Waves — NCERT Class 11 Physics

Q1. 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 a transverse jerk is struck at one end, how long does it take to reach the other end?

μ = mass/length = 2.50/20.0 = 0.125 kg/m v = √(T/μ) = √(200/0.125) = √1600 = 40 m/s Time = distance/speed = 20/40 = 0.5 s

Q2. A stone dropped from the top of a tower 300 m high splashes into water. When is the splash heard at the top? g = 10 m/s², speed of sound = 340 m/s.

Time for stone to fall: s = ½gt² → 300 = ½ × 10 × t² → t₁ = √60 = 7.75 s Time for sound to travel up: t₂ = 300/340 = 0.88 s Total time = 7.75 + 0.88 = 8.63 s

Q3. Two waves are described by y₁ = A sin(ωt) and y₂ = A sin(ωt + φ). What is the resultant amplitude when φ = π/2? When φ = π?

Q4. A pipe 20 cm long is closed at one end. Which harmonic mode is resonantly excited by a 430 Hz source? v_sound = 340 m/s.

Fundamental: ν₁ = v/(4L) = 340/(4 × 0.20) = 340/0.8 = 425 Hz Third harmonic: ν₃ = 3 × 425 = 1275 Hz Since 430 Hz ≈ 425 Hz, the source is close to the fundamental mode.

Q1. A string of length L is fixed at both ends. The wave speed is v. Write the frequencies of the first three harmonics.

Fundamental: ν₁ = v/(2L) Second harmonic: ν₂ = 2v/(2L) = v/L Third harmonic: ν₃ = 3v/(2L) = 3v/(2L)

Q2. Given that the wave y(x,t) = 0.05 sin(20πx − 4πt) m, find (a) wavelength (b) frequency (c) amplitude (d) wave speed (e) direction.

Compare with y = A sin(kx − ωt) (a) k = 20π = 2π/λ → λ = 2π/(20π) = 0.1 m = 10 cm (b) ω = 4π = 2πν → ν = 2 Hz (c) A = 0.05 m = 5 cm (d) v = ω/k = 4π/(20π) = 0.2 m/s (e) The (kx − ωt) sign means the wave travels in the +x direction

Q3. A wave travelling along a string is described by y(x,t) = 0.005 sin(80.0 x − 3.0 t) in SI units. Find (a) wavelength (b) frequency (c) speed (d) amplitude (e) maximum velocity of a particle.

k = 80 rad/m, ω = 3 rad/s, A = 0.005 m (a) λ = 2π/k = 2π/80 = 0.0785 m ≈ 7.85 cm (b) ν = ω/(2π) = 3/(2π) = 0.477 Hz (c) v = ω/k = 3/80 = 0.0375 m/s (d) A = 0.005 m = 5 mm (e) v_max(particle) = Aω = 0.005 × 3 = 0.015 m/s = 1.5 cm/s

Q4. A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of 45 Hz. The mass of the wire is 3.5 × 10⁻² kg and its linear mass density is 4.0 × 10⁻² kg m⁻¹. Find (a) speed of the transverse wave (b) tension.

(a) ν₁ = v/(2L) → v = 2Lν₁ Length: L = mass/μ = 3.5 × 10⁻²/(4.0 × 10⁻²) = 0.875 m v = 2 × 0.875 × 45 = 78.75 m/s (b) v = √(T/μ) → T = μv² = 4.0 × 10⁻² × (78.75)² = 4.0 × 10⁻² × 6201.56 = 248.1 N

Q5. A steel wire of diameter 1 mm is under a tension of 50 N. Find the speed of transverse waves. Density of steel = 7.8 g/cm³.

r = 0.5 × 10⁻³ m A = πr² = π × (0.5 × 10⁻³)² = 7.854 × 10⁻⁷ m² ρ = 7800 kg/m³ μ = ρA = 7800 × 7.854 × 10⁻⁷ = 6.126 × 10⁻³ kg/m v = √(T/μ) = √(50/6.126 × 10⁻³) = √(8162) = 90.3 m/s

Q6. A sitar wire is replaced by one of the same material but 1.5 times the original diameter. Will the new frequency be higher or lower?

ν ∝ v ∝ 1/√μ, and μ ∝ A ∝ d² New μ = (1.5)² × μ₀ = 2.25 μ₀ New v = v₀/√2.25 = v₀/1.5 New ν = ν₀/1.5 (lower)

Q7. Two tuning forks A and B produce 4 beats per second. On loading B with wax, the beat frequency becomes 6 per second. If the frequency of A is 256 Hz, what is the frequency of B?

Originally: |ν_B − 256| = 4 → ν_B = 260 Hz or 252 Hz After loading B: ν_B decreases (loading lowers frequency) If ν_B = 260 → after loading, ν_B' < 260, beats = |ν_B' − 256| = 6 → ν_B' = 250 or 262 Since ν_B' < 260, ν_B' = 250, beats = |250 − 256| = 6 ✓ If ν_B = 252 → after loading ν_B' < 252, beats = |ν_B' − 256| = 6 → ν_B' = 250, beats = |250 − 256| = 6 ✓ Both work. But conventionally, ν_B = 260 Hz (since loading decreased frequency, increasing beat frequency from 4 to 6 confirms B was higher than A).

Q8. Two sound waves of frequencies 256 Hz and 254 Hz are sounded together. Find the beat frequency and the time interval between successive maxima.

Beat frequency = |256 − 254| = 2 Hz Time interval = 1/2 = 0.5 s

Q9. A closed organ pipe and an open organ pipe of the same length produce 2 beats per second when sounded together. If the length of each pipe is 0.5 m, find the frequencies. v_sound = 350 m/s.

Open pipe: ν_open = v/(2L) = 350/(2 × 0.5) = 350 Hz Closed pipe: ν_closed = v/(4L) = 350/(4 × 0.5) = 175 Hz But these differ by 175 Hz, not 2 Hz. The question likely compares the fundamental of one with the second harmonic of the other. ν_open = 350 Hz, 3rd harmonic of closed = 3 × 175 = 525 Hz — not matching. Actually, the problem states both produce 2 beats when sounded together. If we consider the first overtone of closed pipe: ν = 3v/(4L) = 525 Hz. Perhaps the problem means: find frequencies that differ by 2 Hz involving the same harmonic number or nearby frequencies. Let me reconsider: perhaps both pipes have different effective lengths or the question refers to a specific situation. With L = 0.5 m: ν_open = 350 Hz, ν_closed = 175 Hz. These don't produce 2 beats. The problem may intend a different setup. Taking the given data at face value: If the closed pipe fundamental is close to some harmonic of the open pipe: 3 × 175 = 525 (3rd harmonic of closed) — no 2 Hz difference with 350. The question likely has a different intended reading. With the given values: ν_closed(fundamental) = 175 Hz, ν_open(fundamental) = 350 Hz.

Q10. A pipe 30 cm long is open at both ends. Which harmonics can be produced at 20°C? v_sound = 340 m/s.

ν_n = nv/(2L) = n × 340/(2 × 0.30) = n × 566.7 Hz Fundamental (n = 1) = 567 Hz n = 2: 1133 Hz n = 3: 1700 Hz All harmonics (n = 1, 2, 3, ...) can be produced since the pipe is open at both ends.

Q11. A 40 cm long organ pipe closed at one end is sounding its third harmonic. What is the frequency? v_sound = 340 m/s.

For closed pipe: ν_n = nv/(4L) where n = 1, 3, 5, ... Third harmonic means n = 3 ν₃ = 3 × 340/(4 × 0.40) = 1020/1.6 = 637.5 Hz

Q12. An observer moving towards a wall at 5 m/s hears a beat frequency of 5 Hz between the sound directly from the source (500 Hz) and the reflected sound. Find the speed of sound.

Source at rest, observer moves towards wall at 5 m/s. For direct sound (observer approaching source): ν₁ = ν(v + v_o)/v For reflected sound (wall acts as a source, observer approaching it): ν₂ = ν(v + v_o)/v Both give the same formula since the observer is moving towards both. Wait — the observer moves towards the wall. The source is behind the observer. For direct: observer moves towards source → ν₁ = ν(v + v_o)/v For reflected: wall is a stationary "source" and observer moves towards it → ν₂ = ν(v + v_o)/v These give the same frequency! Beats = 0. This contradicts 5 Hz beats. The key: for the reflected sound, the wall acts as a source moving at... no, the wall is stationary. Actually, the observer is moving. The source is stationary behind. The direct frequency: ν₁ = ν(v + v_o)/v (observer approaching stationary source) For reflected: The wall receives frequency ν_wall = ν (wall is stationary, source stationary). The wall re-emits at ν_wall. The observer approaches this "source": ν₂ = ν_wall(v + v_o)/v = ν(v + v_o)/v Same as ν₁ — no beats. Something is off. Let me reconsider: the observer is between source and wall. Observer moves towards wall. Direct sound: source behind observer, observer moves towards wall (away from source). ν₁ = ν(v − v_o)/v (observer receding from source) Reflected: observer approaches the wall (acting as source): ν₂ = ν(v + v_o)/v Beat = ν₂ − ν₁ = ν(v + v_o)/v − ν(v − v_o)/v = ν × 2v_o/v = 500 × 10/v 5 = 5000/v → v = 1000 m/s... That seems too high. Actually: 5 = 500 × 2 × 5/v = 5000/v → v = 1000 m/s. Hmm. Let me check: typical speed of sound ≈ 340 m/s. With v = 340: beats = 500 × 10/340 = 14.7 Hz — not 5 Hz. The problem as stated gives v = 1000 m/s which is non-physical. Perhaps the source frequency is different. Using the formula as derived: Beat freq = 2νv_o/v → v = 2νv_o/(beat) = 2 × 500 × 5/5 = 1000 m/s.

Q13. A train standing at the outer signal of a railway station blows a whistle of frequency 400 Hz in still air. What is the frequency heard by a man on the platform when the train (a) approaches at 10 m/s (b) recedes at 10 m/s? v_sound = 340 m/s.

(a) Approaching: ν' = ν × v/(v − v_s) = 400 × 340/(340 − 10) = 400 × 340/330 = 412.1 Hz (b) Receding: ν' = ν × v/(v + v_s) = 400 × 340/(340 + 10) = 400 × 340/350 = 388.6 Hz

Q14. A train approaching a railway crossing at 108 km/h sounds a whistle of frequency 500 Hz. What is the frequency heard by a person standing at the crossing? v_sound = 340 m/s.

v_s = 108 km/h = 30 m/s ν' = ν × v/(v − v_s) = 500 × 340/(340 − 30) = 500 × 340/310 = 548.4 Hz

Q15. A submarine at 100 m depth emits a sonar pulse which returns from a rock in 0.8 s. What is the distance of the rock? Speed of sound in water = 1500 m/s.

Total distance = speed × time = 1500 × 0.8 = 1200 m Distance of rock = 1200/2 = 600 m

Q16. What are stationary (standing) waves? When are they produced?

Q17. Differentiate between progressive and stationary waves.

Q18. A bat emits ultrasonic sound of frequency 1000 kHz. It flies towards a wall at 36 km/h. Find the frequency of the reflected sound as heard by the bat. v_sound = 340 m/s.

v_bat = 36 km/h = 10 m/s Frequency at wall: ν_wall = ν × v/(v − v_s) = 1000 × 340/330 = 1030.3 kHz Bat (now observer) approaches the wall (source at rest re-emitting): ν' = ν_wall × (v + v_o)/v = 1030.3 × (340 + 10)/340 = 1030.3 × 350/340 = 1060.9 kHz

Q19. Explain why a violin and a sitar playing the same note sound different.

Q20. A string 2 m long is fixed at both ends. A 500 Hz tuning fork sets up resonance. Find (a) wavelength (b) which harmonic. v = 350 m/s.

(a) λ = v/ν = 350/500 = 0.7 m (b) For string fixed at both ends: λ_n = 2L/n → n = 2L/λ = 2 × 2/0.7 = 40/7 ≈ 5.71 Since n must be an integer, n = 6 (approximately). Let me recalculate: n = 2L/λ = 4/0.7 = 5.71 — not an integer. Perhaps the exact length or speed gives n = 5 or 6. With the given numbers: 5th harmonic: ν₅ = 5 × 350/4 = 437.5 Hz 6th harmonic: ν₆ = 6 × 350/4 = 525 Hz Neither matches exactly. The closest is the 6th harmonic at 525 Hz. The problem likely uses slightly different values or expects the answer n = 5 or 6.

Q21. Show that standing waves on a string of length L fixed at both ends have wavelengths λ_n = 2L/n.

For a standing wave, nodes must occur at both fixed ends. Distance between consecutive nodes = λ/2 For n such segments in length L: n(λ/2) = L λ = 2L/n, n = 1, 2, 3, ...

Q22. An open organ pipe has fundamental frequency 300 Hz. The first overtone of a closed organ pipe has the same frequency. Find the length ratio.

Open pipe: ν₁_open = v/(2L_open) = 300 Closed pipe first overtone (3rd harmonic): ν₃_closed = 3v/(4L_closed) = 300 From open: v = 600L_open Substituting: 3 × 600L_open/(4L_closed) = 300 1800L_open/(4L_closed) = 300 L_open/L_closed = 300 × 4/1800 = 1200/1800 = 2/3

Q23. A sound wave of frequency 1000 Hz travels in air at 340 m/s. What is (a) the wavelength (b) the time period?

(a) λ = v/ν = 340/1000 = 0.34 m = 34 cm (b) T = 1/ν = 1/1000 = 10⁻³ s = 1 ms

Q24. The speed of sound in air at NTP is 331 m/s. Find the speed at 30°C.

v ∝ √T (absolute temperature) T_NTP = 273 K, T₂ = 303 K v₂ = v₁ × √(303/273) = 331 × √1.11 = 331 × 1.054 = 348.8 m/s

Q25. The speed of sound in oxygen at NTP is 316 m/s. What is the speed in helium at the same temperature and pressure? M_O₂ = 32, M_He = 4.

v ∝ 1/√M v_He/v_O₂ = √(M_O₂/M_He) = √(32/4) = √8 = 2.83 v_He = 316 × 2.83 = 894 m/s

Q26. What is a displacement node? What is a pressure antinode?

Q27. The fundamental frequency of a closed organ pipe is 250 Hz. What is the frequency of its (a) first overtone (b) second overtone?

Closed pipe: only odd harmonics Fundamental (n = 1): 250 Hz (a) First overtone (n = 3): 3 × 250 = 750 Hz (b) Second overtone (n = 5): 5 × 250 = 1250 Hz