RowQ
The Vault
RowQ
The Vault
CBSE Class 11 Physics · 10 questions · 24 marks
A wave carries energy and information from place to place without carrying any matter with it, and the same handful of equations describes ripples on a pond, sound in air and a plucked guitar string. Here you will write the equation of a travelling wave, work out how fast it moves through different media, and see what happens when two waves overlap to produce standing waves and beats. The chapter closes with the Doppler effect, the reason an ambulance siren drops in pitch as it passes you.
A wave travels from air into water, where its speed is greater. Which of the following quantities is unchanged?
Answer
Its frequency — the frequency of a wave is set by the source that generates it, and each cycle arriving at the boundary must produce exactly one cycle on the far side, or crests would pile up at the interface. The speed is a property of the medium and does change, so since v = fλ with f fixed, the wavelength must change in proportion to the speed. The amplitude also changes, because energy is partly reflected at the boundary.
Two tuning forks of frequencies 512 Hz and 518 Hz are sounded together. The number of beats heard per second is:
Answer
6 — the beat frequency is the difference of the two frequencies, f_beat = |f₁ − f₂| = |518 − 512| = 6 Hz, so six maxima of loudness are heard each second. The figure 515 is the mean frequency, which is the pitch actually perceived, not the beat rate; the two must not be confused.
The distance between two consecutive nodes in a standing wave of wavelength λ is:
Answer
λ/2 — nodes occur wherever sin(kx) = 0, that is at x = 0, λ/2, λ, 3λ/2 and so on, so consecutive nodes are half a wavelength apart. Antinodes are likewise λ/2 apart, but the distance from any node to the nearest antinode is λ/4. A common slip is to give λ/4 for the node-to-node spacing.
The speed of sound in air at constant temperature is:
Answer
Independent of the pressure — from Laplace's formula v = √(γP/ρ). Increasing the pressure of a gas at constant temperature increases its density in exactly the same proportion, so the ratio P/ρ is unchanged and the speed stays the same. Using PV = nRT it follows that v = √(γRT/M), showing that the speed depends on the absolute temperature and the molar mass but not on the pressure at all.
Assertion (A): A standing wave transports no net energy along the medium. Reason (R): A standing wave is formed by the superposition of two identical waves travelling in opposite directions, whose energy flows cancel.
Answer
Both A and R are true and R is the correct explanation of A — each of the two component waves carries energy, but in opposite directions and at equal rates, so the net transport past any point is zero. This is confirmed by the existence of nodes, which never move at all: energy cannot cross a point that is permanently at rest. The energy in a standing wave is instead trapped between adjacent nodes, sloshing back and forth between kinetic and potential form within each segment.
A wire of length 1.2 m has a mass of 9.6×10⁻³ kg and is stretched to a tension of 80 N. Find the speed of a transverse wave along it and the frequency of its fundamental note.
Answer
Linear mass density: μ = mass/length = 9.6×10⁻³/1.2 = 8×10⁻³ kg/m. Wave speed: v = √(T/μ) = √(80/8×10⁻³) = √(10⁴) = 100 m/s. Fundamental frequency of a wire fixed at both ends, where the length is half a wavelength: f₁ = v/2L = 100/(2 × 1.2) = 100/2.4 ≈ 41.7 Hz. So the wire sounds a low note of about 42 Hz; tightening it or shortening it would raise the pitch.
A progressive wave is described by y = 0.02 sin(300t − 1.5x), where y and x are in metres and t in seconds. Find (a) the amplitude, (b) the frequency, (c) the wavelength and (d) the speed and direction of travel.
Answer
Compare with the standard form y = A sin(ωt − kx). (a) Amplitude A = 0.02 m = 2 cm. (b) ω = 300 rad/s, so f = ω/2π = 300/(2 × 3.14) ≈ 47.8 Hz. (c) k = 1.5 rad/m, so λ = 2π/k = 6.28/1.5 ≈ 4.19 m. (d) v = ω/k = 300/1.5 = 200 m/s. (Check: v = fλ = 47.8 × 4.19 ≈ 200 m/s, which agrees.) The minus sign between ωt and kx means the wave travels in the positive x-direction. Had the term been (300t + 1.5x) it would have been travelling in the negative x-direction.
(a) Explain the formation of standing waves on a string fixed at both ends and derive the expression y = 2A sin(kx) cos(ωt) by superposition. Locate the nodes and antinodes. (b) Derive the expression for the frequencies of the harmonics of such a string. (c) A sonometer wire of length 0.60 m and linear density 2×10⁻³ kg/m is under a tension of 180 N. Find the frequencies of its first three harmonics.
Answer
(a) Formation and derivation. When a wave travelling along a string reaches a fixed end it is reflected with a phase reversal and travels back along the string. The incident and reflected waves have equal amplitude, frequency and speed but opposite directions, and their superposition produces a standing wave. Let the incident wave travelling in the +x direction be y₁ = A sin(ωt − kx), and the reflected wave travelling in the −x direction be y₂ = A sin(ωt + kx). By the principle of superposition the resultant displacement is y = y₁ + y₂ = A[sin(ωt − kx) + sin(ωt + kx)]. Using the identity sin C + sin D = 2 sin((C + D)/2) cos((C − D)/2), with C = ωt + kx and D = ωt − kx: (C + D)/2 = ωt and (C − D)/2 = kx. Therefore y = 2A sin(ωt) cos(kx), or equivalently, with the incident and reflected waves written in the other order, y = 2A sin(kx) cos(ωt). The important feature is that x and t now appear in separate factors, so there is no term of the form (ωt − kx) and nothing propagates. Every particle oscillates with the same frequency ω but with an amplitude 2A sin(kx) that depends only on its position. Nodes: points of zero amplitude, where sin(kx) = 0, so kx = 0, π, 2π, … giving x = 0, λ/2, λ, 3λ/2, … Consecutive nodes are λ/2 apart and these particles never move at all. Antinodes: points of maximum amplitude 2A, where sin(kx) = ±1, so kx = π/2, 3π/2, … giving x = λ/4, 3λ/4, 5λ/4, … These are also λ/2 apart, and each lies exactly midway between two nodes, a quarter wavelength from each. (b) Harmonic frequencies. Both ends of the string are fixed, so both must be nodes. If the string has length L, the only wavelengths that fit are those for which a whole number of half-wavelengths spans the string: L = n(λ/2), where n = 1, 2, 3, … Hence λ = 2L/n. Since f = v/λ and the wave speed on the string is v = √(T/μ): f_n = v/λ = nv/2L = (n/2L)√(T/μ). For n = 1 this gives the fundamental or first harmonic, f₁ = (1/2L)√(T/μ). For n = 2, 3, 4 … we get the second, third and higher harmonics, which are exact integer multiples of the fundamental. A string therefore produces a complete harmonic series, which is why plucked and bowed instruments sound rich and musical. (c) Numerical work. Wave speed: v = √(T/μ) = √(180/2×10⁻³) = √(9×10⁴) = 300 m/s. First harmonic (fundamental): f₁ = v/2L = 300/(2 × 0.60) = 300/1.2 = 250 Hz. Second harmonic: f₂ = 2f₁ = 500 Hz. Third harmonic: f₃ = 3f₁ = 750 Hz. So the wire sounds a fundamental of 250 Hz with overtones at 500 Hz and 750 Hz.
(a) Explain the Doppler effect and derive the expression for the apparent frequency when a source of sound moves towards a stationary observer. (b) A siren of frequency 640 Hz is mounted on a van moving at 25 m/s along a straight road. Find the frequency heard by a stationary pedestrian as the van approaches and as it recedes, taking the speed of sound as 340 m/s. (c) The pedestrian now cycles towards the approaching van at 5 m/s. Find the new apparent frequency.
Answer
(a) The Doppler effect is the apparent change in the observed frequency of a wave when there is relative motion between the source and the observer along the line joining them. It is not the frequency of the source that changes but the rate at which wavefronts reach the observer. Derivation for a source moving towards a stationary observer. Let the source emit sound of frequency f and true wavelength λ = v/f in still air, where v is the speed of sound. Suppose the source moves towards the observer with speed v_s. In one time period T = 1/f, the source emits one complete wave. In that same time: the first crest has travelled a distance vT away from the emission point, while the source itself has advanced a distance v_sT towards the observer. The next crest is therefore emitted from a point that is v_sT closer, so the two crests are separated not by vT but by λ' = vT − v_sT = (v − v_s)T = (v − v_s)/f. The waves are bunched up ahead of the source. They still travel through the air at the ordinary speed v, since the speed of sound depends only on the medium and not on the motion of the source. The observer therefore receives them at the rate f' = v/λ' = vf/(v − v_s). Since v − v_s is less than v, the apparent frequency f' is greater than f: the approaching source sounds higher in pitch. If the source recedes, v_s changes sign and f' = vf/(v + v_s), which is lower than f. The general formula, with the observer also moving at speed v_o, is f' = f(v ± v_o)/(v ∓ v_s), where the upper signs are used when the motion is such as to bring the two closer together. (b) Van approaching, pedestrian at rest: v_o = 0, v_s = 25 m/s towards. f' = fv/(v − v_s) = 640 × 340/(340 − 25) = 217600/315 ≈ 690.8 Hz. Van receding: v_s = 25 m/s away. f' = fv/(v + v_s) = 640 × 340/(340 + 25) = 217600/365 ≈ 596.2 Hz. So as the van passes, the pitch heard drops abruptly from about 691 Hz to about 596 Hz, a fall of roughly 95 Hz. This sudden drop at the moment of passing is the familiar siren effect. (c) Pedestrian cycling towards the approaching van: now v_o = 5 m/s towards the source and v_s = 25 m/s towards the observer, so both motions are closing and both use the signs that raise the frequency. f' = f(v + v_o)/(v − v_s) = 640 × (340 + 5)/(340 − 25) = 640 × 345/315 = 220800/315 ≈ 701.0 Hz. The pitch rises further, to about 701 Hz, because the cyclist now meets the crowded wavefronts at a greater rate. Note that the observer's motion enters the numerator and the source's motion the denominator, and the two are not interchangeable: moving the source at 5 m/s and the observer at 25 m/s would give a slightly different answer, because the source's motion physically alters the wavelength while the observer's motion only alters the rate of encounter.
Read the following and answer the questions that follow: A student investigates resonance using a glass tube standing in a water reservoir, so that the length of the air column can be varied. A tuning fork of unknown frequency is held over the open top. Resonance is first heard when the air column is 0.16 m long, and again when it is 0.49 m long. The room temperature is such that the speed of sound is 340 m/s. The student then sounds this fork together with a second fork of frequency 500 Hz and hears 8 beats per second. (a) Explain why this apparatus behaves as a closed pipe and state which harmonics it can produce. (b) Use the two resonance lengths to find the wavelength of the sound and hence the frequency of the fork, explaining why using both lengths avoids the end-correction error. (c) Determine which of the two possible frequencies of the second fork is consistent with the beat observation, and describe a test that would settle the matter. (d) The student repeats the experiment on a much warmer day. State and explain what happens to the resonance lengths.
Answer
(a) The tube is open at the top, where the fork vibrates the air freely, and closed at the bottom by the water surface, which cannot move. A closed end must therefore be a displacement node and the open end a displacement antinode. A pipe closed at one end supports only those modes with a node at one end and an antinode at the other, which requires the length to be an odd number of quarter wavelengths: L = λ/4, 3λ/4, 5λ/4 and so on. Its frequencies are f_n = (2n − 1)v/4L, so it produces only the odd harmonics — the fundamental, the third, the fifth and so on. The even harmonics are missing, which is why a closed pipe sounds noticeably hollower than an open one of the same pitch. (b) The first resonance corresponds to L₁ = λ/4 and the second to L₂ = 3λ/4, so the difference between them is exactly half a wavelength: L₂ − L₁ = 3λ/4 − λ/4 = λ/2. 0.49 − 0.16 = 0.33 m = λ/2 λ = 0.66 m. Frequency: f = v/λ = 340/0.66 ≈ 515 Hz. Why this avoids the end correction: the antinode does not sit exactly at the mouth of the tube but a small distance e above it, so the true relations are L₁ + e = λ/4 and L₂ + e = 3λ/4. Subtracting eliminates e entirely, leaving L₂ − L₁ = λ/2 with no unknown correction in it. Had the student used the single measurement λ = 4L₁ = 0.64 m, the answer would have been wrong by a few hertz. As a bonus, subtracting the two equations the other way gives e = (L₂ − 3L₁)/2 = (0.49 − 0.48)/2 = 0.005 m, a 5 mm end correction. (c) Hearing 8 beats per second means the two frequencies differ by 8 Hz, so the second fork is either 515 + 8 = 523 Hz or 515 − 8 = 507 Hz. The stated frequency of 500 Hz matches neither exactly, so the fork's marked value must be taken as approximate; what the beat measurement fixes is the difference, not which fork is higher. Test to settle it: load one prong of the unknown fork with a small blob of wax. This adds mass and therefore lowers that fork's frequency slightly. If the beat rate now increases, the two frequencies have been pushed further apart, meaning the waxed fork was already the lower of the two. If the beat rate decreases, the waxed fork was the higher one and has been brought closer to the other. Filing the prong instead would raise the frequency and give the opposite indication. (d) On a warmer day the speed of sound increases, since v = √(γRT/M) and so v ∝ √T with T in kelvin. The tuning fork's frequency is a property of the steel and is essentially unchanged. Since λ = v/f with f fixed and v larger, the wavelength is longer. The resonance lengths, being λ/4 and 3λ/4, are therefore both greater: the student must lower the water level further to find each resonance. For example, a rise from 20 °C to 40 °C raises the speed by about 3.4%, which lengthens both resonance positions by roughly the same percentage.
RowQ generates fresh questions on Waves, marks your answers, and explains every step.
Start free