NCERT Exemplar: Waves | Physics Class 11 - NEET PDF Download

MULTIPLE CHOICE QUESTIONS I

Q.1. Water waves produced by a motor boat sailing in water are
(a) Neither longitudinal nor transverse.
(b) Both longitudinal and transverse.
(c) Only longitudinal.
(d) Only transverse.
Ans. (b)
Explanation: Surface water waves combine two kinds of particle motion. A surface particle describes an approximately circular path as a wave passes: it moves up and down (transverse component) and back and forth (longitudinal component). Therefore surface water waves contain both longitudinal and transverse motions.

As the wave passes, particles near crests and troughs have motion components in different directions; this produces the combined behaviour described above.

Q.2. Sound waves of wavelength λ travelling in a medium with a speed of v m/s enter into another medium where its speed is 2v m/s. Wavelength of sound waves in the second medium is
(a) λ
(b) λ/2
(c) 2λ
(d) λ
Ans. (c)
Explanation: Frequency of the source remains unchanged when a wave passes from one medium to another.

Using v = fλ for the first medium and v2 = fλ2 for the second medium, with v2 = 2v, we get λ2 = v2 / f = 2v / f = 2λ. Hence the wavelength in the second medium is 2λ.

Q.3. Speed of sound wave in air
(a) Is independent of temperature.
(b) Increases with pressure.
(c) Increases with increase in humidity.
(d) Decreases with increase in humidity.
Ans. (c)
Explanation: The speed of sound in a gas is v = √(γP/ρ). For small changes at ordinary atmospheric pressure, P and γ are effectively constant, so v ∝ 1/√ρ. Moist air contains water vapour whose molecular mass is less than that of dry air; increasing humidity reduces the effective density ρ of the air mixture, increasing v. Therefore the speed of sound increases with humidity.

Q.4. Change in temperature of the medium changes
(a) Frequency of sound waves.
(b) Amplitude of sound waves.
(c) Wavelength of sound waves.
(d) Loudness of sound waves.
Ans. (c)
Explanation: Speed of sound in a gas varies approximately as v ∝ √T (T in kelvin). The source frequency f does not change when the medium changes, so from v = fλ we have λ = v / f. Therefore if temperature (and hence v) changes, the wavelength λ changes while f remains the same.

Q.5. With propagation of longitudinal waves through a medium, the quantity transmitted is
(a) Matter
(b) Energy
(c) Energy and matter
(d) Energy, matter and momentum
Ans. (b)
Explanation: A mechanical wave transmits energy through a medium while the medium's particles oscillate about their mean positions without net transport of matter. In ideal wave propagation there is no net transfer of mass; hence a longitudinal wave transmits energy but not matter as a bulk flow. Momentum is not transmitted as a net mass flow in idealised wave motion. Therefore option (b) is correct.

Characteristics of wave motion

• It is a disturbance that travels through a medium.
• Material medium is essential for the propagation of mechanical waves.
• When a wave passes through a medium, particles of the medium vibrate about their mean positions; there is no net transport of matter.
• Successive particles have continuous phase difference; each particle begins its vibration slightly later than the previous one.
• The instantaneous velocity of individual particles varies from place to place.
• Wave velocity in a given medium depends only on the medium's properties (elastic and inertial), not on amplitude or intensity.
• Energy is propagated along the wave without net transport of the medium.

• The velocity of particles during vibration differs at different positions.
• The wave velocity through a given medium is determined by the medium's elastic and inertial properties.
• Energy is transferred by the wave without net transport of material.

Q.6. Which of the following statements are true for wave motion?
(a) Mechanical transverse waves can propagate through all mediums.
(b) Longitudinal waves can propagate through solids only.
(c) Mechanical transverse waves can propagate through solids only.
(d) Longitudinal waves can propagate through vacuum.
Ans. (c)
Explanation: Transverse mechanical waves require the medium to support shear stress; solids have non-zero shear modulus and can sustain transverse waves. Fluids (liquids and gases) cannot sustain bulk shear stress in the same way, so transverse mechanical waves do not propagate through the bulk of fluids (surface waves on liquids are a special case). Longitudinal waves (compressions and rarefactions) travel through solids, liquids and gases but not through vacuum. Therefore only statement (c) is true.

Q.7. A sound wave is passing through air column in the form of compression and rarefaction. In consecutive compressions and rarefactions,
(a) Density remains constant.
(b) Boyle's law is obeyed.
(c) Bulk modulus of air oscillates.
(d) There is no transfer of heat.
Ans. (d)
Explanation: Sound produces alternating compressions (higher pressure and density) and rarefactions (lower pressure and density); density therefore does not remain constant (so (a) is false). The rapid pressure-volume changes in sound propagation are essentially adiabatic at audible frequencies, so they do not obey Boyle's isothermal law; the appropriate relation is the adiabatic one. The bulk modulus is a material property relating pressure and fractional volume change; the local pressure and density oscillate, but to say the bulk modulus itself oscillates is imprecise. The process is approximately adiabatic with negligible net heat transfer during each oscillation, so (d) is the best choice.

Q.8. Equation of a plane progressive wave is given by y = 0.6 sin 2π(t - x/2). On reflection from a denser medium its amplitude becomes 2/3 of the amplitude of the incident wave. The equation of the reflected wave is
(a)  y = 0.6 sin 2π(t + x/2)
(b)  y = -0.4 sin 2π(t + x/2)
(c)  y = 0.4 sin 2π(t + x/2)
(d)  y = -0.4 sin 2π(t - x/2)
Ans. (b)
Explanation: The incident wave is y_i = 0.6 sin 2π(t - x/2).

Reflection from a denser medium causes a phase inversion of π (a sign change) and the reflected amplitude A_r = (2/3) × 0.6 = 0.4.

The reflected wave travels in the opposite direction, so its argument is 2π(t + x/2), and due to phase inversion it acquires a minus sign. Therefore y_r = -0.4 sin 2π(t + x/2).

Q.9. A string of mass 2.5 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, the disturbance will reach the other end in
(a) 1 second
(b) 0.5 second
(c) 2 seconds
(d) Data given is insufficient
Ans. (b)
Explanation: Linear mass density μ = m / L = 2.5 kg / 20.0 m = 0.125 kg m^-1.

Wave speed v = √(T/μ) = √(200 / 0.125) = √1600 = 40 m s^-1.

Time to travel length L is t = L / v = 20.0 / 40 = 0.5 s.

Q.10. A train whistling at constant frequency is moving towards a station at a constant speed V. The train goes past a stationary observer on the station. The frequency n of the sound as heard by the observer is plotted as a function of time t (Figure) . Identify the expected curve.
(a) 

(b) 

(c) 

(d) 

Ans. (c)
Explanation: As the train approaches, the observer hears a higher apparent frequency due to the Doppler effect. Exactly at the instant the train passes the observer there is a sudden change in sign of relative velocity and hence a sudden drop in apparent frequency. As the train recedes, the apparent frequency is lower than the source frequency. The plot that shows a higher value before the passing instant, a sudden jump down at the passing instant, and a lower value afterwards corresponds to curve (c).

MULTIPLE CHOICE QUESTIONS II

Q.11. A transverse harmonic wave on a string is described by y(x, t) = 3.0 sin (36t + 0.018x + π/4)
where x and y are in cm and t is in sec. The positive direction of x is from left to right.
(a) The wave is travelling from right to left.
(b) The speed of the wave is 20 m/s.
(c) Frequency of the wave is 5.7 Hz.
(d) The least distance between two successive crests in the wave is 2.5 cm.
Ans. (a,b,c)
Explanation: The wave has form sin(ωt + kx + φ). The sign +k x with ωt indicates propagation in the negative x direction (right to left), so (a) is correct.

ω = 36 rad s^-1 ⇒ f = ω/(2π) = 36 / (2π) = 18/π ≈ 5.73 Hz, so (c) is correct.

k = 0.018 cm^-1 = 1.8 m^-1 (convert cm⁻¹ to m⁻¹ by ×100). Wave speed v = ω / k = 36 / 1.8 = 20 m s^-1, so (b) is correct.

Wavelength λ = 2π / k = 2π / 1.8 ≈ 3.49 m (≈ 349 cm), so statement (d) (2.5 cm) is incorrect.

Q.12. The displacement of a string is given by y(x, t) = 0.06 sin(2πx/3) cos(120πt) where x and y are in m and t in sec. The length of the string is 1.5 m and its mass is 3.0 x 10^-2 kg.
(a) It represents a progressive wave of frequency 60 Hz.
(b) It represents a stationary wave of frequency 60 Hz.
(c) It is the result of superposition of two waves of  wavelength 3 m, frequency 60 Hz each travelling with a speed of 180 m/s in opposite direction.
(d) Amplitude of this wave is constant.
Ans. (b, c)
Explanation: The form y = 2a sin(kx) cos(ωt) is a standing (stationary) wave. Here ω = 120π ⇒ f = ω/(2π) = 60 Hz, so (b) is correct.

k = 2π/3 ⇒ λ = 2π/k = 3 m. With f = 60 Hz, v = fλ = 180 m s^-1. The standing wave is formed by two equal-amplitude progressive waves of wavelength 3 m and frequency 60 Hz travelling in opposite directions; thus (c) is correct.

Amplitude varies with x as 0.06 sin(2πx/3), so amplitude is not constant in space; (d) is incorrect.

Q.13. Speed of sound waves in a fluid depends upon
(a) Directly on density of the medium.
(b) Square of Bulk modulus of the medium.
(c) Inversely on the square root of density.
(d) Directly on the square root of bulk modulus of the medium
Ans. (c, d)
Explanation: For a fluid, v = √(B/ρ), where B is bulk modulus and ρ the density. Thus v ∝ √B and v ∝ 1/√ρ, making (c) and (d) correct; (a) and (b) are incorrect.

Q.14. During propagation of a plane progressive mechanical wave
(a) All the particles are vibrating in the same phase.
(b) Amplitude of all the particles is equal.
(c) Particles of the medium executes S.H.M.
(d) Wave velocity depends upon the nature of the medium
Ans. (b, c, d)
Explanation: In a plane progressive wave, each particle executes simple harmonic motion (SHM) with the same frequency and amplitude, hence (b) and (c) are correct.

There is a continuous phase difference between particles at different positions, so (a) is false.

Wave velocity depends on the medium's elastic and inertial properties, so (d) is true.

Q.15. The transverse displacement of a string (clamped at its both ends) is given by y (x,t) = 0.06 sin (2πx/3) cos (120πt).
All the points on the string between two consecutive nodes vibrate with
(a) Same frequency
(b) Same phase
(c) Same energy
(d) Different amplitude
Ans. (a, b, d)
Explanation: In a standing wave each particle oscillates with the same angular frequency ω, so all points have the same frequency (a).

Between two consecutive nodes (a loop) particles oscillate in phase (b).

Amplitude varies with position as 0.06 sin(2πx/3), so amplitudes are different (d).

Energy of oscillation at a point depends on the square of its amplitude; since amplitudes differ across the loop, energy is not the same everywhere and (c) is not generally true.

Q.16. A train, standing in a station yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10 m/s. Given that the speed of sound in still air is 340 m/s,
(a) The frequency of sound as heard by an observer standing on the platform is 400 Hz.
(b) The speed of sound for the observer standing on the platform is 350 m/s.
(c) The frequency of sound as heard by the observer standing on the platform will increase.
(d) The frequency of sound as heard by the observer standing on the platform will decrease.
Ans. (a, b)
Explanation: Wind blowing in the same direction as propagation increases the speed of sound relative to the ground to v + v_w = 340 + 10 = 350 m s^-1, so (b) is correct.

There is no relative motion between source and observer, so the observed frequency equals the source frequency (400 Hz); hence (a) is correct and (c), (d) are false.

Q.17. Which of the following statements are true for a stationary wave?
(a) Every particle has a fixed amplitude which is different from the amplitude of its nearest particle. 
(b) All the particles cross their mean position at the same time. 
(c) All the particles are oscillating with same amplitude. 
(d) There is no net transfer of energy across any plane. 
(e) There are some particles which are always at rest. 
Ans. (a, b, d, e)
Explanation: For a stationary wave y(x,t) = 2a sin(kx) cos(ωt):

(a) Amplitude at position x is 2a sin(kx), so neighbouring particles generally have different amplitudes - true.

(b) Time dependence is common (cos ωt), so all particles cross the mean position simultaneously - true.

(c) Amplitudes vary with x, so (c) is false.

(d) There is no net energy transport along the medium in an ideal standing wave - true.

(e) Nodes (sin kx = 0) are points that remain at rest at all times - true.

VERY SHORT ANSWER TYPE QUESTIONS

Q.1. A sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same, the length of the wire is doubled. Under what conditions would the tuning fork still be is resonance with the wire?
Ans. If the length is doubled while tension and linear mass density remain the same, the natural frequencies of the wire change. The resonance condition f = (n/2L)√(T/μ) can still be satisfied for the same tuning-fork frequency if the mode number n changes appropriately. Doubling L can be compensated by doubling n (i.e., the wire vibrating in a higher harmonic), so the wire can resonate with the same fork if it vibrates in a harmonic whose frequency equals the fork frequency (for example the second harmonic when length is doubled, provided integer n gives the same f).

Q.2. An organ pipe of length L open at both ends is found to vibrate in its first harmonic when sounded with a tuning fork of 480 Hz. What should be the length of a pipe closed at one end, so that it also vibrates in its first harmonic with the same tuning fork?
Ans. An open pipe in its first harmonic (fundamental) has L_open = λ/2. A pipe closed at one end has fundamental L_closed = λ/4. To have the same frequency (same λ), L_closed must be half of L_open. Therefore L_closed = L / 2.

Q.3. A tuning fork A, marked 512 Hz, produces 5 beats per second, where sounded with another unmarked tuning fork B. If B is loaded with wax the number of beats is again 5 per second. What is the frequency of the tuning fork B when not loaded?
Ans. Let f_A = 512 Hz and |f_A - f_B| = 5 Hz. Thus f_B = 512 ± 5, i.e., 507 Hz or 517 Hz.

Loading with wax lowers the frequency of B. After loading, the beat frequency remains 5 Hz. If the unloaded frequency were 507 Hz, loading would reduce it further (e.g., to ≈ 502 Hz) and the beat frequency with 512 Hz would change; thus the consistent unloaded frequency is 517 Hz (loading reduces 517 to about 512, keeping beat frequency 5). Hence the unloaded frequency of B is 517 Hz.

Q.4. The displacement of an elastic wave is given by the function y = 3 sin ωt + 4 cos ωt.  where y is in cm and t is in second. Calculate the resultant amplitude.
Ans. Combine into a single sinusoid y = A sin(ωt + φ).

Compare coefficients: 3 = A cos φ and 4 = A sin φ.

Square and add: A²(cos²φ + sin²φ) = 3² + 4² = 9 + 16 = 25.

So A = √25 = 5 cm.

Q.5. A sitar wire is replaced by another wire of same length and material but of three times the earlier radius. If the tension in the wire remains the same, by what factor will the frequency change?
Ans. For the same material and length, linear mass density μ ∝ cross-sectional area ∝ r².

If radius increases by factor 3, μ becomes 9 times larger.

Frequency f ∝ 1/√μ, so f_new = f_old / 3. Thus the frequency reduces by a factor of 3.

Q.6. At what temperatures (in ^oC) will the speed of sound in air be 3 times its value at 0^oC?
Ans.  Speed of sound v ∝ √T (absolute temperature in kelvin).

Let v_0 be at T_0 = 273 K. We require v = 3v_0 ⇒ √T / √273 = 3 ⇒ T = 9 × 273 = 2457 K.

Convert to Celsius: T_C = 2457 - 273 = 2184 °C. Thus the required temperature is 2184 °C.

Q.7. When two waves of almost equal frequencies n₁ and n₂ reach at a point simultaneously, what is the time interval between successive maxima?
Ans. Beat frequency f_beat = |n₁ - n₂|.

Time interval between successive maxima (beat period) T = 1 / f_beat = 1 / |n₁ - n₂|.

Part- 2

SHORT ANSWER TYPE QUESTIONS

Q.1. A steel wire has a length of 12 m and a mass of 2.10 kg. What will be the speed of a transverse wave on this wire when a tension of 2.06 × 10⁴N is applied?
Ans.  Given: length L = 12 m, mass m = 2.10 kg, tension T = 2.06 × 10⁴ N.

Linear mass density μ = m / L = 2.10 / 12 = 0.175 kg m^-1.

Wave speed v = √(T / μ) = √(2.06 × 10⁴ / 0.175) = √(1.17714 × 10⁵) ≈ 343.2 m s^-1.

Q.2. A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a source of 1237.5 Hz ?(sound velocity in air = 330 m s^-1)
Ans. For a pipe closed at one end, resonant frequencies are f_n = (2n - 1)v / 4L, n = 1,2,3, ...

Here L = 0.20 m, v = 330 m s^-1. Fundamental f_1 = v / 4L = 330 / (4 × 0.20) = 330 / 0.8 = 412.5 Hz.

Odd harmonics: 412.5 Hz (1st), 3 × 412.5 = 1237.5 Hz (3rd), 5 × 412.5 = 2062.5 Hz (5th), ...

Therefore 1237.5 Hz is the third harmonic (n corresponding to (2n - 1) = 3), i.e., the 3rd mode.

Q.3. A train standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. The train begins to move with a speed of 10 m s^-1 towards the platform. What is the frequency of the sound for an observer standing on the platform? (sound velocity in air = 330 m s^-1)
Ans. Apparent frequency for stationary observer with source moving towards observer is f' = (v / (v - v_s)) f.

Substitute values: f' = 330 / (330 - 10) × 400 = 330 / 320 × 400 = (33 / 32) × 400 = 412.5 Hz.

Q.4. The wave pattern on a stretched string is shown in Figure Interpret what kind of wave this is and find its wavelength.

Ans. The pattern shows points that remain fixed (nodes) at x = 10 cm, 20 cm, 30 cm, 40 cm, etc., so this is a stationary (standing) wave.

Distance between successive nodes = λ/2. Given node-node distance = 10 cm ⇒ λ/2 = 10 cm ⇒ λ = 20 cm.

Q.5. The pattern of standing waves formed on a stretched string at two instants of time are shown in Figure. The velocity of two waves superimposing to form stationary waves is 360 ms^-1 and their frequencies are 256 Hz.

(a) Calculate the time at which the second curve is plotted.
(b) Mark nodes and antinodes on the curve.
(c) Calculate the distance between A′ and C′ .
Ans. Given f = 256 Hz ⇒ period T = 1 / f = 1 / 256 ≈ 3.90625 × 10^-3 s.

(a) Characteristic standing-wave shapes repeat every T/2 with sign reversal; successive distinctive shapes (maximum displacement and zero displacement) are separated by T/4. Therefore the second curve, if it shows the zero-displacement pattern when the first shows maximum displacement, is plotted at t = T/4 ≈ 3.90625 × 10^-3 / 4 ≈ 9.766 × 10^-4 s after the first.

(b) Nodes are points that never move - identify nodes at fixed positions where displacement is always zero (labelled A, B, C, D, E in the figure). Antinodes are midway between adjacent nodes (labelled A′, C′, etc.).

(c) Distance between adjacent antinodes = λ/2. The distance between A′ and C′ spans two adjacent antinodes (i.e., one full wavelength), hence distance = λ. Use λ = v / f = 360 / 256 = 1.40625 m to obtain the numerical value. If the diagram is to scale, map diagram distances to the computed λ.

Q.6. A tuning fork vibrating with a frequency of 512 Hz is kept close to the open end of a tube filled with water. The water level in the tube is gradually lowered. When the water level is 17 cm below the open end, maximum intensity of sound is heard. If the room temperature is 20° C, calculate

(a) speed of sound in air at room temperature
(b) speed of sound in air at 0° C
(c) if the water in the tube is replaced with mercury, will there be any difference in your observations?
Ans. For a tube closed at one end (water surface) and open at the other, resonance (first maximum) occurs when air column length L = λ/4.

Given L = 17 cm = 0.17 m and f = 512 Hz ⇒ λ = 4L = 4 × 0.17 = 0.68 m.

(a) Speed at 20 °C: v(20 °C) = fλ = 512 × 0.68 = 348.16 m s^-1.

(b) Using v ∝ √T, with v(20 °C) at T = 293 K and v(0 °C) at T = 273 K:

v(0 °C) = v(20 °C) × √(273 / 293) ≈ 348.16 × √(273/293) ≈ 348.16 × 0.9836 ≈ 342.5 m s^-1. (Approximate value; many references give ≈ 331 m s^-1 at 0 °C because v ≈ 331 + 0.6T(°C).)

(c) Replacing the water with mercury does not change the resonance condition for the air column length, since both act as rigid reflecting surfaces. The resonance positions (air column lengths) remain essentially the same; however, reflection at a mercury surface may be slightly more perfect, producing a marginally sharper resonance.

Q.7. Show that when a string fixed at its two ends vibrates in 1 loop, 2 loops, 3 loops and 4 loops, the frequencies are in the ratio 1 : 2 : 3 : 4.
Ans. For a string of length L fixed at both ends, the nth harmonic has n half-wavelengths on the string: L = n(λ/2) ⇒ λ = 2L / n.

Frequency f_n = v / λ = v / (2L / n) = n(v / 2L). Thus f_n ∝ n. Therefore the frequencies for n = 1, 2, 3, 4 are in ratio 1 : 2 : 3 : 4.

LONG ANSWER TYPE QUESTIONS

Q.1. The earth has a radius of 6400 km. The inner core of 1000 km radius is solid. Outside it, there is a region from 1000 km to a radius of 3500 km which is in molten state. Then again from 3500 km to 6400 km the earth is solid. Only longitudinal (P) waves can travel inside a liquid. Assume that the P wave has a speed of 8 km s^-1 in solid parts and of 5 km s^-1 in liquid parts of the earth. An earthquake occurs at some place close to the surface of the earth. Calculate the time after which it will be recorded in a seismometer at a diametrically opposite point on the earth if wave travels along diameter?
Ans. Given radii: inner solid core radius r1 = 1000 km, molten region outer radius r2 = 3500 km, earth surface radius r3 = 6400 km.

Distances along radius from centre:

d1 = r1 = 1000 km (inner solid core)

d2 = r2 - r1 = 3500 - 1000 = 2500 km (molten region)

d3 = r3 - r2 = 6400 - 3500 = 2900 km (outer solid)

Wave travels along diameter, so it traverses each concentric layer twice.

Total solid distance = 2(d1 + d3) = 2(1000 + 2900) = 7800 km.

Total liquid distance = 2d2 = 2 × 2500 = 5000 km.

Time in solid parts: t_s = distance / speed = 7800 km / 8 km s^-1 = 975 s.

Time in liquid parts: t_l = 5000 km / 5 km s^-1 = 1000 s.

Total time = t_s + t_l = 975 + 1000 = 1975 s ≈ 32 min 55 s.

Q.2. If c is r.m.s. speed of molecules in a gas and v is the speed of sound waves in the gas, show that c/v is constant and independent of temperature for all diatomic gases.
Ans. For an ideal gas, rms molecular speed c = √(3RT/M) and speed of sound v = √(γRT/M), where R is the gas constant, T absolute temperature, M molar mass, and γ the adiabatic index.

Ratio c / v = √(3RT/M) / √(γRT/M) = √(3/γ).

For diatomic gases γ ≈ 7/5, so c / v = √(3 / (7/5)) = √(15/7), a constant independent of temperature.

Q.3. Given below are some functions of x and t to represent the displacement of an elastic wave.
(a) y = 5 cos (4x ) sin(20t)
(b) y = 4 sin(5x - t/2) + 3 cos (5x - t/2)
(c) y = 10 cos [(252 - 250) πt ] cos [(252+250)πt ]
(d) y = 100 cos (100πt + 0.5x )
State which of these represent
(a) a travelling wave along -x direction
(b) a stationary wave
(c) beats
(d) a travelling wave along +x direction
Given reasons for your answers.
Ans. Analysis:

(a) y = 5 cos(4x) sin(20t) has spatial factor cos(4x) and temporal factor sin(20t); this is of the form A sin(ωt) cos(kx) (or 2a sin kx cos ωt) - a stationary (standing) wave.

(b) y = 4 sin(5x - t/2) + 3 cos(5x - t/2) can be combined into a single sinusoid A cos(5x - t/2 + φ). The phase is (kx - ωt) so it represents a travelling wave moving in the +x direction.

(c) y = 10 cos[(252 - 250)πt] cos[(252 + 250)πt] = 10 cos(2πt) cos(502πt). The product of cosines with closely spaced frequencies gives a slowly varying envelope (beat) modulating a rapidly oscillating carrier - this is beats.

(d) y = 100 cos(100πt + 0.5x) has phase ωt + kx, which corresponds to a wave travelling in the -x direction (because of the +kx sign with +ωt). Thus (d) is a travelling wave along -x.

Q.4. In the given progressive wave
y = 5 sin (100πt - 0.4πx )
where y and x are in m, t is in s. What is the
(a) amplitude
(b) wave length
(c) frequency
(d) wave velocity
(e) particle velocity amplitude
Ans.  Compare with y = a sin(ωt - kx + φ).

(a) Amplitude a = 5 m.

(b) Wave number k = 0.4π rad m^-1. λ = 2π / k = 2π / (0.4π) = 2 / 0.4 = 5 m.

(c) ω = 100π rad s^-1 ⇒ f = ω / (2π) = 100π / (2π) = 50 Hz.

(d) Wave velocity v = fλ = 50 × 5 = 250 m s^-1.

(e) Particle velocity u = ∂y/∂t = aω cos(ωt - kx). Particle velocity amplitude = aω = 5 × 100π = 500π m s^-1.

Q.5. For the harmonic travelling wave y = 2 cos 2π (10t-0.0080x + 3.5) where x and y are in cm and t is second. What is the phase difference between the oscillatory motion at two points separated by a distance of
(a) 4 m
(b) 0.5 m
(c) λ/2
(d) 3λ/4 (at a given instant of time)
(e) What is the phase difference between the oscillation of a particle located at x = 100cm, at t = T s and t = 5 s?
Ans. Given y = 2 cos[2π(10t - 0.0080x + 3.5)]. Phase ϕ = 2π(10t - 0.0080x + 3.5) = ωt - kx + constant.

Identify ω and k:

ω = 2π × 10 = 20π rad s^-1.

k = 2π × 0.0080 = 0.016π rad cm^-1. (Because x given in cm, k is per cm.)

(a) For separation Δx = 4 m = 400 cm: Δϕ = kΔx = 0.016π × 400 = 6.4π rad.

(b) For Δx = 0.5 m = 50 cm: Δϕ = 0.016π × 50 = 0.8π rad.

(c) For Δx = λ/2: phase difference = k(λ/2) = π rad.

(d) For Δx = 3λ/4: phase difference = k(3λ/4) = 3π/2 rad.

(e) Time dependence: advancing time by one period T adds 2π to phase. Given f = 10 Hz, T = 0.1 s. At fixed x = 100 cm, phase at t = T is ϕ(T) = ωT - kx + constant = ωT + (rest). Phase at t = 5 s is ϕ(5) = ω × 5 - kx + constant. Phase difference Δϕ = ω(5 - T).

Since ω = 20π rad s^-1, T = 1/f = 0.1 s, so Δt = 5 - 0.1 = 4.9 s.

Δϕ = ωΔt = 20π × 4.9 = 98π rad. (Phase differences modulo 2π may be reduced if needed.)