunamis are fast-moving waves often generated by underwater earthquakes. In the deep ocean their amplitude is barely noticeable, but upon reaching shore, they can rise up to the astonishing height of a six-story building. One tsunami, generated off the Aleutian islands in Alaska, had a wavelength of 750 km and traveled a distance of 3700 km in 5.3 h. (a) What was the speed (in m/s) of the wave? For reference, the speed of a 747 jetliner is about 250 m/s. Find the wave’s (b) frequency and (c) period. REASONING The speed of a Tsunamis is equal to the distance x it travels divided by the time t it takes for the wave to travel that distance. The frequency f of the wave is equal to its speed divided by the wavelength λ, f = v/λ (Equation 16.1). The period T of the wave is related to its frequency by Equation 10.5, T = 1/f. SOLUTION a. The speed of the wave is (in m/s) v= b. x 3700 × 103 m  1 h  =   = 190 m/s t 5.3 h  3600 s  The frequency of the wave is f = c. v 190 m/s = = 2.5 × 10− 4 Hz λ 750 × 103 m The period of any wave is the reciprocal of its frequency: (16.1) T= 1 1 = = 4.0 × 103 s − 4 f 2.5 × 10 Hz (10.5) WebAssign Problem 3: A steel cable of cross-sectional area is kept under a tension of . The density of steel is 7860 kg/m3 (this is not the linear density). At what speed does a transverse wave move along the cable? REASONING AND SOLUTION The speed of the wave on the cable is v= F = (m / L) F = ρA 1.00 × 104 N = 21.2 m/s (7860 kg/m3 )(2.83 × 10− 3 m 2 ) WebAssign Problem 4: The drawing shows two graphs that represent a transverse wave on a string. The wave is moving in the + x direction. Using the information contained in these graphs, write the mathematical expression (similar to Equation 16.3 or 16.4) for the wave. REASONING AND SOLUTION We find from the graph on the left that λ = 0.060 m – 0.020 m = 0.040 m and Α = 0.010 m. From the graph on the right we find that Τ = 0.30 s – 0.10 s = 0.20 s. Then, f = 1/(0.20 s) = 5.0 Hz. Substituting these into Equation 16.3 we get 2π x   y = A sin  2π f t –  λ   and y = ( 0.010 m ) sin ( 10π t – 50π x ) WebAssign Problem 5: A monatomic ideal gas (γ = 1.67) is contained within a box whose volume is 2.5 m3. The pressure of the gas is . The total mass of the gas is 2.3 kg. Find the speed of sound in the gas. REASONING AND SOLUTION The speed of sound in an ideal gas is given by text Equation 16.5: γ kT v= m (16.2) where γ = cp/cv, k is Boltzmann's constant, T is the Kelvin temperature of the gas, and m is the mass of a single gas molecule. If mTOTAL is the mass of the gas sample and N is the total number of molecules in the sample, then the above equation can be written as v= γ kT (mTOTAL / N ) = γ N kT mTOTAL For an ideal gas, PV = NkT, so that Equation (1) becomes, v= γ PV (1.67)(3.5×105 Pa)(2.5 m3 ) = = 8.0 ×102 m/s mTOTAL 2.3 kg WebAssign Problem 6: Suppose that sound is emitted uniformly in all directions by a public address system. The intensity at a location 22 m away from the sound source is . What is the intensity at a spot that is 78 m away? REASONING AND SOLUTION The intensity of the sound falls off according to the square of the distance according to Equation 16.9. Therefore, the intensity at the new location will be I 78  22 m  =    78 m  2 ( 3.0 × 10− 4 W/m2 ) = 2.4 × 10 –5 W/m 2 WebAssign Problem 7: You are riding your bicycle directly away from a stationary source of sound and hear a frequency that is 1.0% lower than the emitted frequency. The speed of sound is 343 m/s. What is your speed? REASONING You hear a frequency fo that is 1.0% lower than the frequency fs emitted by the source. This means that the frequency you observe is 99.0% of the emitted frequency, so that fo = 0.990 fs. You are an observer who is moving away from a stationary source of sound. Therefore, the Doppler-shifted frequency that you observe is specified by Equation 16.14, which can be solved for the bicycle speed vo. SOLUTION Equation 16.14, in which v denotes the speed of sound, states that v   f o = fs  1 − o  v   Solving for vo and using the fact that fo = 0.990 fs reveal that (1)   f  0.990 fs  vo = v  1 − o  = ( 343 m/s )  1 −  = 3.4 m/s    f f s  s    WebAssign Problem 8: Interactive LearningWare 16.2 at www.wiley.com/college/cutnell provides some pertinent background for this problem. A convertible moves toward you and then passes you; all the while, its loudspeakers are producing a sound. The speed of the car is a constant 9.00 m/s, and the speed of sound is 343 m/s. What is the ratio of the frequency you hear while the car is approaching to the frequency you hear while the car is moving away? REASONING This problem deals with the Doppler effect in a situation where the source of the sound is moving and the observer is stationary. Thus, the observed frequency is given by Equation 16.11 when the car is approaching the observer and Equation 16.12 when the car is moving away from the observer. These equations relate the frequency fo heard by the observer to the frequency fs emitted by the source, the speed vs of the source, and the speed v of sound. They can be used directly to calculate the desired ratio of the observed frequencies. We note that no information is given about the frequency emitted by the source. We will see, however, that none is needed, since fs will be eliminated algebraically from the solution. SOLUTION Equations 16.11 and 16.12 are  1  f oApproach = fs   1 − v / v  s   (16.11)  1  f oRecede = fs   1 + v / v  s   (16.12) The ratio is f oApproach f oRecede  fs   1− =  fs   1+ 1   vs / v  9.00 m/s 343 m/s = 1.054 = = 9.00 m/s 1  1 − vs / v 1 −  343 m/s vs / v  1 + vs / v 1+ As mentioned in the REASONING, the unknown source frequency fs has been eliminated algebraically from this calculation. WebAssign Problem 9: A 3.49-rad/s ( rpm) record has a 5.00-kHz tone cut in the groove. If the groove is located 0.100 m from the center of the record (see drawing), what is the wavelength in the groove? REASONING AND SOLUTION First find the speed of the record at a distance of 0.100 m from the center: (8.9) v = rω = (0.100 m)(3.49 rad/s) = 0.349 m/s The wavelength is, then, –5 λ = v/f = (0.349 m/s)/(5.00 × 103 Hz) = 6.98 × 10 m (16.1) Practice conceptual problems: 6. A rope of mass m is hanging down from the ceiling. Nothing is attached to the loose end of the rope. A transverse wave is traveling on the rope. As the wave travels up the rope, does the speed of the wave increase, decrease, or remain the same? Give a reason for your choice. REASONING AND SOLUTION A rope of mass m is hanging down from the ceiling. Nothing is attached to the loose end of the rope. A transverse wave is traveling up the rope. The tension in the rope is not constant. The lower portion of the rope pulls down on the higher portions of the rope. If we imagine that the rope is divided into small segments, we see that the segments near the top of the rope are being pulled down by more weight than the segments near the bottom. Therefore, the tension in the rope increases as we move up the rope. The speed of a transverse wave on the rope is given by Equation 16.2: vwave = F /( m / L) . From Equation 16.2 we see that, as the tension F in the rope increases, the speed of the wave increases. Therefore, as the transverse wave travels up the rope, the speed of the wave increases. 13. Some animals rely on an acute sense of hearing for survival, and the visible part of the ear on such animals is often relatively large. Explain how this anatomical feature helps to increase the sensitivity of the animal’s hearing for low-intensity sounds. REASONING AND SOLUTION Animals that rely on an acute sense of hearing for survival often have relatively large external ear parts. Sound intensity is defined as the sound power P that passes perpendicularly through a surface divided by the area A of that surface: I = P / A . For low intensity sounds, the power per unit area is small. Relatively large outer ears have a greater area than smaller outer ears. Hence, large outer ears intercept and direct more sound power into the auditory system than smaller outer ears do. 17. A source of sound produces the same frequency underwater as it does in air. This source has the same velocity in air as it does underwater. The observer of the sound is stationary, both in air and underwater. Is the Doppler effect greater in air or underwater when the source (a) approaches and (b) moves away from the observer? Explain. REASONING AND SOLUTION According to Table 16.1, the speed of sound in air at 20 °C is v = 343 m/s, while its value in water at the same temperature is v = 1482 m/s. These values influence the Doppler effect, because of which an observer hears a frequency fo that is different from the frequency fs that is emitted by the source of sound. For purposes of this question, we assume that fs = 1000 Hz and that the speed at which the source moves is vs = 25 m/s. Our conclusions, however, will be valid for any values of fs and vs . a. The Doppler-shifted frequency when the source approaches the observer is given by Equation 16.11 as f o = fs  1 − (vs / v)  . Applying this equation for air and water, we find Air fo = 1000 Hz = 1079 Hz 1 − (25 m s )/(343 m s ) Water fo = 1000 Hz = 1017 Hz 1 − (25 m s )/(1482 m s ) The change in frequency due to the Doppler effect in air is greater. b. The Doppler-shifted frequency when the source moves away from the observer is given by Equation 16.12 as f o = fs  1 + (vs / v)  . Applying this equation for air and water, we find 1000 Hz fo = = 932 Hz Air 1 + (25 m s )/(343 m s ) Water fo = 1000 Hz = 983 Hz 1 + (25 m s )/(1482 m s ) The change in frequency due to the Doppler effect in air is greater.