Physics Mechanics - How to Solve: Doppler Effect in Sound (Source/Observer Moving, Wind) – IIT JEE Guide

How to Solve: Doppler Effect in Sound (Source/Observer Moving, Wind) – IIT JEE Guide

Introduction

Mastering the Doppler Effect in sound unlocks 3-5 marks in IIT JEE (Main + Advanced)—often as a standalone question or part of a wave/optics problem. It’s also the key to real-world applications like radar speed guns, medical ultrasound, and astronomy. If you can solve this, you’re not just scoring marks—you’re beating the curve.

WHAT YOU NEED TO KNOW FIRST

Before diving in, ensure you understand: 1. Wave frequency vs. wavelength – Frequency is the number of waves per second; wavelength is the distance between two crests. 2. Relative velocity – How the speed of the source and observer affects the wave they perceive. 3. Effect of medium (wind) – How wind alters the effective speed of sound.

KEY TERMS & FORMULAS

Key Terms

1. Source (S) – The object emitting sound waves.
2. Observer (O) – The object detecting sound waves.
3. Apparent frequency (f') – The frequency heard by the observer.
4. Actual frequency (f) – The frequency emitted by the source.
5. Speed of sound (v) – Speed of sound in air (~340 m/s at 25°C).
6. Wind speed (w) – Speed of wind (affects effective speed of sound).

Formulas

1. General Doppler Effect (No Wind)

MEMORISE THIS [ f' = f \left( \frac{v \pm v_o}{v \mp v_s} \right) ]

• ( f' ) = Apparent frequency (heard by observer)
• ( f ) = Actual frequency (emitted by source)
• ( v ) = Speed of sound in air (~340 m/s)
• ( v_o ) = Speed of observer
• ( v_s ) = Speed of source

Sign Convention: - Top sign (+ for ( v_o ), – for ( v_s )) → Observer/source moving toward each other. - Bottom sign (– for ( v_o ), + for ( v_s )) → Observer/source moving away from each other.

2. Doppler Effect with Wind

MEMORISE THIS [ f' = f \left( \frac{v \pm w \pm v_o}{v \pm w \mp v_s} \right) ]

• ( w ) = Wind speed (positive if wind is in the same direction as sound, negative if opposite).
• Signs for ( v_o ) and ( v_s ) follow the same rule as above.

3. Special Cases (Given on Exam Sheet)

• Observer at rest, source moving toward observer: [ f' = f \left( \frac{v}{v - v_s} \right) ]
• Observer moving toward source, source at rest: [ f' = f \left( \frac{v + v_o}{v} \right) ]
• Both moving toward each other: [ f' = f \left( \frac{v + v_o}{v - v_s} \right) ]

STEP-BY-STEP METHOD

Step 1: Identify the Scenario

• Is the source moving? If yes, note its speed (( v_s )) and direction.
• Is the observer moving? If yes, note its speed (( v_o )) and direction.
• Is there wind? If yes, note its speed (( w )) and direction.

Step 2: Assign Signs Correctly

• Observer moving toward source? → Use +( v_o ) in numerator.
• Observer moving away from source? → Use –( v_o ) in numerator.
• Source moving toward observer? → Use –( v_s ) in denominator.
• Source moving away from observer? → Use +( v_s ) in denominator.
• Wind in same direction as sound? → +( w ) in both numerator and denominator.
• Wind opposite to sound direction? → –( w ) in both numerator and denominator.

Step 3: Plug into the Correct Formula

• No wind? Use: [ f' = f \left( \frac{v \pm v_o}{v \mp v_s} \right) ]
• With wind? Use: [ f' = f \left( \frac{v \pm w \pm v_o}{v \pm w \mp v_s} \right) ]

Step 4: Simplify and Solve for ( f' )

• Substitute all known values.
• Simplify the fraction.
• Calculate the final apparent frequency.

Step 5: Check Units and Reasonableness

• Ensure all speeds are in m/s.
• If ( f' > f ), the observer should be moving toward the source (or source toward observer).
• If ( f' < f ), the observer should be moving away (or source away).

WORKED EXAMPLES

Example 1 – Basic (No Wind)

Problem: A car (source) emits a sound of frequency 500 Hz while moving toward a stationary observer at 20 m/s. Speed of sound = 340 m/s. What frequency does the observer hear?

Solution: 1. Identify scenario:
- Source moving toward observer (( v_s = 20 \, \text{m/s} )).
- Observer stationary (( v_o = 0 )).
- No wind.

1. Assign signs:
2. Source moving toward observer → –( v_s ) in denominator.

3. Observer stationary → ( v_o = 0 ).

4. Formula:
   [ f' = f \left( \frac{v}{v - v_s} \right) ]

5. Substitute values:
   [ f' = 500 \left( \frac{340}{340 - 20} \right) ]
   [ f' = 500 \left( \frac{340}{320} \right) ]
   [ f' = 500 \times 1.0625 ]
   [ f' = 531.25 \, \text{Hz} ]

6. Check:

7. ( f' > f ) (expected, since source is moving toward observer).

What we did and why: We used the source-moving-toward-observer formula because the car is approaching the observer. The denominator decreases, increasing the frequency.

Example 2 – Medium (Observer Moving, Wind Present)

Problem: A train (source) emits a 600 Hz sound while moving away from an observer at 15 m/s. The observer is moving toward the train at 10 m/s. Wind is blowing toward the observer at 5 m/s. Speed of sound = 340 m/s. What frequency does the observer hear?

Solution: 1. Identify scenario:
- Source moving away (( v_s = 15 \, \text{m/s} )).
- Observer moving toward source (( v_o = 10 \, \text{m/s} )).
- Wind toward observer (( w = +5 \, \text{m/s} )).

1. Assign signs:
2. Observer moving toward source → +( v_o ) in numerator.
3. Source moving away → +( v_s ) in denominator.

4. Wind in same direction as sound (toward observer) → +( w ) in both numerator and denominator.

5. Formula:
   [ f' = f \left( \frac{v + w + v_o}{v + w + v_s} \right) ]

6. Substitute values:
   [ f' = 600 \left( \frac{340 + 5 + 10}{340 + 5 + 15} \right) ]
   [ f' = 600 \left( \frac{355}{360} \right) ]
   [ f' = 600 \times 0.9861 ]
   [ f' = 591.67 \, \text{Hz} ]

7. Check:

8. ( f' < f ) (expected, since source is moving away, but observer is moving toward it—net effect is slight decrease).

What we did and why: We accounted for both observer and source motion, plus wind effect. The wind increases the effective speed of sound, but the source moving away dominates, reducing the frequency.

Example 3 – Exam-Style (Disguised Problem)

Problem: A police siren emits a 1000 Hz sound. A thief is running away from the police car at 12 m/s, while the police car is chasing at 20 m/s. Wind is blowing against the police car at 8 m/s. Speed of sound = 340 m/s. What frequency does the thief hear?

Solution: 1. Identify scenario:
- Source (police car) moving toward thief (( v_s = 20 \, \text{m/s} )).
- Observer (thief) moving away from source (( v_o = 12 \, \text{m/s} )).
- Wind against police car (opposite to sound direction) → ( w = -8 \, \text{m/s} ).

1. Assign signs:
2. Observer moving away → –( v_o ) in numerator.
3. Source moving toward observer → –( v_s ) in denominator.

4. Wind opposite to sound → –( w ) in both numerator and denominator.

5. Formula:
   [ f' = f \left( \frac{v - w - v_o}{v - w - v_s} \right) ]

6. Substitute values:
   [ f' = 1000 \left( \frac{340 - (-8) - 12}{340 - (-8) - 20} \right) ]
   [ f' = 1000 \left( \frac{340 + 8 - 12}{340 + 8 - 20} \right) ]
   [ f' = 1000 \left( \frac{336}{328} \right) ]
   [ f' = 1000 \times 1.0244 ]
   [ f' = 1024.4 \, \text{Hz} ]

7. Check:

8. ( f' > f ) (expected, since source is moving toward observer faster than observer is moving away).

What we did and why: This is a realistic exam problem—it disguises the scenario as a "police chase" but tests the same Doppler principles. We carefully assigned signs for wind, source, and observer motion.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is the Doppler Effect in 60 seconds. First, identify who’s moving: source, observer, or both? If the source is moving toward you, the frequency increases—so denominator gets smaller (( v - v_s )). If the observer is moving toward the source, the frequency increases—so numerator gets bigger (( v + v_o )). Wind? If it’s with the sound, add ( w ) to ( v ). If it’s against, subtract ( w ). Always double-check signs—one wrong sign, and your answer is garbage. Memorise the general formula, but for stationary cases, use the simpler versions. And if you get a negative frequency? You messed up the signs. Now go crush it!