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Physics 2020 Spring 2009 Stephan LeBohec COVER SHEET - EXAM 2 SEAT # Constants: k= 1 9 2 −2 =8.99×10 N⋅m ⋅C 4 0 −19 e=1.602×10 Waves: =c⋅T = c 1 as f = f T In a gas c Sound = Doppler effect: f o= f s c String = Q Proton =e & Q Electron=−e C m F with = L 5 P with = for a monoatomic ideal gas 3 c±v o c∓v s Circular opening diffraction angle sin =1.22 / D Standing wave frequency (string or open pipe) c f n=n 2L Electric force and field: Force on a point charge in an electric field: =q E F Standing wave frequency in pipe open at one end c f n=n 4L Electric field generated by a point charge: ∣=k q ∣E r2 Force between two point charges: q q ∣=k 1 2 ∣F 2 r Triangle geometry: a (ABD) and (ECD), similar triangles AB DB AD = = EC DC ED c µ E b c 2=a 2b2 cos =b/ c sin =a /c tan =a/b A D C B Physics 2020 Spring 2009 Stephan LeBohec Name:_____________________________________ TA (circle one): Michael A) 1 EXAM 2 Sarah Student ID #:___________________________ Adam Isaac [36 points, 3 points per question] For each statement, circle the option you find appropriate. You do not need to show your work. 1) Two sound waves with the same frequency are propagating in the same direction in a pipe. There might be BEATS A STANDING WAVE INTERFERENCE 3) Two sound waves with different frequencies are propagating in opposite directions in a pipe. There might be BEATS A STANDING WAVE 4) Two sound waves with the same frequency are propagating in opposite directions in a pipe. There might be BEATS A STANDING WAVE 5) A wave described by y x , t = A sin 2 f t 2 x / propagates toward INCREASING X DECREASING X 6) Two harmonic waves with the same frequency propagate on a string in the same direction. One has an amplitude A1=2mm and the other A2=3mm . According to the superposition principle, the resulting wave may have a maximal amplitude of 1mm 2mm 7) and a minimal amplitude of 1mm 2mm 3mm 4mm 5mm 6mm 3mm 4mm 5mm 6mm 8) A taut string has a fundamental frequency 25Hz 50Hz 75Hz f 1 =25Hz . A standing wave with 4 anti-nodes has a frequency: 100Hz 125Hz 150Hz 150Hz ° 9) A loud speaker produces a f 1 =440Hz acoustic wave with a 25 diffraction angle. When the same speaker produces a f 1 =880Hz acoustic wave the diffraction angle is ° ° SMALLER THAN 25 LARGER THAN 25 10) A source moves toward and observer and the observer moves away from the source at the same speed. Compared to the source frequency, the observed frequency is SMALLER EQUAL GREATER 11) Two loud speakers are working in phase, producing waves with a wavelength of 1.44m. A listener is standing half way between the two speakers. In order to get to a destructive interference point, he needs to move toward one of the speakers by a distance of 0.18m 0.36m 0.72m 1.44m 2.88m 12) In order to double the fundamental frequency of a taut string, the tension must be multiplied by 1/ 2 1/ 2 2 2 4 Physics 2020 Spring 2009 Stephan LeBohec Name:_____________________________________ TA (circle one): Michael B) 2 EXAM 2 Sarah Adam Student ID #:___________________________ Isaac [15 points] A 0.80m long string under a 300N tension is observed to carry a standing wave with 5 antinodes and a 220Hz frequency . What is the mass of the string? Since there are 5 anti-nodes, the standing wave is the harmonic n=5 for which f n=n c with 2L L=0.8m . With T where m is the mass of the string and m/ L the linear mass density. In total we m/ L T n 2 n T⋅L have f n= in which only m is unknown. Solving for m we get m= . Numerically, L 2f 2L m 2 300N 5 −2 m= =4.84×10 kg=48.4g . With N ≡kg⋅m⋅s−2 we check that the units workout. 0.8m 2×220Hz T =300N we have c= C) [17 points] Two speakers are operating in phase at a frequency known to be between 100Hz and 160Hz . A point located 25.8m from one speaker and 42.5m from the other is observed to be the site of destructive interferences. The speed of sound is c=334m⋅s−1 What are the possible frequencies produced by the speakers? For the interference to be destructive, the difference ∣d 1−d 2∣=42.5m−25.8m=16.7m between the distance to the 1 so ∣d 1−d 2∣=n with 2 −1 1 c 1 c 1 334m⋅s 1 =n =n 20Hz . from which f = n n=0,1,2,3,⋯ or ∣d 1−d 2∣=n 2 f 2 ∣d 1−d 2∣ 2 16.7m 2 Staring with n=1 The frequencies are 30Hz, 50Hz, 70Hz, 90Hz, 110Hz, 130Hz, 150Hz, 170Hz, ⋯ . The only frequencies in the range from 100Hz to 160Hz are 110Hz, 130Hz and 150Hz. two speakers must be an multiple of the wavelength augmented by half Physics 2020 Spring 2009 Stephan LeBohec Name:_____________________________________ TA (circle one): Michael D) 1) 3 EXAM 3 Sarah Adam Student ID #:___________________________ Isaac [24 points] Two charges q 1=1.0 C , and q 2=−2.0 C are positioned as indicated on the figure with w=0.60m and h=0.80m . Calculate the magnitude and the direction (as specified by the angle between the electric field and the x direction) of the electric field at point M. Staring with E1 , 2 ∣E1∣=k q 1 /d 1M =9×10 9×1×10−6 /0.6 20.82 =0.9×104 N⋅C −1 w E 1x =∣ E1∣ =0.9×10 4×0.6 /1.0=0.54×10 4 N⋅C−1 d 1M h E 1y=−∣E1∣ =−0.9×10 4×0.8 /1.0=−0.72×104 N⋅C −1 d 1M E2 E 2x=−k q2 / w =−9×10 9×2×10−6 /0.62=−5.0×104 N⋅C −1 E 2y =0 E2 Θ E1y E1 Then dealing with 2 EM E1x EM = E1 E2 so E Mx =0.54×104 −5.0×104=−4.46×10 4 N⋅C−1 E My =−0.72×104 N⋅C −1 Then we have and ∣EM∣= E 2Mx E 2My = 0.72×10 4 2 4.46×104 2=4.52×104 N⋅C −1 and −1 −1 ° =−cos E Mx /∣EM∣=−cos −4.46/ 4.52=−170 2) [8 points] An electric charge q=−10 C is now placed at point M. What is the magnitude and direction of the electric force acting on this charge? points ∣=∣q∣⋅∣E M∣=10×10−6 ×4.52×104 =0.452N . Since q0 , F ∣F ° ° ° making an angle ' =180=180 −170 =10 with the x direction. The magnitude of the electric force is in a direction opposite to EM