Waves PDF

APNI KAKSHA 1  WAVES Introduction Till now we studied about the objects oscillating in isolation. We will study the same oscillation in a system (collection of such objects) Elastic forces bind the constituents to each other and, therefore, the m...

APNI KAKSHA 1  WAVES Introduction Till now we studied about the objects oscillating in isolation. We will study the same oscillation in a system (collection of such objects) Elastic forces bind the constituents to each other and, therefore, the motion of one affect that of the other The patterns, which move without the actual physical transfer or flow of matter as a whole, are called WAVES They transport energy and the pattern of disturbance has information that propagate from one point to another. Types of Waves 1. Electromagnetic Waves Does not require medium for propagation. E.g., – Light Waves, Radio Waves, etc. 2. Mechanical Waves Requires Medium for its Propagation. Involve oscillations of constituent particles and depend on the elastic properties of the medium. E.g., – Water Waves, Seismic Waves, Sound Waves, Water Waves, etc. 3. Matter Waves Associated with constituents of matter: electrons, protons, neutrons, atoms and molecules They have already found applications in several devices basic to modern technology; matter waves associated with electrons are employed in electron microscopes. In this chapter our focus will be on Mechanical Waves which require material medium for their propagation. Propagation of Sound Waves Wave passing through air, it compresses or expands a small region of air This causes a change in the density of that region, say δρ, this change induces a change in pressure, δρ, in that region. The quantity similar to extension or compression of the spring is the change in density. If a region is compressed, the molecules in that region are packed together, and they tend to move out to the adjoining region, thereby increasing the density or creating compression in the adjoining region. If a region is comparatively rarefied the surrounding air will rush in making the rarefaction move to the adjoining region The compression or rarefaction moves from one region to another, making the propagation of a disturbance possible in air APNI KAKSHA 2 TRANSVERSE AND LONGITUDINAL WAVES If the constituents of the medium oscillate perpendicular to the direction of wave propagation, we call the wave a Transverse wave. As the wave propagates, each element of the medium undergoes a shearing strain Transverse waves can be propagated only in those media, which can sustain shearing stress, such as solids and not in fluids. Transverse waves on water are due to Surface Tension of Water Pulse: Resulting from a single up and down jerk to one end of the string Harmonic Wave: The external agent gives a continuous periodic sinusoidal up and down jerk to one end of the string. Above oscillations are normal to the direction of wave motion along the string, so this is an example of transverse wave. Waves can be looked by: 1. Fixing an instant of time and picture the wave in space. This will give us the shape of the wave as a whole in space at a given instant. 2. Fixing a location i.e., fix our attention on a particular element of string and see its oscillatory motion in time. If the constituents of the medium oscillate along the direction of wave propagation, we call the wave a Longitudinal wave. Fluids and solids can sustain compressive strain; therefore, longitudinal waves can be propagated in all elastic media In medium like steel, both transverse and longitudinal waves can propagate, while air can sustain only longitudinal waves. A long pipe filled with air has a piston at one end. A single sudden push forward and pull back of the piston will generate a pulse of condensations (higher density) and rarefactions (lower density) in the medium (air). If the push-pull of the piston is continuous and periodic (sinusoidal), a sinusoidal wave will be generated propagating in air along the length of the pipe. This is clearly an example of longitudinal waves APNI KAKSHA 3  Q. Given below are some examples of wave motion. State in each case if the wave motion is transverse, longitudinal or a combination of both: (A) Motion of a kink in a longitudinal spring produced by displacing one end of the spring sideways. (B) Waves produced in a cylinder containing a liquid by moving its piston back and forth. (C) Waves produced by a motorboat sailing in water. [NCERT Exercise] (D) Ultrasonic waves in air produced by a vibrating quartz crystal. Sol. (A) Transverse and longitudinal (B) Longitudinal (C) Transverse and longitudinal (D) Longitudinal DISPLACEMENT RELATION IN A PROGRESSIVE WAVE A function of both position x and time t This function at every instant should give the shape of the wave at that instant It should describe the motion of the constituent of the medium at that location If we wish to describe a sinusoidal travelling wave the corresponding function must also be sinusoidal A sinusoidal travelling wave is then described by- y(x, t) = asin⁡(kx − ωt + ϕ) y(x, t) : displacement as a function of position x and time t a : amplitude of a wave ω : angular frequency of the wave k : angular wave number kx − ωt + ϕ : initial phase angle (a + x = 0, t = 0) Crest: The point of maximum positive displacement Trough: The point of maximum negative displacement. Amplitude and Phase a represents the maximum displacement of the constituents of the medium from their equilibrium position Displacement y may be positive or negative, but a is positive. It is called the Amplitude of the wave Quantity (kx–ωt + φ) appearing as an argument of Sine function is called Phase of the wave Given the amplitude a, the phase determines the displacement of the wave at any position and at any instant. APNI KAKSHA 4 Wavelength and Angular Wave Number The minimum distance between two points having the same phase is called the wavelength of the wave, usually denoted by λ The wavelength is then the distance between two consecutive crests or troughs in a wave 2π 2π λ= or k = k λ K – Angular wave number Period, Angular Frequency and Frequency Time Period: Time it takes for an element to complete one full oscillation. 2π ωT = 2π or ω = T Where  is called Angular Frequency Frequency: The number of oscillations per second. 1 ω v= = T 2π Q. A wave travelling along a string is described by, 𝐲(𝐱, 𝐭) = 𝟎. 𝟎𝟎𝟓⁡𝐬𝐢𝐧⁡(𝟖𝟎. 𝟎𝐱 − 𝟑. 𝟎t). in which the numerical constants are in SI units (𝟎. 𝟎𝟎𝟓 𝐦, 𝟖𝟎. 𝟎𝐫𝐚𝐝⁡𝐦−𝟏 , and 𝟑. 𝟎𝐫𝐚𝐝⁡𝐬 −𝟏 ). Calculate (a) the amplitude, (b) the wavelength, and (c) the period and frequency of the wave. Also, calculate the displacement 𝐲 of the wave at a distance 𝐱 = 𝟑𝟎. 𝟎 𝐜𝐦 and time 𝐭 = 𝟐𝟎 𝐬? [NCERT Exercise] Sol. On comparing this displacement equation with the standard equation. y(x, t) = asin⁡(kx − ωt).⁡we find (a) the amplitude of the wave is 0.005 m = 5 mm. (b) the angular wave number k and angular frequency ω are k = 80.0 m−1 and ω = 3.0 s −1 We, then, relate the wavelength λ to k. 2π λ = 2π/k = = 7.85cm 80.0m−1 2𝜋 (c) Now, we relate T to ω by the relation T = 2𝜋/𝜔 = 3.0s−1 = 2.09s and frequency, v = 1/T = 0.48 Hz The displacement y at x = 30.0 cm and time t = 20 s is given by y = (0.005 m) sin(80.0 × 0.3 − 3.0 × 20) = (0.005 m) sin(−36 + 12π) = (0.005 m) sin(1.699) = (0.005m)sin⁡(97∘ ) = 5mm APNI KAKSHA 5 THE SPEED OF A TRAVELLING WAVE To determine the speed of propagation of a travelling wave, it is convenient to look at the motion of the crest of the wave. A in figure gives the shape of the wave at two instants of time, which differ by a small time internal ∆t. In particular, the crest shown by a dot ( ) moves a distance ∆x in time ∆t. The speed of the wave is then ∆x/∆t. The motion of a fixed phase point on the wave is given by (kx − ωt) = constant Thus, as time t changes, the position x of the fixed phase point must change so that the phase remains constant. Thus, kx − ωt = k(x + Δx) − ω(t + Δt) or kΔx − ωΔt = 0 Taking Δx, Δt vanishingly small, this gives dx ω = =v dt k Relating ω to T and k to λ, we get 2πv λ v= = λv = 2π/λ T It should be noted that the speed of a mechanical wave is determined by the Inertial (linear mass density for strings, mass density in general) and Elastic Properties (Young’s modulus for linear media/ shear modulus, bulk modulus) of the medium Speed of a Transverse Wave on Stretched String The speed of a mechanical wave is determined by the restoring force setup in the medium when it is disturbed and the inertial properties (mass density) of the medium. The speed is expected to be directly related to the former and inversely to the latter Using Newton’s Laws of Motion, an exact formula for the wave speed on a string can be derived, but this derivation is outside the scope of this book The speed of transverse waves on a stretched string is given by T v=√ μ Few important points: Depends on the property of the medium T and µ (T is a property of the stretched string arising due to an external force) It does not depend on wavelength or frequency of wave itself In higher studies, you will come across waves whose speed is not independent of frequency of the wave. Of the two parameters λ and ν the source of disturbance determines the frequency of the wave generated. APNI KAKSHA 6 Speed of a Longitudinal Wave (Speed of Sound) In a longitudinal wave, the constituents of the medium oscillate forward and backward in the direction of propagation of the wave. We have already seen that the sound waves travel in the form of compressions and rarefactions of small volume elements of air. The elastic property that determines the stress under compressional strain is the bulk ΔP modulus of the medium defined by B = − ΔV/V The general formula for longitudinal waves in a medium is B v=√ ρ The speed of longitudinal waves in a solid bar is given by Liquids and solids generally have higher speed of sound than gases. This happens because they are much more difficult to compress than gases and so have much higher values of bulk modulus Solids and liquids have higher mass densities (ρ) than gases. But the corresponding increase in both the modulus (B) of solids and liquids is much higher. This is the reason why the sound waves travel faster in solids and liquids. For an ideal gas. the pressure P, volume V and temperature T are related by PV = NκB T where N is the number of molecules in volume V, k B is the Boltzmann constant and T the temperature of the gas (in Kelvin). Therefore, for an Isothermal Change it follows from that ΔP VΔP + PΔV = 0 or − ΔV/V = P Hence, we have B = P Therefore, from the speed of a longitudinal wave in an ideal P gas is given by, v = √ρ This relation was first given by Newton and is known as Newton's formula. For adiabatic processes the ideal gas satisfies the relation PV γ = constant i. e. Δ(PV γ ) = 0 or ⁡PγV γ−1 ΔV + V γ ΔP = 0 where γ is the ratio of two specific heats, Cp /Cv ⋅ Thus, for an ideal gas the adiabatic bulk modulus is given by, ΔP Bad = − ΔV/V = γP γP The speed of sound is, therefore, given by, v = √ ρ This modification of Newton's formula is referred to as the Laplace correction. APNI KAKSHA 7  Q. Estimate the speed of sound in air at standard temperature and pressure. The mass of 1 mole of air is 𝟐𝟗. 𝟎 × 𝟏𝟎−𝟑 𝐤𝐠. [NCERT Exercise] Sol. We know that 1 mole of any gas occupies 22.4 litres at STP. Therefore, density of air at STP is: ρo = (mass of one mole of air) / (volume of one mole of air at STP) 29.0 × 10−3 kg = 22.4 × 10−3 m3 = 1.29 kg m−3 According to Newton's formula for the speed of sound in a medium, we get for the speed of sound in air at STP. 1 1.01 × 105 N m−2 2 v=[ ] 1.29 kg m−3 = 280 m s −1 THE PRINCIPLE OF SUPERPOSITION OF WAVES When the pulses overlap, the resultant displacement is the algebraic sum of the displacement due to each pulse. This is known as the principle of superposition of waves. According to this principle, each pulse moves as if others are not present. The constituents of the medium, therefore, suffer displacements due to both and since the displacements can be positive and negative, the net displacement is an algebraic sum of the two Let y1 (x, t) and y2 (x, t) be the displacements due to two wave disturbances in the medium. If the waves arrive in a region simultaneously, and therefore, overlap, the net displacement y (x, t) is given by y(x, t) = y1 (x, t) + y2 (x, t) If we have two or more waves moving in the medium the resultant waveform is the sum of wave functions of individual waves APNI KAKSHA 8 According to the two waves are described by the functions: y1 (x, t) = asin⁡(kx − ωt) and y2 (x, t) = asin⁡(kx − ωt + ϕ) The net displacement is then, by the principle of superposition, given by y(x, t) = asin(kx − ωt) + asin(kx − ωt + ϕ) (kx − ωt) + (kx − ωt + ϕ) ϕ = a [2sin⁡ [ ] cos⁡ ] 2 2 where we have used the familiar trigonometric identity for sin⁡ A + sin⁡ B. We then have ϕ ϕ y(x, t) = 2acos⁡ sin⁡ (kx − ωt + ) 2 2 is also a harmonic travelling wave in the positive direction of x-axis, with the same frequency and wavelength. However, its initial frequency and wavelength. However, its initial frequency and wavelength. However, its initial phase angle is ϕ. 2 The significant thing is that its amplitude is a function of the phase difference ϕ between the constituent two waves: A(ϕ) = 2acos⁡ 1/2ϕ For ϕ = 0, when the waves are in phase, y(x, t) = 2asin⁡(kx − ωt)⁡i.e., the resultant wave has amplitude 2a, the largest possible value for A. For ϕ = π, the waves are completely, out of phase and the resultant wave has zero displacement everywhere at all times y(x, t) = 0 refers to the so-called constructive interference of the two waves where the amplitudes add up in the resultant wave is the case of destructive interference where the amplitudes subtract out in the resultant wave. APNI KAKSHA 9 REFLECTION OF WAVES We considered waves propagating in an unbounded medium. What happens if a pulse or a wave meets a boundary? If the boundary is rigid, the pulse or wave gets reflected. The phenomenon of echo is an example of reflection by a rigid boundary If the boundary is not completely rigid or is an interface between two different elastic media, the situation is somewhat complicated A part of the incident wave is reflected and a part is transmitted into the second medium. If a wave is incident obliquely on the boundary between two different media the transmitted wave is called the refracted wave The incident and refracted waves obey Snell’s law of refraction, and the incident and reflected waves obey the usual laws of reflection Figure shows a pulse travelling along a stretched string and being reflected by the boundary. Assuming there is no absorption of energy by the boundary, the reflected wave has the same shape as the incident pulse but it suffers a phase change of π or 1800 on reflection. This is because the boundary is rigid and the disturbance must have zero displacement at all times at the boundary. By the principle of superposition, this is possible only if the reflected and incident waves differ by a phase of π, so that the resultant displacement is zero. This reasoning is based on boundary condition on a rigid wall. If on the other hand, the boundary point is not rigid but completely free to move (such as in the case of a string tied to a freely moving ring on a rod), the reflected pulse has the same phase and amplitude (assuming no energy dissipation) as the incident pulse. The net maximum displacement at the boundary is then twice the amplitude of each pulse. An example of non- rigid boundary is the open end of an organ pipe. mathematically, let the incident travelling wave be y2 (x, t) = asin⁡(kx − ωt) At a rigid boundary, the reflected wave is given by yr (x, t) = asin⁡(kx − ωt + π) = −asin⁡(kx − ωt) At an open boundary, the reflected wave is given by yr (x, t) = asin⁡(kx − ωt + 0) = asin⁡(kx − ωt) Clearly, at the rigid boundary, y = y2 + yr = 0 at all times. APNI KAKSHA 10 Standing Waves and Normal Modes In a string, a wave travelling in one direction will get reflected at one end, which in turn will travel and get reflected from the other end. This will go on until there is a steady wave pattern set up on the string. Such wave patterns are called standing waves or stationary waves. Mathematically, consider a wave travelling along the positive direction of x-axis and a reflected wave of the same amplitude and wavelength in the negative direction of x-axis. From Equation of Superposition of waves, with ϕ = 0, we get: y1 (x, t) = asin⁡(kx − ωt) ;⁡y2 (x, t) = asin⁡(kx + ωt) The resultant wave on the string is, according to the principle of superposition: y(x, t) = y1 (x, t) + y2 (x, t) = a[sin⁡(kx − ωt) + sin⁡(kx + ωt)] Using the familiar trigonometric identity Sin(A + B) + Sin(A − B) = 2 sin A cos B we get, y(x, t) = 2asinkxcosωt The amplitude of this wave is 2a sin kx In this wave pattern, the amplitude varies from point-to-point, but each element of the string oscillates with the same angular frequency ω or time period The wave pattern is neither moving to the right nor to the left. The amplitude is fixed at a given location but it is different at different locations The points at which the amplitude is zero (i.e., where there is no motion at all) are Nodes; the points at which the amplitude is the largest are called Antinodes The system cannot oscillate with any arbitrary frequency (contrast this with a harmonic travelling wave), but is characterized by a set of natural frequencies or normal modes of oscillation. The positions of nodes (where the amplitude is zero) are given by sin⁡ kx = 0. which implies kx = nπ; ⁡n = 0,1,2,3, … Since, k = 2π/λ, we get nλ x= ; n = 0,1,2,3, … 2 λ Clearly, the distance between any two successive nodes is 2. The positions of antinodes (where the amplitude is the largest) are given by the largest value of sin kx : |sin⁡ kx| = 1 which implies kx = (n + 1/2)π; n = 0,1,2,3, … APNI KAKSHA 11  With k = 2π/λ, we get λ x = (n + 1/2) ; n = 0,1,2,3, … 2 λ Again the distance between any two consecutive antinodes is 2. Above concept can be applied to the case of a stretched string of length L fixed at both ends. Taking one end to be at x = 0, the boundary conditions are that x = 0 and x = L are positions of nodes. The x = 0 condition is already satisfied. The x = L node condition requires that the λ length L is related to λ by L = n 2 ; ⁡n = 1,2,3, … Thus, the possible wavelengths of stationary waves are constrained by the relation 2L λ= ; ⁡n = 1,2,3, … n nv with corresponding frequencies v = 2 L , for n = 1,2,3, We have thus obtained the natural frequencies - the normal modes of oscillation of the system. The lowest possible natural frequency of a system is called its fundamental mode or the first harmonic. For the stretched string fixed satisfied. If the other end x = L is an antinode, 1 λ L = (n + 2) 2, for n = 0,1,2,3, … The possible wavelengths are then restricted by the relation: 2L λ= , for n = 0,1,2,3, … (n + 1/2) The normal modes - the natural frequencies of the system are 1 v v = (n + ) ; n = 0,1,2,3, … 2 2L v The fundamental frequency corresponds to n = 0, and is given by 4L. The higher frequencies are odd harmonics, i.e., odd multiples of the fundamental frequency: v v 3 4L , 5 4L, etc. Figure shows the first six odd harmonics of air column with one end closed and the other open. For a pipe open at both ends, each end is an antinode. It is then easily seen that an open-air column at both ends generates all harmonics. The systems above, strings and air columns, can also undergo forced oscillations. If the external frequency is close to one of the natural frequencies, the system shows resonance. APNI KAKSHA 12 BEATS ‘Beats’ is an interesting phenomenon arising from interference of waves. When two harmonic sound waves of close (but not equal) frequencies are heard at the same time, we hear a sound of similar frequency (the average of two close frequencies), but we hear something else also. We hear audibly distinct waxing and waning of the intensity of the sound, with a frequency equal to the difference in the two close frequencies. Let us consider 2 sound waves - s1 = acos⁡ ω1 t and s2 = acos⁡ ω2 t Here we have replaced the symbol y by s, since we are referring to longitudinal not transverse displacement. Let ω1 be the (slightly) greater of the two frequencies. The resultant displacement is, by the principle of superposition, s = s1 + s2 = a(cos⁡ ω1 t + cos⁡ ω2 t) Using the familiar trigonometric identity for cos A + cos B, we get (ω1 − ω2 )t (ω1 + ω2 )t = 2 acos cos 2 2 where (ω1 − ω2 ) (ω1 + ω2 ) ωb = ⁡and⁡ωa = 2 2 Now if we assume |ω1 − ω2 | ≪ ω1 , which means ωa ≫ ωb , we can interpret the equation as follows. The resultant wave is oscillating with the average angular frequency ωa ; however, its amplitude is not constant in time, unlike a pure harmonic wave. The amplitude is the largest when the term cos⁡ ωb t takes its limit +1 or −1. In other words, the intensity of the resultant wave waxes and wanes with a frequency which is 2ωb = ω1 − ω2. Since ω = 2πv, the beat frequency vbeat , is given by vbeat = v1 − v2 Figure illustrates the phenomenon of beats for two harmonic waves of frequencies 11 Hz and 9 Hz. The amplitude of the resultant wave shows beats at a frequency of 2 Hz. APNI KAKSHA 13 DOPPLER EFFECT It is an everyday experience that the pitch (or frequency) of the whistle of a fast-moving train decreases as it recedes away. When we approach a stationary source of sound with high speed, the pitch of the sound heard appears to be higher than that of the source As the observer recedes away from the source, the observed pitch (or frequency) becomes lower than that of the source. This motion-related frequency change is called Doppler effect. We shall analyze changes in frequency under three different situations: 1. Observer is stationary but the source is moving 2. Observer is moving but the source is stationary 3. Both the observer and the source are moving Source Moving: Observer Stationary (a) If the source is moving towards the listener but the listener is at rest, then vs is positive and vL = 0 (figure a). Therefore, v v′ = × v i.e. v ′ > v v − vs (b) If the source is moving away from the listener, but the listener is at rest, then vs is negative and vL = 0 (figure b). Therefore, v v′ = v v − (−vs ) v = v i.e. v ′ < v v + vs Observer Moving: Source Stationary (c) If the source is at rest and listener is moving away from the source, the vs = 0 and vL is positive (figure c). Therefore, (v − vL ) v′ = v⁡i. e. v ′ < v v (d) If the source is at rest and listener is moving towards the source, then vs = 0 and vL is negative (figure d). Therefore, v − (−vL ) v′ = v V v + vL = v i.e. v ′ > v V APNI KAKSHA 14 Both Source and Observer Moving (e) If the source and listener are approaching each other, then vs is positive and vL is negative (figure e). Therefore, v − (−vL ) v + vL v′ = v=( ) v i.e. v ′ > v v − vs v − vs (f) If the source and listener are moving away from each other, then vs is negative and vL is positive, (figure f). Therefore, v − vL v − vL v′ = v= v i.e. v ′ < v v − (−vs ) v + vs (g) If the source and listener are both in motion in the same direction and with same velocity, then vs = vL = v ′ (say) (figure g). Therefore, (v − v ′ ) v′ = v i.e. v ′ = v (v − v ′ ) It means, there is no change in the frequency of sound heard by the listener. Apparent wavelength heard by the observer is v − vs λ′ = v Note: If case the medium is also moving, the speed of sound with respect to ground is considered. i.e. v ⃗ +v ⃗m APNI KAKSHA 15 NCERT Practice Questions Q.1. A uniform rope of length L and mass m1 hangs vertically from a rigid support. A block of mass m2 is attached to the free end of the rope. A transverse pulse of wavelength λ1 is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is λ2. The ratio λ2 /λ1 is m1 +m2 m m1 +m2 m (A) √ (B) √m2 (C) √ (D) √m1 m2 1 m1 2 Sol. (A) According to question, we have v Wavelength of trans phase pulse λ = f T (v = velocity of the wave ; f = frequency of the wave) As we know v = √μ ( T = tension in the spring μ = mass per unit length of the rope) 1 T From above Equations, we get λ = f √μ ⇒ ⁡λ ∝ √T λ2 T2 m1 +m2 So, for two different cases, we get =√ =√ λ1 T1 m2 Q.2. A wave travelling in the positive x-direction having displacement along y-direction as 1 1 m, wavelength 2πm and frequency of π Hz is represented by (A) y = sin⁡(x − 2t) (B) y = sin⁡(2πx − 2πt) (C) y = sin⁡(10πx − 20πt) (D) y = sin⁡(2πx + 2πt) Sol. (A) Given, a = 1 m 2π 1 As y = asin⁡(kx − ωt) = sin⁡ ( − 2π × t) = sin⁡(x − 2t) 2π π Q.3. Two waves are represented by the equations y1 = asin⁡(ωt + kx + 0.57)m and y2 = acos⁡(ωt + kx)m, where x is in metre and t in second. The phase difference between them is (A) 1.25rad (B) 1.57rad (C) 0.57rad (D) 1rad Sol. (D) According to question, y1 = asin⁡(ωt + kx + 0.57) and y2 = acos⁡(ωt + kx) π or ⁡y2 = asin⁡ (2 + ωt + kx) As phase difference, π Δϕ = ϕ2 − ϕ1 = 2 − 0.57 = 1.57 − 0.57 = 1rad APNI KAKSHA 16 Q.4. The length of the string of a musical instrument is 90 cm and has a fundamental frequency of 120 Hz. Where should it be pressed to produce fundamental frequency of 180 Hz? (A) 75 cm (B) 60 cm (C) 45 cm (D) 80 cm Sol. (B) Length of string of musical instrument, I = 90 cm = 0.9 m Fundamental frequency, f1 = 120 Hz, f2 = 180 Hz 1 f I f1 l1 120×0.9 2 ∴ We know that f ∝ I ⇒ ⁡ f1 = I2 ⇒ I2 = = = 3 × 0.9 = 0.6 m = 60 cm 2 2 1 f2 180 Q.5. In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency 6 Hz. When tension in B is slightly decreased, the beat frequency increases to 7 Hz. If the frequency of A is 530 Hz, the original frequency of B will be (A) 524 Hz (B) 536 Hz (C) 537 Hz (D) 523 Hz Sol. (A) In case of beats formation unknown frequency (vB ) = vA ± beats where, vA = 530 Hz and beats = 6 Hz ⇒ vB = 530 ± 6 = 536 or 524 Hz When tension in B is slightly decreased, then beats frequency increases to 7 Hz. This is possible if we take original frequency of B as 524 Hz. Hence, correct option is (a). Q.6. A tuning fork with frequency 800 Hz produces resonance in a resonance column tube with upper end open and lower end closed by water surface. Successive resonance are observed at length 9.75 cm, 31.25 cm and 52.75 cm. The speed of sound in air is (A) 500 m/s (B) 156 m/s (C) 344 m/s (D) 172 m/s Sol. (C) For vibrating tuning fork over a resonance tube, the first resonance is obtained at the λ length I1 = 4 …(i) and for second resonance, λ λ 3λ I2 = 4 + 2 = …(ii) 4 From Eq. (i) and (ii), we get 3λ λ λ I2 − I1 = − = ⇒ ⁡λ = 2(I2 − l1 ) 4 4 2 v ⇒ ⁡v = 2f(I2 − l1 ) … (iii) (∵ λ = f ) Here, f = 800 Hz1 I1 = 9.75 cm, I2 = 31.25 cm Substituting the given values in Eq. (iii), we get ⇒ ⁡v = 2 × 800(31.25 − 9.75) = 34400 cm/s = 344 m/s APNI KAKSHA 17 Q.7. A car is moving towards a high cliff. The car driver sounds a horn of frequency f. The reflected sound heard by the driver has a frequency 2f. If v be the velocity of sound, then the velocity of the car, in the same velocity units, will be v v v v (A) (B) 3 (C) (D) 2 √2 4 Sol. (B) When the sound is reflected from the cliff, it approaches the driver of the car. Therefore, the driver acts as an observer and both the source (car) and observer are moving. Hence, apparent frequency heard by the observer (driver) is given by v + v0 f′ = f ( ) v − vs where, v = velocity of sound, v0 = velocity of car = vs Thus, Eq. (i) becomes v + v0 ∴ ⁡2f = f ( ) v − v0 v or 2v − 2v0 = v + v0 or ⁡3v0 = v or v0 = 3 Q.8. An observer moves towards a stationary source of sound with a speed 1/5 th of the speed of sound. The wavelength and frequency of the source emitted are λ and f respectively. The apparent frequency and wavelength recorded by the observer are respectively (A) f, 1.2λ (B) 0.8f, 0.8λ (C) 1.2f, 1.2λ (D) 1.2f, λ Sol. (D) When an observer moves towards a stationary source of sound, then apparent frequency heard by the observer increases. The apparent frequency heard in this situation v + v0 f′ = ( )f v − vs As source is stationary hence, vs = 0 v + v0 ∴ ⁡f ′ = ( )f v v Given, ⁡v0 = 5 Substituting in the relation for f ′ , v+v/5 6 we have f ′ = ( ) f = 5 f = 1.2f v Motion of observer does not affect the wavelength reaching the observer; hence, wavelength remains . APNI KAKSHA 18 Q.9. A vehicle, with a horn of frequency n is moving with a velocity of 30 m/s in a direction perpendicular to the straight line joining the observer and the vehicle. The observer perceives the sound to have a frequency n + n1. Then (If the sound velocity in air is 300 m/s) (A) n1 = 10n (B) n1 = 0 (C) n1 = 0.1n (D) n1 = −0.1n Sol. (B) When velocity of source (vehicle) is perpendicular to the line joining the observer and source, then there is no Doppler effect of sound as the component of velocity either towards or away from the observer is zero. So, there is no change in apparent frequency. Therefore, n1 = 0. Q.10. A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end? Sol. Given: Mass of the string, 𝑀 = 2.50 kg Tension in the string, 𝑇 = 200 N Length of the string, 𝑙 = 20.0 m Then, mass per unit length (linear mass density) of the string M 2.50 μ= = = 0.125⁡kg⁡m − 1 l 20 The transverse wave speed (𝑣) in the string is determined by the relation: T 200 v=√ =√ = √1600 = 40ms −1 μ 0.125 Hence, time taken by the disturbance to reach the other end, l 20 t= = = 0.5s v 40 Q.11. A stone dropped from the top of a tower of height 300 m high splashes into the water of a pond near the base of the tower. When is the splash heard at the top giventhat the speed of sound in air is 340 m s−1? (𝑔 = 9.8 m s−2) Sol. Given: Height of the tower, ℎ = 300 m Initial velocity of the stone, 𝑢 = 0 Acceleration, 𝑎 = 𝑔 = 9.8 m s−2 Speed of sound in air = 𝑣 = 340 m s−1 APNI KAKSHA 19  Let 𝑡1 be the time taken by stone to reach the pond’s surface. 1 There we use, s = ut + 2 at 2 to calculate t1 by substituting s = h, t = t1 , a = g and u = 0 1 h = 0 × t1 + gt12 2 Therefore, 2h 2 × 300 t1 = √ =√ = 7.82s g 9.8 Also, let t2 be the time taken by sound to reach top of the tower of height 300 m. h 300 Then, t 2 = v = 340 = 0.88s Hence, the total time after which the sound of splash is heard = t1 + t 2 ⁡ = 7.82 + 0.88 = 8.7⁡s Q.12. A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speedof sound in dry air at 20o C = 343 m s−1. Sol. Given: Mass of the steel wire, 𝑀 = 2.10 kg Speed of the transverse wave, 𝑣 = 343 ms−1 Length of the steel wire, 𝑙 = 12.0 m Then, mass per unit length (linear mass density) of the steel wire, M 2.10 μ= = = 0.175⁡kg⁡m−1 l 12.0 T v = √μ, where T is the tension in the wire Squaring on both the sides, we get T v2 = ⇒ T = μv 2 μ Hence, T = 0.175 × 3432 = 20,588.575 ≈ 2.06 × 104 ⁡N γP Q.13. Use the formula v = √ ρ to explain why the speed of sound in air (A) is independent of pressure, (B) increases with temperature, (C) increases with humidity. APNI KAKSHA 20  γP Sol. (A) Given: v = √ ρ … (i) 𝑀 = Molecular weight of gas 𝑉 = Volume of gas γPV So, equation (i) reduces to: v = √ M … (ii) Using the ideal gas equation for 𝑛 = 1: 𝑃𝑉 = 𝑅𝑇 For constant 𝑇, 𝑃𝑉=𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Since both 𝑀 and 𝛾 are constants, 𝑣 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Hence, the speed of sound in air is not dependent on pressure at a constant temperature. γP (B) Given: v = √ … (i) ρ Mass M where, ρ = Density = Volume = V ⇒ ρV = M … (ii) 𝑀 = Molecular weight of gas 𝑉 = Volume of gas γPV So, equation (i) reduces to: v = √ M Using the ideal gas equation for n = 1: PV = RT RT P= … (iii) V Putting (iii) in (i), we get γRT γRT v=√ =√ … (iv) ρV M Since 𝑀, 𝛾 and 𝑅 are all constants, v ∝ √T. Hence, the speed of sound in air increases with increase in temperature. (C) Let 𝑣𝑑 and 𝑣𝑚 be the speeds of sound in dry air and moist air respectively. Let 𝜌𝑑 and 𝜌𝑚 be the densities of dry air and moist air respectively. γP We know the formula: v = √ ρ Therefore, the speed of sound in moist air is: γP vm = √ρ … (i) m And the speed of sound in dry air is: γP vd = √ … (ii) ρ d On dividing equations (i) and (ii) we get: vm γP ρ vd = √ρ × γPd … (iii) m However, the presence of water vapour reduces the density of air, i.e., ρd < ρm ∴ vm > vd Hence, the speed of sound in moist air is greater than it is in dry air. Thus, in a gaseous medium, the speed of sound increases with humidity. APNI KAKSHA 21 Q.14. You have learnt that a travelling wave in one dimension is represented by a function 𝑦 = 𝑓(𝑥, 𝑡) where 𝑥 and 𝑡 must appear in the combination 𝑥 − 𝑣𝑡 or 𝑥 + 𝑣𝑡, i.e., 𝑦 = 𝑓(𝑥 ± 𝑣𝑡). Is the converse true? Examine if the following functions for 𝑦 canpossibly represent a travelling wave: x+vt 1 (A) (𝑥 − 𝑣𝑡)2 (B) log⁡ [ ] (C) (x+vt) xo Sol. The converse is not true. An obvious requirement for an acceptable function for a travelling wave is that it should be finite everywhere and at all times. (A) No, the function does not represent a travelling wave as it is not havingfinite value everywhere and at all times. Example: For 𝑥 = ∞ and 𝑡 = ∞, function has infinite value. (B) By substituting 𝑥 = 0 and 𝑡 = 0, the given function tends to infinity. So,the function cannot possibly represent a travelling wave. (C) By substituting 𝑥 = 0 and 𝑡 = 0, the given function tends to infinity. So,the function cannot possibly represent a travelling wave. APNI KAKSHA 22