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- Iowa State University
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- Physics 221
- Prell
- Wr 12 solution.pdf

Jeff N.

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Phys 221 – Spring 2009 Written Homework – Set 12 (Due Tuesday April 14 2009) 1. A wire is made out of a special steel alloy with 11 2.00 10 PaY =× , density 3 kg m 7800 , breaking stress 9 1.00 10 Pa× and coefficient of linear thermal expansion 51 1.20 10 Kα −− =× . The initial temperature of the wire is 50°F and the wire is measured to be exactly 5.000 m long. The wire is then secured at one end to the ceiling of a room and hangs vertically. A 10 kg mass is then attached to the end of the wire and it elongates by 1cm. (a) If the room is warmed up to 50°C, how much is the wire then elongated? (b) Suppose that the lower end of the wire is now securely to a bolt holding to the length calculated in part (a). What is the tension of the wire? (c) If you pluck the wire at the center, what will be the dominant frequency of the sound produced (i.e. the fundamental frequency)? (d) The room is now cooled to 32°F with the end still secured. What is the tension in the wire? 2. (a) In which of these scenarios to you hear a higher pitched sound? Explain. (i) A police car with siren sounding is stationary and you approach it with a speed v 0 . (ii) You are stationary and the same police car with a siren sounding approaches you at speed v 0 . (b) What is the difference in pitch between the above two scenarios if mi 0 hr 90v = , the frequency of the police siren is 200 Hz and the air temperature is 20°C? (c) What is the difference in pitch in (b) if (hypothetically) v 0 is half the speed of sound? 3. Two identical violin strings are tuned to a frequency of 440 Hz. (a) If the tension of one of the strings is increases by 1% and the two strings are plucked together, what beat frequency will be heard? (b) If, instead, a musician places his finger on one of the strings in such a way as to reduce its length by 1%, what beat frequency will be heard if the two strings are plucked together? Phys 221 – Spring 2009 Written Homework – Set 12 (Due Tuesday April 14 2009) 4. The figure below shows two speakers at points A and B. Both of these speakers radiate sound uniformly in all directions in air at 20°C. The speakers are in phase and the acoustic power output from each of the speakers is 4 1.00 10 W − × . The wavelength of the sound is 4mλ = . Find the ratio between the pressure amplitudes in the following cases: (a) For P and Q, if only speaker A is turned on .Ie, (amplitude pressure at P):(amplitude pressure at Q). (b) For P and Q, if only speaker B is turned on. (c) For P when both speakers A and B are on versus when only speaker A is on (i.e. max max (P from A&B) : (P from A)pp). (d) Same as (c) for point Q. 1 m 1 m ● B ● A ● P ● Q Phys 221 – Spring 2009 Written Homework – Set 12 (Due Tuesday April 14 2009) Solutions 1. (a) The initial temperature is 50°F=10°C and the final temperature is 50°C so 40 KTΔ= . The elongation is therefore, 51 (5.01m)(1.20 10 K )(40K) 2.4 mm LL Tα −− Δ= Δ= × = So the further elongation due to the temperature increase is 2.4mm. (b) The tension in the wire depends will be the same as the force due to the weight hanging on the end since the bolt must pull with the same force to stretch the wire to this new length. Thus the tension is F=(10kg)(9.8 2 m s )=98N. (c) The fundamental frequency for a string fixed at both ends is given by eqn. 15.35: 1 1 2 F f L μ = The tension is given by part (b); to get the linear mass density, we need to use the definition of Young’s modulus and the initial elongation. 11 72 (98N)(5m) (.01m)(2.00 10 Pa) 2.5 10 m FL A LY − == Δ× =× so the linear mass density is 3 kg72 m kg3 m (2.5 10 m )(7800 ) 1.9 10 ADμ − − ==× =× Plugging this in 1 kg 3 m 1198N 22(5m)1.910 23 Hz F f L μ − == × = (d) The temperature is 32°F=0°C so in comparison to the initial room temperature of 10°C, the change in temperature is 10 KTΔ =− . If the wire had no tension on it, the change in the length compared to the initial length of 5.000m would be: 5 (5m)(1.20 10 K)( 10K) 0.6mm thermal LLTα − Δ=Δ= × − =− The final length, however, deviates by a total of 12.4mmLΔ = from the initial, 10°C length of 5.000m. The total elongation is given by: Phys 221 – Spring 2009 Written Homework – Set 12 (Due Tuesday April 14 2009) 12.4mm ( 0.6mm)=13mm thermal tension tension thermal LL L LLL Δ=Δ +Δ ∴Δ=Δ−Δ =−− Again, using the definition of Young’s modulus, 11 7 2 (2.00 10 Pa)(2.5 10 m )(0.013m) 5m 130 N YA L F L − Δ× × == = 2. (a) It is important to realize that there is a distinction between these two scenarios since sound waves travel at a constant speed through air, not relative to the source of the sound. There will therefore be a difference between these two cases. In particular, in scenario (i) where the police car is stationary, the wavelength of the sound is the same regardless of how you move, the Doppler effect comes about because the relative velocity of the sound with respect to your moving car is v+v 0 rather than just v. Thus, in this case, the frequency is fractionally increases by v 0 /v. In contrast in scenario (ii) where the police car is moving and you are stationary, the sound waves are passing you at speed v but their wavelength has been fractionally shortened by the same amount, v 0 /v. Since wavelength is inversely related to frequency, this means that the frequency in (ii) will always be higher for a non-zero v 0 (positive or negative in fact). The easiest way to see this is to draw a graph of / L f f in the two cases: v 0 /v f L /f S Scenario ii Scenario i Phys 221 – Spring 2009 Written Homework – Set 12 (Due Tuesday April 14 2009) As can be seen from the graph, in scenario (i) the frequency ratio is linear in the velocity ratio while in scenario (ii) the curve is a hyperbola which always curves upwards where the scenario (i) line is tangent at the point v 0 =0. The frequency of scenario (ii) thus always exceeds the frequency of (i). We can obtain the same result in terms of the formalism of section 16.8 of the text, the heard frequency in case (i) is 0 () 1 L iS v f f v ⎛⎞ =+ ⎜⎟ ⎝⎠ while case (ii) is () 0 1 S Lii f f v v = ⎛⎞ − ⎜⎟ ⎝⎠ This follows from equation 16.29 of the text. Note that in case (ii) the convention of the text is that v S <0. To make clear the ordering of the two cases, let us Taylor expand ()L ii f in terms of v 0 /v (which is small in this case): 2 00 () 0 1 ... 1 S Lii S fvv f f v vv v ⎛⎞ ⎛⎞ ==+++ ⎜⎟ ⎜⎟ ⎛⎞ ⎝⎠ ⎝⎠ − ⎜⎟ ⎝⎠ so () ()L ii L i f f> . (b) Doing the numbers for the given case, mi m 0 hr s 90 40v = = and the speed of sound is m s 344v = therefore 0 0.116 v v = . Thus 0 () 1 1.116(440 Hz) 491 Hz Li S v ff v ⎛⎞ =+ = = ⎜⎟ ⎝⎠ () 0 440 Hz 498 Hz 0.884 1 S Lii f f v v === ⎛⎞ − ⎜⎟ ⎝⎠ So the pitch of scenario (ii) is indeed higher. (c) In this case, 0 () 1 1.5(440 Hz) 660 Hz Li S v ff v ⎛⎞ =+ = = ⎜⎟ ⎝⎠ () 0 440 Hz 880 Hz 0.5 1 S Lii f f v v === ⎛⎞ − ⎜⎟ ⎝⎠ Phys 221 – Spring 2009 Written Homework – Set 12 (Due Tuesday April 14 2009) 3. (a) The fundamental harmonic of a string, fixed at both ends, is 1 1 2 F f L μ = The dependence on the tension is 1 2 1 f F∝ . Thus increasing the tension by 1% will increase this frequency by a factor of 1 2 (1 0.01)+ . The frequency of the tightened string is thus: (440 Hz) 1.01 442.2 Hz tight f = = The beat frequency is therefore 1 2.2 Hz beat tight fff=− = Note that we could also reason that the ½ power dependence on the tension implies that a 1% tightening will give roughly a 0.5% increase in frequency so the beat frequency will be ~0.5% of f 1 , i.e. 2.2 Hz. (b) The dependence of the fundamental on the length of the string is 1 1 f L − ∝ . The shortening of the string by 1% will therefore change the tone to 1 (440 Hz)(0.99) 444.4 Hz short f − = = The beat frequency is therefore 1 4.4 Hz beat short fff=− = Again it is expected because of the -1 power dependence that a 1% decrease in length will raise the pitch by about 1% so the beat frequency will be ~1% of f 1 , i.e. 4.4 Hz. Phys 221 – Spring 2009 Written Homework – Set 12 (Due Tuesday April 14 2009) 4 (a) The intensity of sound, I, from a source varies with distance from a single source like 2 1 I r ∝ while 2 max I p∝ therefore the variation of pressure amplitude with distance from a single source is max 1 pI r ∝∝ In this case then, max max 11 (P from A): (Q from A) : : QA PA PA QA p prr rr == From the geometry of the diagram, 5m and 3m PA QA rr==so max max (P from A): (Q from A) : 3: 5 1.34:1 1:0.75 QA PA p prr= == = (b) In this case, 5m and 5m PB QB rr==so, using the same logic as part (a) max max (P from B): (Q from B) : 5: 5 5 :1 2.24:1 1:0.45 QB PB p prr= === = (c) Because the distance from each of the speakers is the same and the speakers are in phase, there will be no phase difference between the sound waves from each of the speakers alone. Furthermore the speakers have the same power so the pressure amplitude from each sound alone is the same. By superposition, the pressure amplitude at P with both speakers will be twice that with one speaker. Thus, max max (P from A&B) : (P from A)pp=2:1. (d) The distance from Q to A is 3m while the distance from Q to B is 5m. The difference between these two distances is 2m, which is half a wavelength. The phase difference between the sound from B and the sound from A will thus be 180° (π radians) so when the pressure amplitudes are superimposed, they will subtract. From the discussion in part (a) we see that: 3 max max5 (Q from B) (Q from A)pp= Therefore max max max 3 max max5 2 max5 (Q from A&B) (Q from A) (Q from B) (Q from A) (Q from A) (Q from A) p pp p =− =− = Therefore 2 max max 5 (Q from A&B) : (Q from A) :1 2:5 pp= = Paula Microsoft Word - Written_12_221_2009_Sp

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