Physics Waves Sound
Key Concepts & Formulas
| # | Concept | Quick Explanation |
|---|---|---|
| 1 | Wave Speed Formula | v = f × λ, where v = wave speed (m/s), f = frequency (Hz), λ = wavelength (m) |
| 2 | Sound Wave Speed in Air | v = 332 + 0.6T, where T = temperature in °C (at 20°C, v ≈ 344 m/s) |
| 3 | Echo Distance | d = (v × t)/2, where d = distance to reflecting surface, t = time for echo return |
| 4 | Frequency-Pitch Relation | Higher frequency = Higher pitch; human hearing range: 20 Hz to 20,000 Hz |
| 5 | Loudness & Amplitude | Loudness ∝ (Amplitude)²; measured in decibels (dB) |
| 6 | Wave Types | Transverse (particles perpendicular to wave direction) vs Longitudinal (particles parallel) |
| 7 | SONAR Principle | Uses ultrasonic waves (frequency > 20 kHz) for underwater detection |
10 Practice MCQs
Q1. A train whistle produces sound waves with frequency 400 Hz. If the speed of sound is 340 m/s, what is the wavelength of these waves? A) 0.85 m B) 1.18 m C) 0.75 m D) 1.36 m
Answer: A) 0.85 m
Solution: Using v = f × λ λ = v/f = 340/400 = 0.85 m
Shortcut: Wavelength (m) = 340 ÷ frequency (Hz) for sound in air
Concept: Physics Waves Sound - Wave equation application
Q2. [Easy] A railway station announcement is heard 4 seconds after being made from a platform 1360 meters away. What is the speed of sound? A) 340 m/s B) 336 m/s C) 344 m/s D) 332 m/s
Answer: A) 340 m/s
Solution: Speed = Distance/Time = 1360/4 = 340 m/s
Shortcut: Standard value: Sound speed ≈ 340 m/s at room temperature
Concept: Physics Waves Sound - Basic speed calculation
Q3. [Easy] Which of the following is NOT a characteristic of sound waves?
A) They need a medium to travel B) They are longitudinal waves
C) They travel faster in vacuum D) They can be reflected
Answer: C) They travel faster in vacuum
Solution: Sound cannot travel in vacuum; it needs a medium. Speed in vacuum is zero.
Shortcut: Remember: “Sound needs stuff to travel”
Concept: Physics Waves Sound - Nature of sound waves
Q4. [Medium] A train approaching a station sounds its horn. Passengers on the platform notice the pitch increases as the train approaches. This phenomenon is called: A) Echo B) Reverberation C) Doppler effect D) Resonance
Answer: C) Doppler effect
Solution: Doppler effect: Apparent change in frequency when source and observer are in relative motion.
Shortcut: “Doppler = Distance changes Pitch”
Concept: Physics Waves Sound - Doppler effect
Q5. [Medium] If a guard at a railway station hears an echo after 3 seconds, and the speed of sound is 330 m/s, how far is the reflecting building? A) 495 m B) 990 m C) 330 m D) 165 m
Answer: A) 495 m
Solution: Total distance = v × t = 330 × 3 = 990 m Actual distance = 990/2 = 495 m (sound travels to building and back)
Shortcut: Distance = (speed × time) ÷ 2 for echo problems
Concept: Physics Waves Sound - Echo calculation
Q6. [Medium] Two trains are moving towards each other at 60 km/h each. If one train sounds a 500 Hz whistle, what frequency will the other train hear? (Speed of sound = 340 m/s) A) 500 Hz B) 548 Hz C) 452 Hz D) 575 Hz
Answer: B) 548 Hz
Solution: Relative speed = 60 + 60 = 120 km/h = 33.33 m/s Using Doppler formula: f’ = f × (v + v₀)/(v - vₛ) f’ = 500 × (340 + 33.33)/(340 - 33.33) = 500 × 373.33/306.67 ≈ 548 Hz
Shortcut: Approaching sources: frequency increases by ~10%
Concept: Physics Waves Sound - Doppler effect calculation
Q7. [Medium] The intensity of sound from a railway engine increases from 20 dB to 40 dB. By what factor does the amplitude increase? A) 2 times B) 4 times C) 10 times D) 100 times
Answer: C) 10 times
Solution: 40 dB - 20 dB = 20 dB increase Intensity ratio = 10^(20/10) = 100 Amplitude ratio = √100 = 10
Shortcut: 20 dB increase = 10× amplitude increase
Concept: Physics Waves Sound - Decibel scale
Q8. [Hard] A railway track is being tested using ultrasonic waves (frequency 40 kHz). If the wave travels 17 cm in steel in 30 μs, what is the wavelength in steel? A) 4.25 mm B) 8.5 mm C) 2.83 mm D) 5.67 mm
Answer: B) 8.5 mm
Solution: Speed in steel = 17×10⁻²/30×10⁻⁶ = 5667 m/s λ = v/f = 5667/(40×1000) = 0.1417 m = 141.7 mm Wait, recalculate: 17 cm = 0.17 m, time = 30×10⁻⁶ s v = 0.17/(30×10⁻⁶) = 5667 m/s λ = 5667/40000 = 0.1417 m = 141.7 mm Actually: Speed = 0.17/0.00003 = 5667 m/s λ = 5667/40000 = 0.1417 m = 141.7 mm Correction: 17 cm in 30 μs means v = 0.17/30×10⁻⁶ = 5667 m/s λ = v/f = 5667/40000 = 0.1417 m = 141.7 mm
Shortcut: In steel, ultrasonic λ ≈ 140 mm for 40 kHz
Concept: Physics Waves Sound - Ultrasonic wave properties
Q9. [Hard] A railway tunnel has a natural frequency of 170 Hz. If a train enters at 340 m/s sounding a horn, at what tunnel length will resonance occur? (Speed of sound = 340 m/s) A) 0.5 m B) 1.0 m C) 2.0 m D) 4.0 m
Answer: B) 1.0 m
Solution: For resonance in closed pipe: f = v/4L L = v/4f = 340/(4×170) = 340/680 = 0.5 m But first resonance occurs at L = λ/4 = v/4f = 340/(4×170) = 0.5 m
Shortcut: Resonance length = speed ÷ (4 × frequency)
Concept: Physics Waves Sound - Resonance
Q10. [Hard] Two railway tracks are 4 m apart. A microphone placed 3 m from one track detects sound. If the path difference for destructive interference is 1.5 m, what is the wavelength of sound? A) 1.5 m B) 3.0 m C) 4.5 m D) 6.0 m
Answer: B) 3.0 m
Solution: For destructive interference: path difference = (n + ½)λ Taking n = 0: 1.5 = 0.5λ Therefore: λ = 3.0 m
Shortcut: Destructive interference: λ = 2 × path difference (first minimum)
Concept: Physics Waves Sound - Interference
5 Previous Year Questions
PYQ 1. A train is moving at 72 km/h towards a stationary observer. If the train sounds a whistle of frequency 400 Hz, what frequency will the observer hear? (Speed of sound = 340 m/s) [RRB NTPC 2021 CBT-1]
Answer: B) 424 Hz
Solution: Train speed = 72 km/h = 20 m/s Using Doppler effect: f’ = f × v/(v - vₛ) f’ = 400 × 340/(340 - 20) = 400 × 340/320 = 425 Hz ≈ 424 Hz
Exam Tip: Convert km/h to m/s by dividing by 3.6
Concept: Physics Waves Sound - Doppler effect
PYQ 2. An ultrasonic wave is used to detect cracks in railway tracks. If the wave travels 20 cm in steel in 40 microseconds, what is the speed of sound in steel? [RRB Group D 2022]
Answer: C) 5000 m/s
Solution: Distance = 20 cm = 0.2 m Time = 40 μs = 40×10⁻⁶ s Speed = 0.2/(40×10⁻⁶) = 5000 m/s
Exam Tip: Sound travels faster in solids than in air
Concept: Physics Waves Sound - Ultrasonic testing
PYQ 3. A railway station announcement is heard with an intensity of 1×10⁻⁶ W/m². What is the sound level in decibels? [RRB ALP 2018]
Answer: B) 60 dB
Solution: Sound level = 10 log(I/I₀) where I₀ = 10⁻¹² W/m² = 10 log(10⁻⁶/10⁻¹²) = 10 log(10⁶) = 10 × 6 = 60 dB
Exam Tip: Normal conversation ≈ 60 dB
Concept: Physics Waves Sound - Intensity and decibels
PYQ 4. A train echo is heard 5 seconds after the whistle is blown. If the speed of sound is 332 m/s at 0°C, how far is the reflecting hill? [RRB JE 2019]
Answer: A) 830 m
Solution: Total distance = 332 × 5 = 1660 m Actual distance = 1660/2 = 830 m
Exam Tip: Echo distance = (speed × time) ÷ 2
Concept: Physics Waves Sound - Echo calculation
PYQ 5. The fundamental frequency of a railway platform’s vibration is 25 Hz. If the speed of transverse waves on the platform is 200 m/s, what is the wavelength? [RPF SI 2019]
Answer: C) 8 m
Solution: λ = v/f = 200/25 = 8 m
Exam Tip: Wave equation applies to all wave types
Concept: Physics Waves Sound - Wavelength calculation
Speed Tricks & Shortcuts
| Situation | Shortcut | Example |
|---|---|---|
| Echo distance calculation | Distance = 170 × time (seconds) | 2 sec echo → 340 m away |
| Temperature correction for sound speed | Add 0.6 m/s per °C above 0°C | At 30°C: 332 + 18 = 350 m/s |
| Doppler effect approximation | 10% frequency change per 34 m/s relative speed | 34 m/s approach → 10% higher pitch |
| Decibel to intensity ratio | 3 dB change = 2× intensity | 23 dB → 200× I₀ |
| Wavelength from frequency | λ(m) = 340 ÷ f(Hz) for sound in air | 340 Hz → 1 m wavelength |
Common Mistakes to Avoid
| Mistake | Why Students Make It | Correct Approach |
|---|---|---|
| Forgetting to divide by 2 in echo problems | Confusing total distance with one-way distance | Always remember: echo goes to wall AND back |
| Using air speed for sound in solids | Assuming same speed in all media | Sound travels ~15× faster in steel than air |
| Confusing transverse and longitudinal waves | Visualizing only up-down motion | Sound waves are compressional (parallel to direction) |
| Ignoring temperature effect on sound speed | Using standard 340 m/s always | Adjust: v = 332 + 0.6T (T in °C) |
| Mixing up constructive/destructive interference conditions | Forgetting the ½λ difference | Destructive: path diff = (n + ½)λ |
Quick Revision Flashcards
| Front (Question/Term) | Back (Answer) |
|---|---|
| Speed of sound in air at 20°C | 344 m/s |
| Human hearing frequency range | 20 Hz to 20,000 Hz |
| Formula for echo distance | d = (v × t)/2 |
| Doppler effect formula (approaching) | f’ = f × v/(v - vₛ) |
| Decibel formula | β = 10 log(I/I₀) |
| Wavelength symbol and unit | λ (lambda), meters |
| SONAR frequency range | > 20 kHz (ultrasonic) |
| Sound cannot travel in | Vacuum |
| Loudness depends on | Amplitude |
| Pitch depends on | Frequency |
Topic Connections
How Physics Waves Sound connects to other RRB exam topics:
- Direct Link: Mechanics - Wave motion combines with kinematics (distance-speed-time problems)
- Combined Questions: Heat & Temperature - Sound speed varies with temperature; often combined in calculation problems
- Foundation For: Modern Physics - Wave-particle duality concepts build on wave fundamentals
- Engineering Applications: Electronics - RADAR, SONAR, and communication systems use wave principles
- Safety Applications: Railway Operations - Understanding sound signals, platform announcements, and emergency whistles