Waves and sound-05-OBJECTIVE SOLUTION (LEVEL-I)

OBJECTIVE SOLUTION (LEVEL-I) 1. Determinant of equation must be equal to zero for single root thus Ans. (c) 2. Ans (d) 3. hence Ans. (a) 4. . Ans. (b) 5. . Ans. (b) 6. Speed of sound depends on atmospheric conditions. Ans. (a) 7. = 36% . Ans. (b) 8. . Ans. (d) 9. Intensity varies as so answer will be (d) only Ans. (d) 10. & m. Ans. (c) 11. . Ans. (d) 12. . Ans. (b) 13. Ans. (d) 14. . Ans. (a) 15. & Hz cm/s. Ans. (a) OBJECTIVE SOLUTION (LEVEL-II) 1. should be zero for a maximum / minimum where . Ans. (b) 2. leads since it is +ive valued when is zero. lags since it is –ive valued when is zero. Ans. (a) 3. sec. = 200 m/s. Ans. (c) 4. 5 = 10 . Ans. (a) 5. Ans. (b) 6. Ans. (a) 7. , . Ans. (a) 8. Since beats is increased then . Ans. (c) 9. Since net amplitude is zero, so energy will be purely kinetic. Ans. (b) 10. Ans. (a) 11. = 240 Hz. 12. Path difference when the waves are in phase . Ans. (c) 13. . Since hence = 0.072 m. Ans. (b) 14. And (b) 15. , . SUBJECTIVE SOLUTION (LEVEL-I) 1. . 2. (a) mm. (b) m. (c) sec., Hz. 3. . 4. Rod behaves as a closed organ pipe . 5. . If both ends are open No. 6. Minimum value of would be 1. hence m/s. 7. … (i) At tension increased the frequency of , beats will decreases as given. ____ equation (i) must be in the form of Hz. 8. Hz. 9. (a) Hz (b) similarly = 388.3 Hz. 10. (i) Travelling, and m/s (ii) cm (iii) Hz (iv) m. 11. cm Now phase difference path difference (i) rad. rad. Similarly (ii) rad. (iii) rad (iv) rad. 12. rad, rad cm (i) cm/s cm/s (ii) cm/s2 cm/s2. 13. The particle displacements are given by where a = 0.1 m. We are given - 0.05 = 0.1 sin …(i) and + 0.05 = 0.1 sin …(ii) Eqs. (i) and (ii) give and Subtracting these equations, we get m. Now . Therefore, time taken is . 14. Let be the frequency of the source and that of the th harmonic of a closed pipe. The source will resonantly excite that harmonic mode of the pipe for which for any value of p = 1, 2, 3, …… Now for a closed pipe, we know that Therefore, for resonance, . 15. Since the source of sound is stationary, the wavelength remains unchanged by the motion of the observer. Therefore, m The frequency of sound heard by the observer is Hz. SUBJECTIVE SOLUTION (LEVEL-II) 1. On loading frequency will decrease and also beats increases thus Hz. 2. . 3. cm . and cm cm cm’s. 4. As calculated by , the velocity of sound is equal thus the only possible figure so that the standing waves will be developed in the ‘system’ is Thus min frequency = 180 Hz. Alternative : Since & are integers thus . . 5. and since for minimum is 3 is integer number) Hence . 6. (a) 500 Hz, since no relative motion (b) Hz. (c) 500 Hz Hz where Velocity of observer w.r.t. to medium m/s and m/s Hz Hz. 7. Let where so by principle of superposition where for to be max or min i.e., either or, maximum at ; Intensity is maximum (with two different values) i.e. within the second we got two maxi and two minima so beat frequency is 2. 8. Observer will listen max./min. frequency when source velocity is pointed at observer. . . 9. m. m/s. Apparent . m/s Hz. 10. Acceleration Now, where etc. SUBJECTIVE SOLUTION (LEVEL-III) 1. (a) (b) (c) . (d) and and . 2. … (i) and … (ii) from (i) and (ii) hence (from (i)) . 3. as tension decrease, it will decrease the frequency, , so beats will be decreased thus . 4. . . 5. Path difference, (a) For minimum intensity . (b) For maximum intensity (c) when so 2 maximum. 6. 7. (i) The frequency heard after reflection will be the same as heard directly from the source so beat frequency will be zero in this case. (ii) beats. (iii) beats. 8. and . 9. (a) loops. (b) . (c) where (d) . 10. where . . 1. Frequency received in case I and in case II Obviously Beat frequency or, . 2. Frequency of second overtone of the closed pipe (Given) . Substituting speed of sound in air = 330 m/s . (b) Open end is displacement antinode. Therefore, it would be a pressure node or, Pressure amplitude at , can be written as Therefore, pressure amplitude at or (15/32) m will be . (c) Open end is a pressure nod i.e., Hence Mean pressure (d) Closed end is a displacement node or pressure antipode. Therefore, and . 3. Tension Amplitude of incident wave Mass per unit length of wire is and mass per unit length of wire is (a) Speed of wave in wire is and speed of wave in wire is Time taken by the wave pulse to reach from to is – . (b) The expressions for reflected and transmitted amplitudes and in terms of and are as follows : and Substituting the values we get i.e. the amplitude of reflected wave will be 1.5 cm. Negative sign of indicates that there will be a phase of in reflected wave. Similarly i.e. the amplitude of transmitted wave will be 2.0 cm. 4. Velocity of sound in water is Frequency of sound in water will be . (a) Frequency of sound detected by receiver (observer) at rest would be ¬– . (b) Velocity of sound in air is = 344 m/s. Frequency does not depend on the medium. Therefore, frequency in air is also . Frequency of sound detected by receiver (observer) in air would be . 5. (a) Frequency of second harmonic in pipe frequency of third harmonic in pipe or, (as ) or, or, (as ) or, . (b) Ratio of fundamental frequency in pipe and in pipe is: (as ) (as ) substituting from part (a), we get . 6. Fundamental frequency, speed of sound or, m/s. 7 or, The amplitude at a distance from is given by Total mechanical energy at of length is or, … (i) Here and . Substituting these values in equation (i) and integrating it from to , we get total energy of string, .