Prashant Patil (99709774) PRASHANT PATIL PHYSICS CLASSES NEET/JEE(Main) Date : 9/07/07 TEST ID: Time : 00:45:00 PHYSICS Marks : 80 5. STATIONARY WAVES Single Correct Answer Type. Stationary waes are set up in air column. Velocity of sound in air is 330 m/s and frequency is 65 Hz. Then distance between the nodes is a) m b) m c) 0.5 m d) 4 m. An open pipe is in resonance in nd harmonic with frequency Now one end of the tube is closed and frequency is increased to such that the resonance again occurs in nth harmonic. Choose the correct option. a) n = 3, = 3 4 _ b) n = 3, 5 4 c) n = 5, = 5 4 d) n = 5, = 3 4 3. A tuning fork of frequency 330 Hz resonates with an air column of length 0 cm in a cylindrical tube, in the fundamental mode. When water is slowly poured in it, the minimum height of water required for obsering resonance once again is (elocity of sound 330ms ) a) 75 cm b) 60 cm c) 50 cm d) 45 cm 4. In the fundamental mode, time taken by the wae to reach the closed end of the air filled pipe is 0.0s. the fundamental frequency is a) 5 b).5 c) 0 d) 5 5. A closed Prgan pipe and an open organ pipe of same length produce beats/second while ibrating in their fundamental modes. The length of the open organ pipe is haled and that of closed pipe is doubled. Then the number of beats produced per second while ibrating in the fundamental mode is a) b) 6 c) 8 d) 7 6. A piston fitted in cylindrical pipe is pulled as shown in the figure. A tuning fork is sounded at open end and loudest sound is heard at open length 3cm, 4 cm and 69 cm, the frequency of tuning fork if elocity of sound is 350ms is a) 50 Hz b) 65 Hz c) 47 Hz d) 75 Hz 7. An organ pipe, open at both ends produces 5 beats/s when ibrates with a source of frequency 00 Hz. The second of the same pipe produces 0 beats/s with a source of frequency 40 Hz. The frequency of source is a) 95 Hz b) 05 Hz c) 90 Hz d) 0 Hz 8. A string ibrates according to the equation y = 5 sin ( πx ) cos 0πt where x and y are in cm 3 and t in second. The distance between two adjacent nodes is a) 3 cm b) 4.5 cm c) 6 cm d).5 cm 9. A ibrating string of certain length I under a tension T resonates with a mode corresponding to the second oertone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generate 4 beats/s when excited along with a tuning fork of frequency n. now when the tension of the string also generate 4 beats/s when excited along with a tuning fork of frequency n. now when the tension of the string is slightly increased the number of beats reduces per second. Assuming the elocity of sound in air to be 340ms, the frequency n of the tuning fork in Hz is a) 344 b) 336 c) 7.3 d) 09.3 0. If man were standing unsymmetrical between parallel cliffs, claps his hands and starts hearing a series of echoes at a interals of s. If speed of sound in air is 340ms, the distance between two cliffs would be a) 340m b) 50m c) 70m d) 680m. Two instruments haing stretched strings are being played in union. When the tension of one of the instruments is increased by %, 3 beats P a g e
are produced in s. the initial frequency of ibration of each wire is a) 300 Hz b) 500 Hz c) 000 Hz d) 400 Hz. The equation of a stationary wae along a stretched string is gien by y = 4 sin πx cos40πt where x and y are in cms and t is in sec. The separation between two adjacent nodes is a) 3 cm b).5 cm c) 6 cm d) 4 cm 3. Distance between nodes on a string is 5 cm. elocity of transerse wae isms. Then the frequency is a) 5 Hz b) 0 Hz c) 0 Hz d) 5 Hz 4. If is the speed of sound in air then the shortest length of the closed pipe which resonates to a frequency, is a) b) c) 4 d) 4 5. In meld s experiment in the transerse mode, the frequency of the tuning fork and the frequency of the waes in the string are in the ratio a) : b) 4: c) : d) : 6. A hollow cylinder with both sides open generates a frequency f in air. When the cylinder ertically immersed into water by half its length the frequency will be a) f b) f c) f/ d) f/4 7. In Melde s experiment, the string ibrates in 4 loops when a 50 g weight is placed in the pan of weight 5 g. To made the string ibrate in 6 loops, the weight that has to be remoed from the pan in approximately a) 7 g b) 36 g c) g d) 9 g 8. Fundamental frequency of pipe is 00 Hz and other two frequencies are 300 Hz and 500 Hz, then a) Pipe is open at both the ends b) Pipe is closed at both the ends c) One end is open and another end is closed d) None of the aboe 9. An organ pipe open at one end is ibrating in first oertone and is in resonance with another pipe open at both ends and ibrating in third harmonic. The ratio of length of two pipe is a) 3:8 b) 8:3 c) : d) 4: 0. If,, 3 are the waelengths of the waes giing resonance with the fundamental, first and second oertones respectiely of a closed organ pipe, then the ratio of,, 3 is a) :3:5 b) ::3 c) 5:3: d) : 3 : 5. In an open organ pipe wae is present. a) Transerse standing b) Longitudinal standing c) Longitudinal moing d) Transerse moing. The stationary wae y = a sin kx cos ωt in a closed organ pipe is the result of the superposition of y = a sin(ωt kx) and a) y = a cos(ωt + kx) b) y = a sin(ωt + kx) c) y = a sin(ωt + kx) d) y = a cos(ωt + kx) 3. What is minimum length of a tube, open at both ends, that resonates with tuning fork of frequency 350 Hz? [elocity of sound in air = 350 m/s] a) 50 cm b) 00 cm c) 75 cm d) 5 cm 4. A column of air of length 50 cm resonates with a stretched string of length 40 cm. The length of the same air column which will resonates with 60 cm of the same string at a the same tension is a) 00 cm b) 75 cm c) 50 cm d) 5 cm 5. Find the fundamental frequency of a closed pipe, if the length of the air column is 4 m. (speed of sound in air = 33 m/sec) a) Hz b) 4 Hz c) 7 Hz d) 9 Hz 6. A stretched string of length l fixes at both ends can sustain stationary waes of waelength, gien by a) =ln b) = ι n c) = ι d) = n n ι 7. Two open organ pipes gies 4 beats/sec when sounded together in their fundamental nodes. If the length of the pipe are 00 cm and 0.5 cm respectiely, then the elocity of sound is : a) 496 m/s b) 38 m/s c) 40 m/s d) 60 m/s 8. In a resonance tube, using a tuning fork of frequency 35 Hz, two successie resonance lengths are obsered as 5.4 cm and 77.4 cm respectiely. The elocity of sound in air is a) 338 ms b) 38 ms c) 330 ms d) 30 ms 9. A glass tube of length.0 m is completely filled with water. A ibrating tuning fork of frequency 500 Hz is kept oer the mouth of the tube and water is drained out slowly at the P a g e
bottom of tube. If elocity of sound in air is 330 ms, then the total number of resonance that occur will be a) b) 3 c) d) 5 30. Two closed pipes produce 0 beats per second when emitting their fundamental nodes. If their lengths are in ratio of 5 : 6. Then their fundamental frequency in Hz, are a) 70, 80 b) 60, 70 c) 60, 50 d) 60, 80 3. A hollow cylinder with both sides open generates a frequency in air. When the cylinder ertically immersed into water by half its length the frequency will be a) V b) c) / d) /4 3. A wire under tension ibrates with a fundamental frequency of 600 Hz. If the length of the wire is doubled, the radius is haled and the wire is made to ibrate under one-ninth the tension. Then the fundamental frequency will become a) 400 Hz b) 600 Hz c) 300 Hz d) 00 Hz 33. In m long open pipe what is the harmonic of resonance obtain with a tuning fork of frequency 480 Hz? a) First b) Second c) Third d) Fourth 34. The harmonic which are present in a pipe open at one end are a) Odd harmonics b) Een harmonics c) Een as well as odd harmonics d) None of these 35. If L and L are the lengths of the first and second resonating air columns in a resonance tube, then the waelength of the note produced is a) (L + L ) b) (L L ) c) (L L ) d) (L + L ) 36. A 5.5 m length of string has a mass of 0.035 kg. If the tension in the string is 77 N, the speed of a wae on the string is a) 0ms b) 65m c) 77ms d) 0ms 37. n Is the frequency of the pipe closed at one end and n is the frequency of the pipe open at both ends. If both are joined end to end, find the fundamental frequency of closed pipe so formed n n n n a) b) n + n n + n c) n + n d) n + n n n n n 38. Two wires made up of the same material are of equal length but their radii are in the ratio of :. On stretching each of these two strings by the same tension, the ratio between the fundamental frequencies is a) :4 b) 4: c) : d) : 39. An open tube is in resonance with string (frequency of ibration of tube is n 0 ). If tube is dipped in water so that 75% of length of tube is inside water, then the ratio of the frequency of tube to string now will be a) b) c) d) 3 3 40. The extension in a string obeying Hook s law is x. the speed of sound in the stretched string is. if the extension in the string is increased to.5x, the speed of sound will be a). b) 0.6 c).50 d) 0.75 4. An air column in a pipe, which is closed at one end, will be in resonance with a ibrating body of frequency 66 Hz, if the length of the air column is a).00 m b).50 m c).00 m d) 0.50 m 4. A wae is represented by the equation y=a cos (kx-ωt) is superposed with another wae to form a stationary wae such that the point x=0 is a node. The equation of the other wae is a) a sin (kx+ωt) b) - a sin (kx-ωt) c) -a cos (kx+ωt) d) a cos (kx+ωt) 43. An open organ pipe has fundamental frequency 00 Hz. What frequency will be produced if its one end is closed? a) 00,00,300 b) 50,50,50 c) 50,00, 00,300. d) 50,00,50,00 44. If the temperature increases, then what happens to the frequency of the sound produced by the organ pipe a) Increases b) Decreases c) Unchanged d) Not definite 45. An open tube is in resonance with string. If tube is dipped in water, so that 75% of length of tube is inside water, then ratio of the frequency ( 0 ) of tube to string is a) 0 b) 0 c) 3 0 d) 3 0 P a g e 3
Prashant Patil (99709774) PRASHANT PATIL PHYSICS CLASSES NEET/JEE(Main) Date : 9/07/07 TEST ID: Time : 00:45:00 PHYSICS Marks : 80 5. STATIONARY WAVES ) b ) c 3) d 4) a 5) d 6) b 7) b 8) d 9) a 0) b ) a ) b 3) c 4) b 5) c 6) a 7) b 8) c 9) c 0) d ) c ) b 3) a 4) b : ANSWER KEY : 5) a 6) b 7) b 8) a 9) b 30) c 3) a 3) d 33) c 34) a 35) b 36) a 37) a 38) c 39) b 40) a 4) d 4) c 43) b 44) a 45) b : HINTS AND SOLUTIONS : Single Correct Answer Type (b) = 330 m/s; n = 65 Hz. Distance between two successie nodes = = n = 330 65 = m (c) = l ( nd harmonic of open pipe) Here, n is odd and > It is possible when n=5 Because with n=5 = 5 4 ( l ) > 4 (a) In the fundamental mode. frequency n = 4ι n = n = t 4ι 0.0 4 n = ( = t ) n = 5 5 (d) Gien, f 0 f c = (i) Frequency of fundamental mode for a closed organ pipe, f c = 4L c Similarly frequency of fundamental mode an open orgen pipe, f 0 = L 0 Gien L c = L 0 f 0 = f c (ii) From Eqs. (i) and (ii), we get f 0 = 4Hz And f c = Hz When the length of the open pipe is haled, its frequency of fundamental mode is f 0 = [ L 0 ] = f 0 = 4Hz = 8Hz When the length of the closed pipe is doubled, its frequency of fundamental mode is f 0 = 4(L c ) = f c = = Hz Hence, number of beats produced per second is f 0 = f = 8 = 7 6 (b) P a g e 4
In a closed organ pipe in which length of aircolumn can be increased or decreased, the first resonance occurs at /4 and second resonance occurs at 3/4. Thus, at first resonance = 3 (i) 4 And a second resonance 3 = 4 (ii) 4 8 (d) The nodes and antinodes are formed in a standing wae pattern as a result of the interface of two waes. Distance between two nodes is half waelength () Subtracting Eq.(i) from Eq.(ii), we hae 3 4 = 4 3 4 = 8 = 56 cm Hence, frequency of tuning fork = = 350 = 365 Hz 56 0 7 (b) Since, an open pipe produces both een and odd harmonics, hence frequency of pipe = 00 ± 5 =95 Hz or 05 Hz. Frequency of second harmonic of pipe =. Now, the number of beats=0 = 40 ± 0 = 40 Hz or 430 Gz =05 Hz or 5Hz. Standerd equation of standing wae is y = a sin πx πt cos γ γ Where a is amplitude, the wawlength 9 (a) With reflection in tension, frequency of ibrating string will increase. Since number of beats are decreasing. Therefore, frequency of ibrating string or third harmonic frequency of closed pipe should be less than the frequency of tuning fork by 4. frequency of tuning fork = Third harmonic frequency of closed pipe+4 = 3 ( 4ι ) + 4 = 3 ( 340 ) + 4 = 344 Hz 4 0.75 0 (b) Let x be distance of person from one cliff and y be distance of person from nd cliff. Let y > x. x + x = t = 340 = 340 x = 70 m P a g e 5
y + y = t = 340 = 680 y = 340 m. frequency is the same as that of the fork. Hence, the required ratio is :. Distance between two cliffs = x + y = 70 + 340 = 50m (a) Frequency 6 (a) = l ( T m ) + 3 = l 0 T ( 00 m ) =.005 l ( T m ) +.5=.005 =300 Hz (b) Compare the gien equation with the standard form of stationary wae equation y = r sin πx We get πx πt cos, = πt 3 = 3cm Separation between two adjacent nodes = =.5cm 3 (c) Distance between modes = 5 cm =0 cm=0.n Frequency = = 0. = 0Hz 4 (b) In close organ pipe = 4l So, l = 4 5 (c) In transerse arrangement the tuning fork is placed such that the ibration of the prongs is in direction perpendicular to the length of the string as shown in figure. As the tuning fork completes one ibration, the one ibration of wae on string is completed. Thus, in transerse mode, its (i) = l = l = l 7 (b) (ii) = l 4 = 4l = l =, the same frequency l As p T = constant T T = 6 36 T = 6 65 = 9 36 = p T p = 4 6 Weight to be remoed = 65 9 = 36 g 8 (c) For closed organ pipe n, : n : n 3. = : 3: 5:. 9 (c) For an organ pipe open at one end, 0 (d) Frequency of st oertone n = 3 4l For the organ pipe open at both ends, Frequency of 3rd harmonic,n = 3 l As n = n 3 4l = 3 l or l l = 4 = P a g e 6
As is clear from figure l = 4, = 4l l = 3 4, = 4l 3 l = 5 3 4, 3 = 4l 5 : : 3 = : 3 : 5 (b) In closed organ pipe. If y incident = a sin(ωt kx) then y reflected = a sin(ωt + kx + π) = a sin(ωt + kx) Superimposition of these two waes gie the required stationary wae 3 (a) Fundamental frequency n = l 350 = 350 l = m = 50cm l 4 (b) As l = l l l 60 = l, 40 50 l = 50 60 40 = 75 cm 8 = [ 0.975] = 8 38 m/s 0.05 8 (a) Velocity of sound in air = n(l l ) = 35(77.4 5.4) = 650/5 00 =338 m/s 9 (b) Frequency n = 4l or l = 4 l = = 300 4 4 500 = 0.65m; l = 3 4 = 3ι = 0.495 m l 3 = 5 4 = 5ι = 0.85m and l 4 = 7 = 7l 4 =.55 > m Therefore, number of resonance =3 30 (c) n n = 0 (i) Using n = and n 4l = 4l n = l = 6 (ii) n l 5 After soling these equation n = 60 Hz, n = 50 Hz 3 (a) (i) Here, = l = l So, = l (ii)and 4 = l = 4 = l =, the same l (original)is the frequency, 5 (a) For closed pipe n = 4l 6 (b) As we know that n l = I or = n 7 (b) n = 33 4 4 = Hz n = n n 4 = l l = [.00.05 ] 3 (d) fundamental frequency f = rι T πρ P a g e 7
f f = ι ι r r 600 f = T T/9 f = 00Hz 33 (c) For an open pipe of length l, the frequency is gien by =. l Where is elocity of sound, the oertone. Gien, =450 Hz, l=m, = 330 ms = (l) = 480 =.3 3 330 37 (a) Frequency of closed pipe n = ι 4ι = 4n Frequency of open pipe, n = ι ι = n When both pipes are joined then length of closed pipe ι = ι + ι 4n = + 4n n Or n = + n n Or n = n + n n n Or n = n n n + n 38 (c) The fundamental frequency of a wire is gien by = l T m Hence, this is the second oertone or third harmonic. 34 (a) In closed pipe only odd harmonics are present 35 (b) if L and L are the first and second resonances, then we hae L + e = 4 and L + e = 3 4 L L = = (L L ) 36 (a) m = 0.035 5.5 kg m, T = 77N = T/m = 77 5.5 0.035 = 0ms Where lis length of wire, T the tension and m the mass per unit length. mass of wire m = length of wire = πr L density = πr d L = l T πr d = rl T πd = r = r 39 (b) For open tube, n 0 = l For closed tube length aailable for resonance is l = l 5 = l Fundamental frequency of 00 4 water filled tube n = 4l = 4 (l/4) = l = n 0 n = n 0 40 (a) Speed of sound in a stretched string = T μ (i) P a g e 8
Where T is the tension and μ is mass per unit length. According to Hooke s law, F x T x (ii) From Eqs. (i) and (ii) x =.5 =. 4 (d) For closed pipe n = l = = 33 = 0.5m 4l 4n 4 66 4 (c) Equation of gien wae is y = a cos(kx ωt).. (i) Let equation of other wae be y=-a cos (kx+ωt)..(ii) And y=a cos (kx+ωt) (iii) If Eq. (i) propagates with Eq. (ii), then from the principle of superposition, we hae Eq.(i)+Eq.(ii) y=a cos (kx-ωt)-a cos (kx+ωt) Y= a [cos (kx-ωt)-a cos (kx+ωt)] Using A + B B A cos A cos B = sin. sin, We get Y=a sin kx sin ωt (i) Similarly when Eqs. (i) propagates with Eq. (iii), we get Y=a cos kx cos ωt After putting x=0,in Eq.(i) and (),weget Y=o and y=a cos ωt Hence Eq. (ii) is an equation of unknown wae. 43 (b) When one end is closed, n = 00 = 50 Hz n = 3n = 50Hz, n 3 = 5n = 50 Hz and so on. 44 (a) Due to rise in temperature, the speed of sound increases. Since n = and remains unchanged, hence n increases 45 (b) When open tube is dipped in water, it becomes a tube closed at one end. Fundamental frequency for open tube is 0 = l Length aailable for resonance of closed tube is 0.5l c = 4(0.5l) = l = 0 P a g e 9