Waves and sound-02-SUBJECTIVE SOLVED

SOLVED SUBJECTIVE PROBLEMS Problem 1. One end of each of two identical springs, each of force constant 0.5 N/m are attached on the opposite sides of wooden block of mass 0.01 kg. The other ends of the springs are connected to separate rigid supports such that the springs are unscratched and are collinear in a horizontal plane. To the wooden piece is fixed a pointer which touches a vertically moving plane paper. The wooden piece kept on a smooth horizontal table is now displaced by 0.02 m along the line of springs and released. If the speed of the paper, perpendicular to the springs’ length, is 0.1 m/s, find the equation of the path traced by the pointer on the paper and the distance between two consecutive maxima on this path. Solution : The effective force constant of the spring system is 2 k (since they constitute a parallel combination). The angular frequency of simple harmonic oscillation is The amplitude m The speed of the paper may be assumed as the speed of wave-propagators and the curve traced on the paper can be represented by the equation …(i) The wavelength …(ii) Substituting the value in (i), we get the required equation of the path The distance between the two consecutive maxima is the wavelength . Problem 2. A uniform rope of mass 0.1 kg and length 2.45 m hangs from the ceiling. (a) Find the speed of transverse wave in the rope at a point 0.5 m distance from the lower end. (b) Calculate the time taken by a transverse wave to travel the full length of the rope. Solution : (a) The speed of the transverse wave is given by where tension and linear mass density At a distance x from the free end of the rope, the tension …(i) Here (b) The speed of the wave at a distance x from the free end is {from (i)} or, Integrating over the whole length of the rope, Problem 3. A wave pulse starts propagating in the +x direction along a non-uniform wire of length 10 m with mass per unit length given by and under a tension of 100 N. Find the time taken by a pulse to travel from the lighter end (x = 0) to the heavier end. Solution : The speed of the wave pulse substituting the values, we get . Problem 4. When a train is approaching the observer, the frequency of the whistle is 100 cps. When it has passed the observer, it is 500 cps. Calculate the frequency when the observer moves with the train. Solution : When the source (the train) moves towards the observer, the apparent frequency is …(i) while the source is moving away, the apparent frequency …(ii) or or …(iii) When the observer moves with the train, there is no relative motion between the source and the observe and hence he will listen the true frequency . Substituting (iii) in (i), we get or 66.6 Hz. Problem 5. When 0.98 m long metallic wire is stressed, an extension of 0.02 m is produced. An organ pipe 0.5 m long and open at both ends, when sounded with this stressed metallic wire, produces 8 beats in its fundamental mode. By decreasing the stress in the wire, the frequency of beats is found to decrease. Find the Young’s modulus of the wire. The density of metallic wire is 104 kg/m3 and speed of sound in air is 292 m/s. Solution : Frequency of the transverse vibration of the stretched string in …(i) Here or and …(ii) when the stress in the wire is decreased, will decrease, consequently the beat frequency will decrease if or Substituting the value, we get Simplifying, we get Problem 6. The fundamental frequency of a sonometer wire increases by 6 Hz if its tension is increased by 44% keeping the length constant. Find the change in the fundamental frequency of the sonometer wire when the length of the wire is increased by 20% keeping the original tension in the wire. Solution : Fundamental frequency of sonometer wire is when length L is constant, or Differentiations, Given that =0.44 Next, when T is constant and length is changed, Differentiating, = -5.5 Hz The fundamental frequency of sonometer wire will decrease by 5.5 Hz Problem 7. Two metallic strings A and B of different materials are connected in series forming a joint. The strings have equal cross-sectional area. The length of A is and that of B is . One end of the combined string is tied with a support rigidly and the other end is loaded with a block of mass m passing over a frictionless pulley. Transverse waves are set up in the combined string using an external sources of variable frequency. Calculate (a) Lowest of frequency for which standing waves are observed such that the joint is a node, (b) The total number of antinodes at this frequency. The densities of A and B are 6.3 103 kg/m3 and 2.8 103 kg/m3 respectively. Solution : (a) Let and be the number of loops formed in A and B and the ratio of the number of loops for the lowest frequency, this means that A will have 3 loops and B will have 5 loops. Hence the required minimum frequency or . (b) the total number of antinodes at this minimum frequency is 3+5 = 8 Problem 8. A source S emitting sound of 300 Hz is fixed on block A which is attached to free end of a spring as shown in the figure. The detector D fixed on block B attached to the free end of spring SB detects this sound. The blocks A and B are simultaneously displaced towards each other through a distance of 1.0 m and then left to vibrate. Find the maximum and minimum frequencies of sound detected by D of the vibrational frequency of each block is 2 Hz. (velocity of sound = 340 m/s) Solution : The motion of A and B is synchronized. Hence the maximum and minimum frequencies detected by D will be (when S and D are approaching each other while at their equilibrium position.) Similarly, the apparent frequency will be minimum when the source and the detector recede from each other with maximum speed ( = A) =278.6 Hz Problem 9. Determine the speed of sound in a gas in which two wavelengths of 1m and 1.01 m respectively produce 10 beats in 3 s. Solution : Let the speed of sound = V. The corresponding frequencies are and the beat frequency (or the number of beats per second) will be or or = 336.67 m/s Problem 10. The frequency of sound produced by a bell is 500 Hz. The velocity of the source relative to still air is 60 m/s. An observer moves at 30 m/s along the same line as the source. Calculate the frequency of sound wave measured by the observer. Consider all the possible cases (speed of sound v = 340 m/s) Solution : Case (i) let the observer at right side of the source. Both source and observe is moving in right direction. There will be apparent increase in frequency because Case (ii) Source is moving right and observer is moving left There will be apparent increase in frequency since both are moving towards each other. (more than that as in case (i)) Case (iii) Both source and observer is receding from each other. There will be an apparent decrease in frequency = 387.5 Hz. Case (iv) Both source and observer is moving towards left. = 462.5 Hz. You should repeat the example assuming that the observer is to the left to the source. What conclusions can be derived from the this example? Do you get a different set of four apparent frequencies Problem 11. An un-stretched spring has a length of 1.0 m and a mass of 0.2 kg. When a body of mass 2 kg is hung from the spring it stretches 3 cm. Determine the velocity of longitudinal wave along the spring. Solution : The spring constant or force constant of the spring Hence the velocity of the longitudinal wave along the spring in Problem 12. Given the equation for a wave on a string where y and x are in meters and t is in seconds. (a) At what are the values of the displacement at x = 0, 0.1 m, 0.2 m and 0.3 m ? (b) At what are the values of the displacements at t = 0, 0.1s and 0.2 s ? (c) What is the equation for the velocity of oscillation of the particles of the string ? (d) What is the maximum velocity of oscillation ? (e) what is the velocity of propagation of the wave ? Solution : (a) At for , x = 0, sin (0) = 0 Similarly for m, and for m. (b) At y = 0.03 sin (0.3 – 2t) At y = 0.03 sin (0.3 radian) = 8.87  10-3m m m (c) Velocity of particle, (d) Maximum velocity or the velocity amplitude m/s (e) Comparing with the wave equation we have and . Problem 13. A train approaching a railway crossing at a speed of 120 km/h sounds a short whistle at frequency 640 Hz when it is 300 m away from the crossing. The speed of sound in air is 340 m/s. What will be the frequency heard by a person standing on a road perpendicular to the track through the crossing at a distance of 400 m from the crossing? Solution: The observer is at rest with respect to the air and the source is traveling at a velocity of 120 km/h i.e., . As is clear from the figure, the person receives the sound of the whistle in a direction making an angle with the track where . The component of the velocity of the source (i.e., of the train). Along this direction is m/s = 20 m/s. As the source is approaching the person with this component, the frequency heard by the observer is Hz = 680 Hz. Problem 14. A wave reached the boundary of media 1 and 2, at which it is partly reflected into medium 1 and partly transmitted into medium 2. Call these waves and , respectively. Assuming that at the boundary the displacement arising from the transmitted wave, show that . Further, assuming that the slope of the displacement wave at the boundary of medium 1 is equal to the slope of the displacement wave at the boundary of medium 2, show that Solution: The wave in medium 1 is given by and that in medium 2 is At the boundary or At the boundary or or . Problem 15. A metallic rod of length 1 m is rigidly clamped at its mid-point. Longitudinal stationary waves are set up in the rod in such a way that there are two nodes on either sides of mid-point. The amplitude of antinode is m. Write the equation of motion at a point 2 cm from the mid-point and those of the constituent waves in the rod. (Young’s modulus = , density = 8000 kg ). Solution: Figure shows the longitudinal displacement y as a function of x from x = 0 to x = L, where L is length of the rod. We know that the separation from a node and the next antinode is and that between two consecutive nodes or antinodes is . Therefore Here is the wavelength of either of the interfering waves. Thus m The speed of longitudinal waves in a metal of Young’s modulus Y and density is given by Frequency Hz. Notice from figure that there is an antinode at the ends and of the rod. Hence the equation of the stationary wave is (here is the amplitude of each of the interfering waves) …(i) At any instant of time, the amplitude is given by …(ii) At , there is an antinode, i.e. (given). Putting in Eq. (ii) we have or . Putting values of and in Eq. (i) we get or …(iii) where x and y are in metre and t in seconds. For a point at a distance of 2 cm from the mid-point, the value of x is x = 50 cm + 2 cm = 52 cm = 0.52 m using this value of x in Eq. (iii), the required equation of motion at a point 2 cm from the mid-point is or The equations of the constituent waves (incident wave and reflected wave) in the rod are and .