Standing Wave in Organ Pipe and Beats

STANDING WAVE IN ORGAN PIPE AND BEATS Reflection of Sound Waves Reflection from denser medium Consider a scenario where a sound wave traverses through a medium and undergoes reflection from a rigid wall, as depicted in the illustration. Immediately preceding the reflection, the particles of the medium adjacent to the wall become compressed, resembling a compression wave in terms of propagation. Following the reflection from the wall particles, the medium's particles rebound in a manner akin to their state just before the reflection. In simpler terms, after reflecting at a fixed end (displacement node), a compression pulse returns as a compression pulse, and a rarefaction pulse returns as a rarefaction pulse. Consequently, the denser medium, representing the fixed end, behaves like a displacement node. Rarer medium Denser medium As the wave pulse retains its nature unchanged upon reflection from the denser medium, there is no introduction of any phase difference in the sound wave pulse after this reflection Reflection from rarer medium Examine a sound wave advancing through a denser medium and undergoing reflection from a rarer medium. In this scenario, as the disturbance progresses through the medium, it reaches the interface precisely in the form of a compression at the moment of reflection, as depicted in the figure. In this context, the interface acts akin to a free end. When the compression encounters the interface, the particles of the medium displace into the rarer medium, reflecting back into the initial medium as a rarefaction. Consequently, the rarer medium (representing the free end) behaves like a displacement antinode, as illustrated in the figure. Denser medium: Rarer medium 99999999 U 8888 Interface In summary, when a sound wave reflects from the rarer medium, the initial compression transforms into a rarefaction. Therefore, it can be concluded that the sound wave pulse undergoes a phase change of t after reflecting from the rarer medium. Denser medium Rarer medium Interface Note For the reflection of sound waves: 1. In the denser medium: (a) A displacement node or pressure antinode will emerge at the interface. (b) A compression pulse will revert as a compression pulse, and a rarefaction pulse will revert as a rarefaction pulse. (c) The reflected pressure wave will undergo no phase change, indicating no inversion. 2. In the rarer medium: (a) A displacement antinode or pressure node will manifest at the interface. (b) A compression pulse will return as a rarefaction pulse, and a rarefaction pulse will return as a compression pulse. (c) The reflected pressure wave will undergo a phase change of t, signifying wave inversion. 3. The total phase difference between sound waves Delta*phi = Delta*phi_{x} + phi_{0} + Delta*phi_{r} where Delta*phi is the phase difference due to the path difference phi_{0} 0 is the initial phase difference or epoch, and Delta*phi_{r} the phase difference due to the reflection of the sound waves. Ex. A source of sound S and a detector D are placed at some distance from one another. A big cardboard is placed near the detector and perpendicular to the line SD as shown in the figure. It is gradually moved away, and it is found that the intensity changes from the maximum to the minimum as the board is moved through a distance of 20 cm. Find the frequency of the sound emitted. The velocity of sound in air is 336 m * s ^ - 1 S D Sol. It's important to highlight that due to the interference between the sound wave originating from the source and the reflected sound wave from the cardboard, the detector registers both maximum and minimum intensities of sound waves as the cardboard progressively moves away. S In addressing this issue, considering that the sound wave reflects off the cardboard, it is reasonable to infer that the sound wave undergoes reflection from a denser medium, thus Delta*phi_{t} = 0 As the incident and reflected waves stem from the identical source, the initial phase (epoch) phi_{0} = 0 Hence, the overall phase difference is, Delta*phi = Delta*phi_{x} Therefore, the phase difference arises exclusively due to the disparity in the path traveled. Consider the distance between the source and the detector as d, and the initial separation between the detector and the cardboard as lo. At this cardboard position, let the path difference between the incident and reflected waves be denoted as Delta*x_{1} It is specified that due to this path difference, the detector registers the maximum intensity. Now, let's assume the cardboard is displaced to a distance x from its original position. Consequently, for this cardboard location, the path difference between the incident wave and the reflected wave becomes: Delta*x_{t} = Delta*x_{t} + 2x and due to this variation in path length, the detector. registers the minimum intensity. The intensity at the detector transitions from maximum to minimum due to the cumulative path difference. (Delta*x_{t} - Delta*x_{i}) this path difference will be equal to lambda/2 Therefore Therefore, the frequency of the sound f= v lambda = 336 ( 4 5 ) Hz Rightarrow f = 84 * 5 = 420Hz lambda = 4x lambda = 4 * 20cm Delta*x_{f} - Delta*x_{i} = lambda/2 lambda = 80cm 2x = lambda/2 lambda = 4/5 Ex. A source S and a detector D are placed at a distance d apart. A big cardboard is placed at a distance sqrt(2d) from the source and the detector as shown in the figure. The source emits a wave of wavelength which is received by the detector after the reflection from the cardboard. It is found to be in phase with the direct wave received from the source. By what minimum distance should the cardboard be shifted away so that the reflected wave becomes out of phase with the direct wave? d/2 Ex. The air column in a pipe closed at one end is made to vibrate in its second overtone by a tuning fork of frequency 440 Hz. The speed of sound in air is 330m * s ^ - 1 The end corrections may be neglected. Let PO denote the mean pressure at any point in the pipe and Delta*P_{0} | the maximum. amplitude of pressure variation. Find the length (L) of the air column. (a) What is the amplitude of pressure variation at the middle of the column? (b) What are the maximum and minimum pressures at the open end of the pipe? (c) What are the maximum and minimum pressures at the closed end of the pipe? (d) Sol Since the pipe is closed at one end, it behaves as a closed organ pipe. Also, we know that for a closed organ pipe, L = (2n - 1) * lambda/4 and f = (2n - 1) * (v/(4L))! Hence, for the second overtone (n = 3) , the length of the air column is L = (5lambda)/4 and the frequency is f_{2} = 5(v/41) Given, The frequency of the tuning fork that sets the air column to vibrate in its second overtone mode is, The speed of the sound in air is, v = 330m * s ^ - 1 Therefore, the wavelength is, lambda = 330/440 * m f_{2} = 440Hz lambda = v/r (a) The length of the air column is, L = 5/4 * 3/4 L = 15/16 * m L = 5(lambda/4) lambda = 7/4 * m (b) Since it is a closed organ pipe, there should be a pressure node at the open end. Suppose that the closed organ pipe is placed in such a way that the open end is at x = 0 The equation of pressure should be, p= Delta*p_{o} * sin(kx) cos(cot). where Delta p sin(kx) is the amplitude of pressure. Now, the angular wave number is given by k= (2pi)/(3/4) ( Since lambda = 3/4 * m ) We also know that the length of the air column is, L = 15/16 * m Therefore, at the middle of the air column, Le.. at x = 1/2 = 15/32 * m the amplitude of the pressure becomes, p = Delta*p_{0} * sin((0pi)/3 * 15/32) p = Delta*p_{0} * sin((5m)/4) p = Delta*p_{0} * sin(pi + pi/4) p = - Delta * p_{0} * sin(pi/4) Since amplitude cannot be negative, the amplitude of the pressure will is, k = (8pi)/3 * m ^ - 1 p = - (Delta*p_{0})/(sqrt(2)) k = (z*pi)/lambda p = Delta*p_{0} * sin(kx) (c) At the open end (pressure node), there will not be any fluctuation in the pressure. So, the amplitude of the pressure is constant at the open end. p = (Delta*p_{0})/(sqrt(2)) (d) At the closed end, the amplitude of the pressure is, p = Delta*p_{0} * sin((theta*pi)/3 * 15/16) p= Delta p 0 sin(kL)