Planning to start your own coaching institute for JEE, NEET, CBSE or Foundation? Here's why not to waste time in developing study material from scratch. Instead use the readymade study material to stay focused on quality of teaching as quality of teaching is the primary job in coaching profession while study material is secondary one. Quality of teaching determines results of your coaching that decides success & future of your coaching while good quality study material is merely a necessary teaching aid that supplements your teaching (video in Hindi)

https://docs.google.com/document/d/1P6vd3pIz8bfyn2vW93Ml8LKCm2jgQ4a4/edit?usp=share_link&ouid=109474854956598892099&rtpof=true&sd=true WAVES AND TYPE OF WAVES WAVE MOTION In simple terms, we can say that wave motion involves transfer of disturbance (energy) from one point to the other with the particles of medium oscillating about their mean positions. That is, the particles of the medium do not themselves travel along with the wave. Instead, they oscillate back and forth about some equilibrium position as the wave passes by. Only the disturbance is propagated. In this chapter, we will limit our discussion to mechanical waves (elastic waves) which require a medium to travel. There are also electromagnetic waves which do not require any medium and can travel in vacuum. TRANSVERSE WAVE In transverse waves the particle of the medium oscillate perpendicular to the direction in which the wave travels. Travelling waves on a tight rope are transverse waves. If one end of the rope is rigidly fixed and the other end is given periodic up and down jerks, the disturbance propagates along the length of the rope but the particles of the rope oscillate up and down. Disturbance travels along the rope in form of crests (upward peaks) and troughs (downward peaks). LONGITUDINAL WAVES In longitudinal waves, the oscillation of the particles is parallel to the direction in which the wave travels. Disturbance travelling in a ring parallel to its length a pressure variation propagating in a liquid, sound waves travelling in a medium are examples of longitudinal waves. Longitudinal waves do not require shearing stress and hence can travel in any elastic medium: solid liquid and gases. eed of longitudinal waves in a liquid = eed of longitudinal waves in a gas = eed of longitudinal waves in solid rod = Where B: Bulk Modulus d: density ρ: density of gas γP: adiabatic bulk modulus Y: Young’s modulus P: pressure WAVE PROPERTIES WAVE EED (C) The eed of a wave is the distance it covers in one second. It should be carefully noted that wave eed is completely different from particle eed. Particle eed is the eed of the vibrating particles in the medium. On the other hand, wave eed is the eed with which the disturbance (or wave) propagates in the medium. WAVE FREQUENCY (υ) The frequency with which the particles of the medium (through which the wave is passing) oscillate is known as wave frequency. In transverse waves, frequency is the number of crests (or troughs) that pass through a point in one second. In longitudinal waves frequency is the number of compressions (or rarefactions) that pass through a point in one second. TIME PERIOD (T) The time period of the oscillation of the particles in the medium is the time period of the wave. AMPLITUDE (A) Amplitude of the wave is same as the amplitude of the oscillating particles. WAVELENGTH (λ) Wavelength is the distance between two consecutive crests (or compression) in a wave. Wavelength, wave eed and frequency are related as follows c = υλ PHASE When a wave passes through a medium, all particles oscillate with same frequency, but they reach the correonding particles in their path at different time instants. For example, in the above figure, the particle at P is at its top extreme; the particle at Q is passing through its mean position; the particle at R is at its bottom extreme. These relative positions represents the phase of motion. (i) If two particles have same position and same velocities at all time instants, they are said to be in same phase (or in–phase). (ii) Two particles are said to be in opposite phase (or exactly out of phase) if their dilacements from mean position and their velocities are equal in magnitude but opposite in direction. the distance between particles in same phase = 0, λ, 2λ, 3λ, …….. the distance between particles in opposite phase = λ/2, 3λ/2, 5λ/2, ….. When we describe the wave equation, we will give a mathematical meaning to the phase of an oscillating particle. EQUATION OF A TRAVELLING WAVE Suppose, man holding a stretched string starts snapping his hand at t = 0 and finishes his job at t = Δt. The vertical dilacement y of the left end of the string is a function of time. It is zero for t < 0, has non-zero value of 0 < t < Δ t and is again zero for t > Δt. Let us represent this function by f(t). Take the left end of the string a the origin and take the X-axis along the string towards right. The function f(t) represents the dilacement of the left end at time t – x / v is f(t – x/v). Hence, y (x, t) = y (x = 0, t – x/v) = f(t – x/v). The dilacement of the particle at x at time t i.e., y(x, t) is generally abbreviated as y and the wave equation is written as y = f(t – x/v) …… (i) Equation (i) represents a wave travelling in the positive x-direction with a constant eed v. such a wave is called a travelling wave or a progressive wave. The function f is arbitrary and depends on how the source moves. The time t and the position x must appear in the wave equation in the combination t – x/v only. For example, etc. are valid wave equations. They represent waves travelling in positive x-direction with constant eed. The equation does not represent a wave travelling eed. If a wave travels in negative s-direction with eed v, its general equation may be written as y = f(t + x/v) …. (ii) The wave travelling in position x-direction (equation (i)) can also be written as or, y = g(x – vt), …. (iii) Where g is some other function having the following meaning. If we put t = 0 in equation (iii), we get the dilacement of various particles at t = 0 i.e., y(x, t = 0) g(x) Thus, g(x) represents the shape of the string at t = 0. If the dilacement of the different particles at t = 0 is represented by the function g(x), the dilacement of the particle at x at time t will be y = g(x – vt). Similarly, if the wave is travelling along the negative x-direction and the dilacement of different particles at t = 0 is g(x), the dilacement of the particle at x at time t will be y = g(x + vt) ….(iv) Thus, the function f in equation (i) and represents the dilacement of the point x = 0 as time passes and g in (iii) and (iv) represents the dilacement at t = 0 of different particles. The travelling wave moving with constant eed v towards positive x direction must satisfy must satisfy the following wave function condition. …. (v) EQUATION OF A SIMPLE HARMONIC PLANE WAVE In case of harmonic wave the dilacement of successive particles of the medium is given by a sine wave or cosine function of position. The dilacement y for different values of x at t = 0 is given by y = A sin kx …. (vi) Where A and k are constants. Suppose this disturbance is propagating along positive x-direction then y = A sin k(x –vt) …. (vii) Since the waveform represented by equation (vi) is based on sine function, it would repeat itself at regular distances. The first repetition would take place when kx = 2π or x = This distance after which the repetition takes place is called the wavelength and denoted by λ. Hence λ = or k = This constant k is called propagation constant or wave number. Now equation (vii) turns into y = A sin (x – vt) At t = 0 y = A sin x Relation Between Wavelength and Velocity of Propagation Time taken for one complete cycle of wave to pass any point is the time period (T). This is also the time taken by the disturbance in propagating a distance λ. v = = fλ where f = frequency (Hz) ω = = 2πf = circular frequency (rad/s) Different Forms of Simple Harmonic Wave Equation y = A sin(ωt – kx – φ) = A sin = A sin Where φ = phase angle. ENERGY OF A PLANE PROGRESSIVE WAVE Consider a plane wave propagative with a velocity v in x-direction across an area S. An element of material medium (density = ρ (Sdx). The dilacement of a particle from its equilibrium position is given by the wave equation y = A sin(ωt – kx) Total energy of this element is dE = (ρSdx) (Aω)2 = ρSdx (2π2f2A2) ⇒ Energy density = = 2π2f2A2ρ (J/m3) Energy per unit length = 2π2f2A2ρS ∴ Power transmitted = 2π2f2A2ρS (Watt = J/s) Intensity of the Wave (I) Intensity of the wave is defined as the power crossing per unit area = 2π2f2A2ρv ….Watt/m2 For wave propagation through a taut string, ρS = μ, the linear density in kg/m ∴ Energy per unit length = 2π2f2A2μ TRANSVERSE WAVE IN A STRETCHED STRING Consider a transverse pulse produced in a taut string of linear mass density μ. Consider a small segment of the pulse, of length Δl, forming an arc of a circle of radius R. A force equal in magnitude to the tension T pulls tangentially on this segment at each end. Let us set an observer at the centre of the pulse which moves along with the pulse towards rights. For the observer any small length dl of the string as shown will appear to move backward with a velocity v. Now the small mass of the string is in a circular path of radius R moving with eed v. Therefore, the required centripetal force is provided by the only force acting, (neglecting gravity) is the component of tension along the radius. The net restoring force on the element is F = 2T sin(Δθ) ≈ T (2Δθ) = T The mass of the segment is Δm = μΔ The acceleration of this element toward the centre of the circle is a = , where v is the velocity of the pulse. Using second law of motion, T = (μΔ) or, v = Laws of Transverse Vibrations of A string: Sonometer The fundamental frequency of vibration of a stretched string fixed at both ends is given by v = . From this equation, one can immediately write the following statements known as “Laws of transverse vibrations of a string” (a) Law of length – the fundamental frequency of vibration of a string (fixed at both ends) is inversely proportional to the length of the string provided its tension and its mass per unit length remain the same. v ∝ 1 / L if T and μ are constants. (b) Law tension – The fundamental frequency of a string is proportional to the square root of its tension provided its length and the mass per unit length remain the same. v ∝ if L and μ are constants. (c) Law of mass – The fundamental frequency of a string is inversely proportional to the square root of the linear mass density, i.e., mass per unit length provided the length and the tension remain the same. v ∝ if L and T are constants. Illustration 1: A transverse wave of wavelength 50 cm is travelling towards +ve X axis along a string whose inner density is 0.05 g/cm. The tension in the string is 450N. At t = 0, the particle through its mean position with an upward velocity. Form an equation describing the wave. The amplitude of the wave is 2.5 cm. Solution: Let the wave be described by: y (x, t) = A sin (kx – +) Where k = and A = 2.5 cm y (0,0) = 0 Velocity of transverse wave in the string is A sin = 0 given by : .... (i) We also have > 0 rad/s. –A cos > 0 ...... (ii) Using all the quantities, the equation is: From I and II, we get Y =2.5 sin (4 x – 1200 t + ) Illustration 2: The vibrations of a string of length 60 cm fixed at both ends are represented by the equation y = 4sin cos where x and y are in cm and t in seconds. (i) What is the maximum dilacement of a point at x = 5 cm? (ii) Where are the nodes located along the string ? (iii) What is the velocity of the particle at x = 7.5 cm and at t = 0.25 s? (iv) Write down the equations of the component waves whose superposition’s give the above wave. Solution: y = 4 sin () cos (96) can be broken up into y = Thus the waves are of the same amplitude and frequency but travelling in opposite directions which thus, superimpose to give a standing wave, (i) At x = 5cm the standing wave equation gives y= 4 sin cos = 4 sin cos (96) = 4 × 1/2 cos (96) Maximum dilacement = 2 (ii) The nodes are the points permanently at rest. Thus they are those points for which sin = 0 i.e. (x/15) = n , n = 0, 1, 2, 3, 4....... x = 15 n i.e. at x = 0, 15, 30, 45 and 60 cm (iii) The particle velocity is equal to = 4sin =384sin sin at x = 7.5 and t = 0.25 we get = –384sin sin = –384sin sin(24) = 0 (iv) The equations of the component waves are y1 = 2 sin and y2 = SUPERPOSITION OF WAVES Two or more waves can traverse the same ace independently of another. Thus the dilacement of any particle in the medium at any given time is simply the vector sum of dilacements that the individual waves would give it. This process of the vector addition of the dilacement of a particle is called superposition. INTERFERENCE When two waves of the same frequency, superimpose each other, there occurs redistribution of energy in the medium which causes either a minimum intensity or maximum intensity which is more than the sum of the intensities of the individual sources. This phenomenon is called interference of waves. Let the two waves be y1 = A1 sin(kx – ωt), y2 = A2 sin(kx – ωt + δ) According to the principal of superposition y = y1 + y2 = A1 sin(kx – ωt) + A2 sin(kx – ωt + δ) = A1 sin(kx – ωt) + A2 sin(kx – ωt) cosδ + cos(kx – ωδ) (A2 sinδ) = sin(kx – ωt) (A1 + A2 cosδ) +cos(kx – ωt) (A2 sinδ) = R sin (kx – ωt + φ) where A1 + A2 cosδ = R cos φ and A2 sin δ = R sin φ and R2 = (A1 + A2 cos δ)2 + (A2 sin δ)2 = If I1 and I2 are intensities of the interfering waves and δ is the phase difference, then the resultant intensity is given by Now, Illustration 3: Two coherent sound sources are at distances x1 = 0.2 m and x2 = 0.48m from a point. Calculate the intensity of the resultant wave at that point if the frequency of each wave is f = 400 Hz and velocity of wave in the medium is v = 448 m/s. The intensity of each wave is Io = 60W/m2. Solution: Path difference, p = x2 − x1 = 0.48 - 0.2 = 0.28 m φ = I = I1 + I2 + 2cosφ or I = Io + Io + 2Io cos(π/2) = 2Io = 2(60) = 120 W/m2. STANDING WAVES A standing wave is formed when two identical waves travelling in the opposite directions along the same line, interfere. On the path of a stationary wave, the amplitude of vibration varies simple harmonically with reect to the distance from one end of the path. On the path of the stationary wave, there are points where the amplitude is zero, they are known as NODES. On the other hand there are points where the amplitude is maximum, they are known as ANTINODES. The distance between two consecutive nodes or two consecutive anitnodes is λ/2. The distance between a node and the next antinode is λ/4. Consider two waves of the same frequency, eed and amplitude which are travelling in opposite directions along a string. Two such waves may be represented by the equations y1 = a sin (kx - wt ) and y2 = a sin (kx + wt) Hence the resultant may be written as y = y1 + y2 = a sin (kx - wt) + a sin (kx + wt) y = 2 a sin kx cos wt This is the equation of a standing wave. Note: (i) In this equation, it is seen that a particle at any particular point ‘x’ executes simple harmonic motion and all particles vibrate with the same frequency. (ii) The amplitude is not the same for different particles but varies with the location ‘x’ of the particle. (iii) The points having maximum amplitudes are those for which 2asinkx, has a maximum value of 2a, these are at the positions, kx = π/2, 3π/2, 5π/2, .............. or x = λ/4, 3λ/4, 5λ/4, ............... These points are called antinodes. The strain is minimum at these points. (iv) The amplitude has minimum value of zero at positions where kx = π, 2π, 3π,........... or x = λ/2, λ, 3λ/2 ,2λ............ These points are called nodes. The strain is maximum at these points. (v) Energy is not tranorted along the string to the right or to the left, because energy can not flow past the nodal points in the string which are permanently at rest. REFLECTION OF WAVES (a) Waves on reflection from a fixed end undergoes a phase change of 180°. (b) While a wave reflected from a free end is reflected without a change in phase. STATIONARY WAVES IN STRINGS A string of length L is stretched between two points. When the string is set into vibrations, a transverse progressive wave begins to travel along the string. It is reflected at the other fixed end. The incident and the reflected waves interfere to produce a stationary transverse wave in which the ends are always nodes. (a) In the simplest form, the string vibrates in one loop in which the ends are the nodes and the centre is the antinode. This mode of vibration is known as the fundamental mode and the frequency of vibration is known as the fundamenal frequency or first harmonic. Since the distance between consecutive nodes is λ/2 L = If f1 is the fundamental frequency of vibration , then the velocity of transverse waves is given as , v = λ1 f1 or v = 2Lf1 . . . (1) (b) The same string under the same conditions may also vibrate in two loops, such that the centre is also the node. If f2 is the frequency of vibrations, then the velocity of transverse waves is given as , v = λ2f2 ∴ v = L X f2 . . . (2) The frequency f2 is known as second harmonic or first overtone. (c) The same string under the same conditions may also vibrate in three segments. If f3 is the frequency in this mode of vibration, then, v = λ3f3 ∴ v = . . . (3) The frequency n3 is known as the third harmonic or second overtone. Thus a stretched string in addition to the fundamental mode, also vibrates with frequencies which are integral multiples of the fundamental frequencies. These frequencies are known as harmonics. The velocity of transverse waves in a stretched string is given as where T = tension in the string. μ = linear density or mass per unit length of string. If the string fixed at two ends, vibrates in its fundamental mode, then v = 2Lf ∴ f = μ = volume of unit length × density = πr2 × 1 × ρ = π × × ρ where D = diameter of the wire, ρ = density. Note: If the ratio of two frequencies is 1 , then they are to be in unison. If = 2 , then f1 is the upper octave of f2 . If = , then f1 is the lower octave of f2 . Illustration 4: A wire of uniform cross-section is stretched between two points 100 cm apart. The wire fixed at one end and a weight is hung over a pulley at the other end. A weight of 9 kg produces a fundamental frequency of 750 Hz. (i) What is the velocity of wave in wire ? (ii) If the weight is reduced to 4 kg, what is the velocity of wave ? What is the wavelength and frequency of wave ? Solution: (i) L = 100 cm, f1 = 750 Hz. v1 = 2Lf1 = 2 × 100 × 750 = 150000 cms-1 = 1500 ms-1 (ii) ∴ v2 = 1000 m s-1 λ2 = wave length = 2L = 200 cm = 2 m f2 = Illustration 5: When a solid block is suended from one end of sonometer wire, it vibrates with fundamental frequency. If the block is immersed completely in water, the length of the wire has to be reduced to the half of its value so that it again vibrates with same frequency. Find the relative density of the solid. Solution: Let and be the initial and final length of the v0 = ...(II) vibrating wire. Let V = volume of block d = density of solid = density of water from (i) & (ii) : When the block is in air tension = weight of block = Vdg v0 = ....(I) When the block is in water Using = , we get tension = apparent weight = Vdg – Vpg Relative density = DILACEMENT WAVE AND PRESSURE WAVE A longitudinal wave in a fluid (liquid or gas) can be described either in terms of the longitudinal dilacement suffered by the particles of the medium or in terms of the excess pressure generated due to the compression or rarefaction. Let us see how the two representations are related to each other. Consider a wave going in the x-direction in a fluid. Suppose that at a time t, the particle at the undisturbed position x suffers a dilacement s in the x-direction. The wave can then be described by the equation. s = s0 sinω(t – x / v) ….(i) Consider the element of the material which is contained within x and x + Δx (figure) in the undisturbed state. Considering a cross-sectional area A, the volume of the element in the undisturbed state is A, the volume of the element in the undisturbed state is AΔx and its mass is ρ AΔx. As the wave passes, the ends at x and x + Δx are dilaced by amounts s and s + Δs according to equation (xv) above. The increase in volume of this element at time t is = As0(–ω/v) cos(t–x/v)Δx, Where Δs has been obtained by differentiating equation (xv) with reect to x. The element is, therefore, under a volume strain. The correonding stress i.e., the excess pressure developed in the element at x at time t is, , Where B is the bulk modulus of the material. Thus, …… (xvi) Comparing with (xv), we see that the pressure amplitude P0 and the dilacement amplitude s0 are related as , Where k is the wave number. Also, we see from (xv) and (xvi) that the pressure wave differs in phase by /2 from the dilacement wave. The pressure maxima occur where the dilacement is zero and dilacement maxima occur where the pressure is at its normal level. The fact that, dilacement is zero where the pressure-change is maximum and vice versa, puts the two descriptions on different footings. The human ear or an electronic detector reonds to the change in pressure and not to the dilacement in a straight forward way. Suppose two audio eakers are driven by the same amplifier and are placed facing each other. A detector is placed midway between them. The dilacement of the air particles near the detector will be zero as the two sources drive these particles in opposite directions. However, both send compression waves and rarefaction waves together. As a result, pressure increases at the detector simultaneously due to both the sources. Accordingly, the pressure amplitude will be doubled, although the dilacement remains zero here. A detector detects maximum intensity in such a condition. Thus, the description in terms of pressure wave is more appropriate than the description in terms of the dilacement wave as far as sound properties are concerned. SOUND WAVES IN SOLIDS Sound waves can travel in solids just like they can travel in fluids. The eed of longitudinal sound waves in a solid rod can be shown to be Where Y is the Young’s modulus of the solid and its density. For extended solids, the eed is a more complicated function of bulk modulus and shear modulus. Table gives the eed of sound in some common materials. Medium eed m/s Medium eed m/s Air (dry o0 C) 332 Copper 3810 Hydrogen 1330 Aluminum 5000 Water 1486 Steel 5200 Effect of Pressure, Temperature and Humidity on the eed of Sound in Air We have stated that for an ideal gas, the pressure, volume and temperature of a given mass satisfy As the density of a given mass is inversely proportional to its volume, the above equation may also be written as Where c is a constant. The eed of sound is …. (xvii) Thus, if pressure is changed but the temperature is kept constant, the density varies proportionally and P/ remains constant. The eed of sound is not affected by the change in pressure provided the temperature is kept constant. If the temperature of air is changed then the eed of sound is also changed. From equation (xvii), At STP, the temperature is 00 C or 273 K. If the eed of sound at 00 C is v0, its value at the temperature T (in Kelvin) will satisfy , Where t is the temperature in 0C. This may be approximated as or, . The density of water vapour is less than dry air at the same pressure. Thus, the density of moist air is less than that of dry air. As a result, the eed of sound increases with increasing humidity. INTENSITY OF SOUND WAVES As a wave travels in a medium, energy is tranorted from one part of the ace to another part. The intensity of a sound wave is defined as the average energy crossing a unit cross-sectional area perpendicular to the direction of propagation of the wave in unit time. It may also be stated as the average power transmitted across a unit cross-sectional area perpendicular to the direction of propagation. The loudness of sound that we feel is mainly related to the intensity of sound. It also depends on the frequency to some extent. Consider again a sound wave travelling along the x-direction. Let the equations for the dilacement of the particles and the excess pressure developed by the wave be given by ….(xviii) Where. Consider a cross-section of area A perpendicular to the x-direction. The medium to the left to it exerts a force pA on the medium to the right along the X-axis. The points of application of this force move longitudinally, that is along the force, with a eed . Thus, the power W, transmitted by the wave from left to right across the cross-section considered, is . By (xviii), W = Ap0 cosω(t – x/v)ωs0 cosω(t – x/v) . The average of cos2 ω(t – x/v) over a complete cycle or over a long time is ½. The intensity I, which is equal to the average power transmitted across unit cross-sectional area is thus, As B = ρv2, the intensity can also be written as We see that the intensity is proportional to the square of the pressure amplitude P0. Loudness Human ear is sensitive for extremely large range of intensity. So a logarithmic rather than an arithmetic scale is convenient. Accordingly, intensity level β of a sound wave is defined by the equation decibel Where I0 = 10–12 W/m2 is the reference or threshold intensity level to which any intensity I is compared. Illustration 6: Calculate the velocity of sound in air at N.T.P. The density of air at N.T.P. is 1.29 gm/ . Assume air to be diatomic with Hence calculate the velocity of sound in air at 270C. Solution: Velocity of sound in air = = = 331.6m/s. Using v2 = v1 We can see that the velocity of sound is proportional to the square root of absolute temperature. Illustration 7: A source of sound emits a total power of 10W in all directions. Find the distances from the source where the sound level is 100 dB. Solution: The intensity level () of a sound wave I = 10–12 ×1010 = 10–2 Wm2 Decibels (dB) and is given by Using I = = 10 og10 10–2 = Where I is the intensity in W/m2 and I0 = 10–12W/m2 is the weakest sound intensity That can be heard. = 100dB r = = 8.92m. 100 = 10 log10 I = I0 × 1010 STATIONARY WAVES IN AIR COLUMN Open pipe: If both ends of a pie are open and a system of air is directed against an edge, standing longitudinal wave can be set up in the tube. The open ends are a dilacement antinodes and pressure nodes a) For fundamental mode of vibrations, …(xxii) b) For the second harmonic or first overtone, …(xxiii) c) For the third harmonic or second overtone, …(xxiv) From (xxii), (xxiii) and (xxiv) we get, f1 : f2 : f3 : ……………. = 1 : 2 : 3 : …………. i.e. for a cylindrical tube, open at both ends, the harmonics excitable in the tude are all integral multiples of its fundamental. ∴ In the general case, Frequency = Closed pipe: If one end of a pipe is closed the reflected wave is 1800 out of phase with the incoming wave. Thus the dilacement of the small volumes elements at the closed end must always be zero. Hence the closed end must be a dilacement node and pressure antinode a) This represents the fundamental mode of vibration, , If f1 is the fundamental frequency, then the velocity of sound waves is given as, ….(xxv) b) This is the third harmonic or first overtone. ….(xxvi) c) This is the fifth harmonic or seconds overtone. ….(xxvii) From (xxv), (xxvi) and (xxvii) we get, f1 : f2 : f3 : …………. = 1 : 3 : 5 : ……….. In general, where n = 0, 1, 2 …….. Velocity of sound = v Frequency = where n = 0, 1, 2 ……… Illustration 8: A tube closed at one end has a vibrating diaphragm at the other end, which may be assumed to be dilacement node. It is found that when the frequency of the diaphragm is 2000 Hz, a stationary wave pattern is set up in which the distance between adjacent nodes is 8 cm. When the frequency is gradually reduced, the stationary wave pattern disappears but another stationary wave pattern reappears at a frequency of 1600 Hz. Calculate (i) the eed of sound in air, (ii) the distance between adjacent nodes at a frequency of 1600 Hz, (iii) the next lower frequencies at which stationary wave patterns will be obtained. Solution: Since the node-to-node distance is /2 /2 = 0.08 or = 0.16 m (i) c = n c = 2000 0.16 = 320 ms–1 (ii) 320 = 1600 or = 0.2 m distance between nodes = 0.2/2 = 0.1 m = 10 m (iii) Since there are nodes at the ends, the distance between the closed end and the membrane must be exact integrals of /2. 0.4 = 2/2 = n’ 0.2/2 When n = 5, n’ = 4 I = n 0.16/2 = 0.4 m = 40cm. (iv) For the next lower frequency n = 3, 2, 1 0.4 = 3/2 or = 0.8/3 Since c = n, Hz Again 0.4 = 1. /2 or = 0.4 m n = 320 / 0.4 = 800 Hz Again 0.4 = 1. /2 or = 0.8 m n = 320 / 0.8 = 400 Hz. BEATS When two interacting waves have slightly different frequencies the resultant disturbance at any point due to the superposition periodically fluctuates causing waxing and waning in the resultant intensity. The waxing and waning in the resultant intensity of two superposed waved of slightly different frequency are known as beats. Let the dilacement produced at a point by one wave b y1 = A sin(2πf1t – φ) And the dilacement produced at the point produced by the other wave of equal amplitude as y2 = A sin(2πf2t – φ2) By the principle of superposition, the resultant dilacement is Y = y1 + y2 = A sin(2πf1t – φ1) + A sin (2πf2t – φ2) Where, The time for one beat is the time between consecutive maximum or minima. First maxima would occur when Then or The time for one beat = Similarly it may also be shown that time between two consecutive minima is . Hence, frequency of beat i.e. number of beats in one second or Beat frequency = f1 ~ f2. Illustration 9: A column of air at 510C and a tuning fork produce 4 beats per sec when sounded together. As the temperature of the air column is gradually decreased to 160C, they again produce 1 beat/s. Find the frequency of the tuning fork. It is found that they again produce 1 beat/s if temperature is further decreased explain. Solution: Let n = frequency of the tuning fork: The number of beats per sec decreases n = 50 Hz as the frequency of the air column decreases. They can again produce 1 beat/s when Hence the frequency of air column is initially air column frequency reduced to [n – 1] (n + 4) at 510C. At 16oC, the frequency of air at a further low temperature T. column can be (n + 1). n + 4 c51 n + 4 c16 Where c is the eed of sound = Illustration 10: A closed air column 32 cm long is in resonance with a tuning fork. Another open air column of length 66 cm is in resonance with another tuning fork. If the two forks produce 8 beats per sec when sounded together, find: (a) the eed of sound in the air (b) the frequencies of the forks. Solution: Let c be the eed of sound and v1, v2 be the As v1 > v2, we have v1 – v2 = 8 Frequencies of tuning forks. v1 = = = v2 = C = = 33792 cm/s Now |v1 – v2| = 8 DOPPLER EFFECT The apparent shift in frequency of the wave motion when the source of sound or light moves with reect to the observer, is called Doppler Effect. Calculation of Apparent Frequency Suppose v is the velocity of sound in air, vs is the velocity of the source of sound(s), vo is the velocity of the observer (O), and f is the frequency of the source. (i) Source moves towards stationary observer. If the source were stationary the f waves sent out in one second towards the observer O would occupy a distance v, and the wavelength would be v/f. If S moves with a velocity vs towards O, the f waves sent out occupy a distance (v - vs) because S has moved a distance vs towards O in 1s. So the apparent wavelength would be λ′ = Thus, apparent frequency f ′ = f ′ = (ii) Source moves away from stationary observer. Now, apparent wavelength λ′ = ∴ Apparent frequency f ′ = v/λ′ or f ′ = f (iii) Observer moves towards stationary source. f ′ = Here, velocity of sound relative to O = v + vO and wavelength of waves reaching O = v/f ∴ f ′ = (iv) Observer moves away from the stationary source. f ′ = (v) Source and observer both moves toward each other. f ′ = (vi) Both moves away from each other. f ′ = (vii) Source moves towards observer but observer moves away from source f = f ′ (viii) Source moves away from observer but observer moves towards source f = f ′ Wind's Effect The above formulae can be modified by taking the wind effects into account. The velocity of sound should be taken as v + vw or v - vw if the wind is blowing in the same or opposite direction as SO (source to observer). Illustration 11: Two trains are moving with speeds 30m/s and 25m/s towards each other. The frequency of the whistle of the faster train is 500 Hz. What is the apparent frequency of this whistle if heard by an observer on the slower train? What will be the apparent frequency after the trains have crossed each other? Solution: (a) The faster train's speed is the source's speed and the other is the observed speed Taking the direction from S to O as +ve, Vs = +30 m/s, V0 = –25 m/s v' = v' = v' = = 547.62 Hz (b) Behind the train: After train have crossed each other, they will move away from each other as shown Now V0 = + 25 m/s, Vs = –30 m/s v' = v = 500 = 426.67 Hz. ASSIGNMENT 1. A wave disturbance in a medium is described by y(x, t) = 0.02 cos cos(10πx) where x and y are in metre and t is in second: (A) A node occurs at x = 0.15 m (B) An antinode occurs at x = 0.3m (C) The speed wave is 5 ms-1 (D) The wave length is 0.3 m Solution: (C) Comparing it with y(x, t) = A cos(ωt + π/2)cos kx. If kx = π/2, a node occurs; ∴10πx = π/2 ⇒ x = 0.05m If kx = π, an antinode occurs ⇒ 10πx = x ⇒ x = 0.1 m Also speed of wave = ω/k = and λ = 2π/k = 2π/10π= 0.2 2. The extension in a string, obeying Hooke’s law, is x. The speed of sound in the stretched string is v. If the extension in the string is increased to 1.5x, the speed of sound will be (A) 1.22v (B) 0.61v (C) 1.50v (D) 0.75v Solution: (A) According to Hooke’s law FR ∝ x [Restoring Force FR =T, tension of ring] Velocity of sound by a stretched string where m is the mass per unit length ∴ ⇒ 3. A whistle giving out 450 Hz approaches a stationary observer at a speed of 33 m/s. The frequency heard by the observer in Hz is (A) 409 (B) 429 (C) 517 (D) 500 Solution: (D) 4. A travelling wave in a stretched string is described by the equation y = A sin (kx – ωt). The maximum particle velocity is (A) Aω (B) ω/k (C) dω/dk (D) x/t Solution: (A) ∴ Vmax = Aω 5. A siren placed at a railway platform is emitting sound of frequency 5kHz. A passenger sitting in a moving train A records a frequency of 5.5 kHz while the train approaches the siren. During his return journey in a different train B he records a frequency of 6.0 kHz while approaching the same siren. The ratio of the velocity of train B to that train A is (A) 242/252 (B) 2 (C) 5/6 (D) 11/6 Solution: (B) and ⇒ 6. An organ pipe P1 closed at one end vibrating in its first harmonic and another pipe P2 open at ends vibrating in its third harmonic are in resonance with a given tuning fork. The ratio of the length of P1 to that of P2 is: (A) 8/3 (B) 3/8 (C) 1/6 (D) 1/3 Solution: (C) for first harmonic for third harmonic ∴ ⇒ 7. Two whistles A and B have frequencies 600Hz and 590Hz reactively. An observer is standing in the middle of the line joining the two sources. Source B and observer are moving towards right with velocity 30m/s and A is standing at left side. If the velocity of sound in air is 300m/s, the number of beats listened by the observer are: (A) 2 (B) 4 (C) 6 (D) 8 Solution: (B) Frequency of A as heard by the observer N1 = 660 = 594Hz. Frequency of B as heard by observer = 594Hz, as there is no relative motion between source and observer. ∴ Beats frequency = (594 − 590)= 4 per/s 8. A cylindrical tube, open at both ends has a fundamental frequency 'f' in air. The tube is vertically dipped in water so that half of it is in water, the fundamental frequency of the air column is (A) f/2 (B) 3f/4 (C) f (D) 2f Solution: (C) For open tube f = f = v/2. On dipping the tube in water, it becomes a closed tube. For closed tube f′ = = = f 9. The amplitude of a wave disturbance propagating in the positive x-direction is given by y =1/ (1 + x2) at time t = 0 and by y = 1/[1 + (x – 1)2 at t = 2 seconds, where x and y are in metres. The shape of the wave disturbance does not change during the propagation. The velocity of the wave is (A) 1 ms−1 (B) 0.5 ms−1 (C) 1.5 ms−1 (D) 2 ms−1 Solution: (B) Writing the general expression for y in terms of x as y = at t = 0, y = 1/ (1 + x)2 At t = 2 s, y = Comparing with the given equation we get 2v = 1 and v = 0.5 m/s. 10. A train has just completed a U-curve in a track which is a semicircle. The engine is at forward end of the semicircular part of the track while the last carriage is at the rear end of the semi-circular track. The driver blows a whistle of frequency 200Hz. Velocity of sound is 340 m/s. Then the apparent frequency as observed by a passenger in the middle of the train, when the need of the train is 30 m/s is (A) 219Hz (B) 288 Hz (C) 200Hz (D) 181Hz Solution: (C) Velocity component of the source in the direction of motion of sound = 30 cos45° along BA. Velocity component of observer in the direction BA = 30 cos45°. ∴ There is no relative motion between the source and the observer, hence no change in real frequency is observed. 11. Two waves represented by y1 =10 sin(2000 πt+2x) and y2=10 sin (2000 π t +2x+π/2) are superposed at any point at a particular instant. The resultant amplitude is: (A) 10 units (B) 20 units (C) 14.1 units (D) zero Solution: (C) The resultant amplitude A of two waves of amplitudes a1 and a2 at a phase difference φ is ( Substituting a1 = 10, a2 = 10 and φ = 900, we get A = 14.1. 12. A transverse wave is described by the equation y = y0 sin 2π (ft – x/a). The maximum particle velocity is equal to four times the wave velocity if a is equal to (A) πy0 / 4 (B) πy0/2 (C) πy0 (D) 2πy0 Solution: (B) The maximum particle velocity of a SHM of amplitude Y0 and frequency f is 2πfY0. The wave velocity is fλ. For 2π fY0 to be equal to 4fλ, λ has to be πY0 / 2 (Here λ = a). 13. A standing wave having three antinodes and four nodes is formed between two atoms having a distance of 2A° between them. The wavelength of the standing wave is (A) 1.33A° (B) 6 A° (C) 1.4 A° (D) 8 A° Solution: (A) 3λ/2 = 2A° λ = (4/3)A° = 1.33 A° 14. The speed of sound through oxygen at T K is (300 m/s). When the temperature is increased to 3T, the molecule dissociates into oxygen atom, now the speed of sound will be (A) 520 m/s (B) 801 m/s (C) 600 m/s (D) 580 m/s Solution: (B) vT = = 300 m/s v3T = = 2.67 v3T = 300 × 2.67 = 801 m/s 15. Standing wave is formed in string between two fixed ends. If separation between two consecutive antinode is 2 mm and length of string is 9 cm. Then harmonics in which string is vibrating (A) 2nd harmonics (B) 9th harmonics (C) 45th harmonics (D) 90th harmonics. Solution: (C) If there exits pth harmonic P× 2 = 90 P = 45 16. A source of sound emitting a note of frequency 200Hz moves towards an observer with a velocity v equal to the velocity of sound. If the observer also moves away from the source with the same velocity v, the apparent frequency heard by the observer is (A) 50Hz (B) 100Hz (C) 150Hz (D) 200Hz Solution: No relative motion between source and observers. Hence (D). 17. A piano string 1.5m long is made of steel of density 7.7103 kg/m3 and Young's modulus 21011 N/m2. It is maintained at a tension which produces an elastic strain of 1% in the string. The fundamental frequency of transverse vibrations of string is (A) 85Hz (B) 170Hz (C) 340Hz (D) 310Hz Solution: Y = =21011A n = n = 170Hz 18. The stationary waves set up on a string have the equation . This stationary wave is created by two identical waves, of amplitude A, each moving in opposite directions along the string. The value of A. (A) A = 2 mm (B) A = 1 mm (C) The smallest length of the string is 60 cm (D) The smallest length of the string is 2 m. Solution : Comparing with the equation Y = 2A sin 2A = 2 mm ⇒ A = 1 mm or L = for n = 1, L = 50 cm ∴ (B) 19. An accurate and reliable audio oscillator is used to standardise a tuning fork. When the oscillator reading is 514, two beats are heard per second. When the oscillator reading is 510, the beat frequency is 6 Hz. The frequency of the tuning fork is (A) 506 (B) 510 (C) 516 (D) 158 Solution: When the oscillator reading is 514, two beats are heard. Hence the frequency f the tuning fork is 514 ± 2 = 516 or 512. When the oscillator reading is 510, the frequency of the tuning fork is 510 ± 6 = 516 or 504. The common value is 516. Hence the frequency is 516 Hz. ∴ (C) 20. A sound wave of wavelength λ travels towards the right horizontally with a velocity v. It strikes and reflects from a vertical plane surface, travelling at acceleration a starting from rest. The number of positive crests striking in a time interval of 5 sec on the wall is (A) 5 (B) 5 (C) 5v/ λ (D) (v – 5a) /5λ Solution : ∴ Distance covered = v.5 + a.52 = 5[v+(5a/2)] Number of positive crests striking per second is same as the frequency. ∴ (A) 21. Two waves represented by y1 =10 sin (2000 πt+2x) and y2=10 sin (2000 π t +2x+π/2) are superposed at any point at a particular instant. The resultant amplitude is (A) 10 units (B) 20 units (C) 14.1 units (D) zero Solution: The resultant amplitude A of two waves of amplitudes a1 and a2 at a phase difference φ is (. Substituting a1 = 10, a2 = 10 and φ = 900, we get A = 14.1. ∴ (C) 22. When pressure increases by 1/2 atm o here and temperature increases by 10C from (100°C), the velocity of sound may, (Take vs = 340 m/s) (A) increases by 0.455 ms−1 (B) decreases by 4.55 ms−1 (C) increases by 0.455 ms−1 (D) decreases by 4.55 ms−1 Solution : (A) V ∝ and the increase of pressure does not change the velocity of sound. ⇒ V′ = V0 ⇒ 340 ⇒ 340.4554 m/s 23. Which of the following wave equations have maximum amplitude at the origin of coordinate system at time t = 0 sec? (A) y = ym sin k (x + vt) (B) y = ym cos k (x + vt) (C) y = ym sin k(x − vt) (D) none of these. Solution : (D) Because y = ym cos(0°) = ym 24. A heavy uniform rope hangs vertically from the ceiling, with its lower end free. A disturbance on the rope travelling upward from the lower end has a velocity v at a distance x from the lower end. (A) (B) v α x (C) (D) Solution: Let m be the mass per unit length of the rope. T = tension in the rope at a distance x from the lower end. ∴ T = mgx ⇒ v α ∴ (C) 25. If x = a sin [ωt + (π/3)] and x′ = a cos ωt, then what is the phase difference between two waves? (A) π/3 (B) π/6 (C) π/2 (D) π Solution: Now sin[ωt + (π/3)] and cosωt = sin[(π/2) + ωt] ⇒ Phase difference = π/2 − π/3 = π/6 ∴ (B) 26. The power of sound from a speaker is raised from 10 mW to 500 mW. What is the power in creased in (decibel) dB as compared to initial original power is (A) 1.6 dB (B) 50 dB (C) 16.9 dB (D) 6.9 dB Solution: P(dB) = 10 log = 10 log = 10log1050 = 10(1.69) = 16.9 dB ∴ (C) CMP : A small here of radius R is arranged to pulsate so that its radius varies in simple harmonic motion between a minimum of R – R and a maximum of R + R with frequency f. This produces sound waves in the surrounding air of density and bulk modulus B (The amplitude of oscillation of the here is the same as that of the air at the surface of the here). 27. Find the intensity of sound waves at the surface of the here (A) (B) (C) (D) Solution: (B) The density is (by definition) the time average value of P (x, t)vy (x, t). For any value of x the average value of function cos2 (Kx – t) over one period\ is half By using the relation and So we get (intensity of sinusoidal wave) The amplitude of oscillation is . Hence, 28. The total acoustic power radiated by the here will be (A) (B) (C) (D) Solution: (D) 29. At a distance d > > R from the centre of the here, find the amplitude. (A) (B) (C) (D) Solution: (A) But Hence 30. At a distance d > > R from the centre of the here, find pressure amplitude (A) (B) (C) (D) Solution: (C) Hence, For Test 1. When we hear a sound, we can identify its source from (A) the frequency of the sound (B) the amplitude of the sound (C) the wavelength of the sound (D) the overtones present in the sound Solution: ∴ (D) 2. A man is riding a motorcycle with velocity v, towards a stationary car sounding the horn at the frequency of 165 Hz. Police jeep is following the motorcycle at 22 m/s and sounding a siren with a frequency of 176 Hz. Find v so that the man on the motorcycle hears no beats. (A) 33 m/s (B) 22 m/s (C) 0 (D) 11 m/s Ans. jeep → motorcycle → car For no beats the person should hear and apparent frequency of 176 Hz. By stationary car ∴ 176 = 165 (vs = velocity of sound) ⇒ v = 22 m/s ∴ (B) 3. In a resonance tube water is filled so that height of air column is 0.1 m when it resonates in its fundamental mode. Now water is removed so that the height of air column becomes 0.35 m and it resonates with next higher frequency. Then end correction is (A) 0.012 (B) 0.025 (C) 0.05 (D) 0.024 Ans. End correction = = 0.025 ∴ (B) 4. The length of nth seconds pendulum on the surface of Earth is 1 m. Its length on the surface of Moon, where g is 1/6th the value of g on the surface of Earth, is (A) 1/6 m (B) 6 m (C) 1/36 m (D) 36 m. Solution : (A) T = 2p ; T is constant. 5. A particle moves according to the equation x = a cos πt. The distance covered by it in 2.5 s is (A) 3a (B) 5a (C) 2a (D) 9a. Solution : (B) clearly, oscillations are completed in 2.5 second. 6. The time period of a mass suspended from a ring is 5s. The ring is cut into four equal parts and the same mass is now suspended from one of its parts. The period is now (A) 5s (B) 2.5s (C) 1.25s (D) . Solution : (B) 7. Two pendulum oscillate with a constant phase difference of 90o and same amplitude. The maximum velocity of one is v. The maximum velocity of the other is v + x. The value of x is (A) 0 (B) (C) (D) v tan 90o. Solution : (A) When, one is at the mean position, the other is at the extreme position. Since phase difference I constant. Therefore ω is same for both pendulum. Also, a is same. So, α ω is same. 8. The displacement of a particle in simple harmonic motion in one time period is (A) A (B) 2A (C) 4A (D) zero. Solution : (D) At the end of one complete vibration, the particle returns to the initial position. 9. If the displacement equation of a particle be represented by y = A sin pt + B cos pt, the particle executes (A) a uniform circular motion (B) a uniform elliptical motion (C) a SHM (D) a rectilinear motion. Solution : (C) y = A sin pl + B cos pt V = cos pt – Bp sin pt Acceleration = –Ap2 sub pt–Bp2 cos pt = –p2 (A sin pt + B cos pt) = –p2y Clearly, motion is SHM. 10. The time period of a mass sounded by a ring is T. If the ring is cut into four equal parts and the mass is again sounded by one of the pieces, the period of vibration is now (A) T/2 (B) (C) (D) 2T. Solution : 11. The need of sound in air is 350 ms–1. The fundamental frequency of an open pipe 50 cm long will be (A) 175 Hz (B) 350 Hz (C) 700 Hz (D) 50 Hz. Solution : (C) = 12. A flat horizontal board moves us and down in SHM of amplitude a. Then the shortest permissible time period of the vibration such that an object placed on the board may not lose contact with the board is (A) 2π (B) 2π (C) 2π (D) . Solution : (B) mαω2 = mg or . 13. A progressive sound wave of frequency 500 Hz is travelling through air with a speed of 350 ms–1. A compression maximum appears at a place at a given instant. The minimum appears at a place at a given instant. The minimum occurs at the same point, is (A) 200 s (B) (C) (D) . Solution : (D) 14. A uniform ring of force constant k is cut into two pieces, the lengths of which are in the ratio 1:2. The ratio of the force constants of the shorter and the longer pieces is (A) 1 : 2 (B) 2 : 3 (C) 2 : 1 (D) 1 : 3. Solution : (C) or or kl = constant or or . 15. A mass on the end of a ring undergoes simple harmonic motion with a frequency of 0.5 Hz. If the attached mass is reduced to one quarter of its value, then the new frequency in Hz is (A) 4.5 (B) 2.0 (C) 0.25 (D) 1.0. Solution : (D) or ν’ = 2 × 0.5 Hz = 1 Hz 16. The motion of particle executing SHM is given by x = 0.01 sin 100 π (t + 0.05), where x is in meter and time is in second. The time period is (A) 0.2 s (B) 0.1 s (C) 0.02 s (D) 0.01 s. Solution : (C) Comparing with x = α sin (ωt + φo), we get ω = 100 π or or T = 17. A pendulum sounded from the ceiling of a train has a period T when the train is at rest. When the train is accelerating with a uniform acceleration, the period of oscillation will (A) increase (B) decrease (C) remain unaffected (D) become infinite. Solution : (B) Comparing with y = Clearly, T’ < T 18. What fraction of the total energy is potential when the displacement is one-half of the amplitude? (A) 200 s (B) (C) (D) . Solution : (C). 19. Consider a stretched string under tension and fixed at both ends. If the tension is doubled and the cross-sectional area halved, then the frequency becomes (A) twice (B) half (C) four times (D) eight times. Solution : (A) But 20. Two organ pipes produce 7 beats per second at 5oC. When the temperature rises to 10oC, the number of beats is (A) = 7 (B) < 8 (C) > 7 (D) Data is inadequate. Solution : (C) 21. Two tuning forks when sounded simultaneously give one beat each 0.4 second. What is the difference of the frequencies? (A) 1 Hz (B) 1.5 Hz (C) 2 Hz (D) 2.5 Hz. Solution : (D) Beat frequency = Difference of frequencies = 2.5 Hz 22. The ratio of intensity of wave and energy density gives (A) momentum (B) total energy (C) propagation constant (D) velocity. Solution : (D) Intensity = Energy density = = 23. Two sound waves travel in the same direction in a medium. The amplitude of each wave is A and the phase difference between the two waves is 120o. The resultant amplitude will be. (A) A (B) 2A (C) 4A (D) . Solution : (A) A’2 = A2 + A2 +2A2 cos 120o = 2A2 + 2A2 24. The intensity ratio of two waves is 1 : 16. The ratio or their amplitudes is (A) 1 : 16 (B) 1 : 4 (C) 4 : 1 (D) 2 : 1. Solution : (B) 25. If the speed of sound at 0oC is ν0, then the velocity at 273oC will be (A) 2ν0 (B) 0.5ν0 (C) 1.4ν0 (D) 0.72ν0. Solution : (C) 26. The equation of transverse wave in stretched string is Where distances are in cm and time in second. The wavelength of wave is (A) 15 cm (B) 10 cm (C) 25 cm (D) 50 cm. Solution : (D) CMP: A person observe two points on a string as a travelling wave passes them. The points are at x1 = 0 and x2 = 1m. The transverse motions of the two points are found to be as follows: y1 = 0.2 sin 3πt y2 = 0.2 sin (3 πt + π/8) 27. What is the frequency in hertz? (A) 1.5 Hz (B) 3 Hz (C) 4.5 Hz (D) 1 Hz Solution: (A) 28. What is the maximum wavelength? (A) 32 m (B) 16 m (C) 8 m (D) 4 m Solution: (B) For maximum wave length Phase diff = path difference i.e., 29. With what maximum speed does the wave travel? (A) 48 m/s (B) 24 m/s (C) 16 m/s (D) 8 m/s Solution: (B) Maximum wave speed = 24 m/s 30. Which way is the travelling? Show how you reach this conclusion. (A) Positive x–axis (B) Negative x–axis (C) Along y–axis (D) Data insufficient Solution: (D) From the given data it is not possible to predict that wave is moving in +ve or –ve direction.