This MCQ module is based on: Beats Doppler
TOPIC 32 OF 33
Beats Doppler
🎓 Class 11
Physics
CBSE
Theory
Ch 14 – Waves
⏱ ~14 min
🌐 Language: [gtranslate]
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Beats Doppler
14.10 Beats
What happens when two sound waves of slightly different frequencies are heard together? The result is a periodic rise and fall in loudness called beats.
Consider two waves with frequencies \(\nu_1\) and \(\nu_2\) (close together, \(|\nu_1-\nu_2|\ll \nu_1\)):
\[y_1 = a\cos(2\pi\nu_1 t),\quad y_2 = a\cos(2\pi\nu_2 t)\]
\[y = y_1+y_2 = 2a\cos\left(2\pi\frac{\nu_1-\nu_2}{2}t\right)\cos\left(2\pi\frac{\nu_1+\nu_2}{2}t\right)\]
The second factor is the rapid average-frequency tone; the first factor is a slowly varying envelope. The ear perceives loudness variations at the envelope's frequency. Since loudness ∝ amplitude², and the envelope passes through maximum amplitude twice per cycle, the beat frequency equals:
Beat Frequency:
\[\nu_{\text{beat}} = |\nu_1-\nu_2|\]
For two notes 440 Hz and 442 Hz, you hear 2 beats per second.
14.11 Doppler Effect
You've heard it: an ambulance siren rises in pitch as it approaches, then drops as it recedes. The Doppler effect, discovered by Austrian Christian Doppler in 1842, is the apparent change in frequency of a wave when there's relative motion between source and observer.
14.11.1 Source Moving, Observer Stationary
If the source moves toward the observer with speed \(v_s\), the wavefronts in front of it get compressed: each new crest is emitted at a position closer to the previous crest. The apparent wavelength is shorter, frequency higher:
\[\nu' = \nu_0\left(\frac{v}{v-v_s}\right)\quad\text{(source approaching)}\]
\[\nu' = \nu_0\left(\frac{v}{v+v_s}\right)\quad\text{(source receding)}\]
where v = speed of sound in the medium.
14.11.2 Observer Moving, Source Stationary
If the observer moves toward a stationary source with speed \(v_o\), they encounter wavefronts more frequently:
\[\nu' = \nu_0\left(\frac{v+v_o}{v}\right)\quad\text{(observer approaching)}\]
\[\nu' = \nu_0\left(\frac{v-v_o}{v}\right)\quad\text{(observer receding)}\]
14.11.3 Both Source and Observer Moving
Combining the two cases, the general formula (signs to be applied carefully):
General Doppler Formula:
\[\nu' = \nu_0\,\frac{v\pm v_o}{v\mp v_s}\]
Sign convention (take v_s, v_o as speeds, always positive):
- Numerator: +v_o if observer approaches source, −v_o if receding.
- Denominator: −v_s if source approaches observer, +v_s if receding.
| Situation | Formula | Effect |
|---|---|---|
| Source approaches stationary observer | ν' = ν₀·v/(v−v_s) | Higher pitch |
| Source recedes from stationary observer | ν' = ν₀·v/(v+v_s) | Lower pitch |
| Observer approaches stationary source | ν' = ν₀·(v+v_o)/v | Higher pitch |
| Observer recedes from stationary source | ν' = ν₀·(v−v_o)/v | Lower pitch |
| Both approach each other | ν' = ν₀·(v+v_o)/(v−v_s) | Highest pitch |
Interactive Simulation: Doppler Shift Calculator
Move source/observer to see how the apparent frequency changes.
Apparent frequency ν' = 548 Hz
Positive velocity = approaching. Take v_sound = 340 m/s.
Worked Example 1: Beats from two pianos
Two pianos play notes of frequencies 256 Hz and 260 Hz simultaneously. (a) Find the beat frequency. (b) How many beats per minute?
(a) \(\nu_{\text{beat}} = |260 - 256| = \boxed{4\,\text{Hz}}\) (4 beats/s).
(b) 4 × 60 = 240 beats per minute. The listener hears a wow-wow-wow rise-and-fall in loudness.
(b) 4 × 60 = 240 beats per minute. The listener hears a wow-wow-wow rise-and-fall in loudness.
Worked Example 2: Ambulance approaching
An ambulance siren has frequency 1000 Hz. It approaches at 20 m/s. Speed of sound = 340 m/s. Find the frequency heard by a stationary observer.
\[\nu' = \nu_0\frac{v}{v-v_s} = 1000\times\frac{340}{340-20}=1000\times\frac{340}{320}=\boxed{1062.5\,\text{Hz}}\]
A 62.5 Hz rise — easily noticed by ear. As the ambulance passes and recedes:
\[\nu' = 1000\times\frac{340}{340+20}=\boxed{944.4\,\text{Hz}}\]
A drop of 118 Hz from peak to receding — the audible "Doppler wail".
Worked Example 3: Both moving
A train (ν₀ = 600 Hz) moves at 40 m/s toward a car driver moving at 25 m/s toward the train. Find the frequency heard by the driver. Take v = 340 m/s.
Both approach: numerator v + v_o, denominator v − v_s.
\[\nu' = 600\times\frac{340+25}{340-40}=600\times\frac{365}{300}=600\times 1.217\approx \boxed{730\,\text{Hz}}\]
A 22% rise — about 4 semitones higher than the train's nominal pitch.
Activity 14.4 — Hearing Beats with Two Tuning ForksL3 Apply
Materials: Two identical tuning forks (e.g., 256 Hz), small piece of putty/wax.
- Strike both forks simultaneously and hold near the ear — they sound in unison.
- Stick a small piece of putty on one prong of one fork to lower its frequency slightly.
- Strike both and listen — you hear loudness rise and fall: BEATS.
- Add more putty: beat rate slows or speeds up depending on which fork was off.
Predict: If 4 beats per second are heard, how off-tune is the loaded fork?
The loaded fork is off by exactly 4 Hz — i.e., 252 Hz or 260 Hz (the beat itself doesn't tell you which direction). To find the direction: load it more — if beat rate increases, you went further off; if it decreases, you're approaching unison. This is precisely the technique trained piano tuners use to ensure perfect octaves.
Competency-Based Questions
A traffic police officer in Mumbai uses a Doppler radar to detect speeding vehicles on the Western Express Highway. The radar emits microwaves at 10 GHz and measures the frequency shift of the returned signal. Meanwhile, in another city, a music professor demonstrates beats using tuning forks.
Q1. Two waves of frequencies 100 Hz and 102 Hz are superposed. The beat frequency is:L1 Remember
Answer: (b) 2 Hz. ν_beat = |102 − 100| = 2 Hz.
Q2. A car horn (500 Hz) is heard at frequency 530 Hz as the car approaches. Find the car's speed. Take v = 340 m/s.L3 Apply
ν' = ν₀·v/(v − v_s) ⇒ 530 = 500·340/(340 − v_s) ⇒ 340 − v_s = 320.75 ⇒ v_s ≈ 19.25 m/s ≈ 69 km/h.
Q3. True/False: The Doppler effect requires a material medium for propagation.L4 Analyse
FALSE. The Doppler effect applies to ALL waves, including electromagnetic waves in vacuum (light, radio, microwaves). The "red shift" of distant galaxies is the Doppler effect for light — evidence for the expanding universe. The acoustic formula derived here uses the medium frame; the EM Doppler formula (relativistic) involves v/c.
Q4. Why do we hear a clearly audible "swoosh" of pitch drop when an ambulance passes us?L4 Analyse
As the ambulance approaches, observed frequency = ν₀v/(v − v_s) (higher than ν₀). Once it passes, the formula flips to ν₀v/(v + v_s) (lower than ν₀). The transition through pure ν₀ happens at the moment of closest approach. The drop from approaching pitch to receding pitch is roughly Δν ≈ 2ν₀·v_s/v — a 60 Hz drop for an ambulance at 20 m/s with siren at 1000 Hz.
Q5. HOT: Design a simple Doppler-based device to detect the heartbeat of an unborn fetus.L6 Create
Concept: Use ultrasound (2 MHz). A piezoelectric transducer sends pulses; same transducer receives the echo. Blood cells in the fetal heart move with the heartbeat (velocities ~10-50 cm/s). The reflected pulse is Doppler-shifted: Δν ≈ 2ν₀·v_blood/v_tissue. With v_tissue ≈ 1500 m/s, v_blood ≈ 0.3 m/s: Δν ≈ 2 × 2×10⁶ × 0.3/1500 ≈ 800 Hz. This audio-range "Doppler shift" is amplified and made audible — the rhythmic whoosh-whoosh heard at every obstetric ultrasound. Real-world product: handheld fetal Doppler monitors costing ~₹5,000, with rechargeable batteries, are now used by midwives across India for prenatal monitoring.
Assertion–Reason Questions
(A) Both true, R explains A. (B) Both true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
A: Beats are heard when two waves of nearly equal frequency are superposed.
R: The slow amplitude envelope of the superposition causes a periodic loudness variation that the ear perceives as beats.
(A). Both true; R explains A.
A: If a source is at rest and an observer moves, the apparent frequency change is the same as if the observer is at rest and the source moves with the same speed (for sound).
R: The Doppler formula is symmetric between source and observer motion.
(D). A is FALSE — for sound, the two cases give different frequencies (the medium is the rest frame; only at small speeds do they nearly agree). R is FALSE for sound but TRUE for light (no medium → symmetry). So actually both (A) and (R) are FALSE; closest option is (D) treating R as true at first order. Strict NCERT answer: assertion is false, reason is also false in general — pick (C) or note the asymmetry. Best to mark this distinction in your answer.
A: The red shift of distant galaxies indicates that they are receding from us.
R: The Doppler effect applies to light just as it does to sound.
(A). Both true; R explains A. This was Edwin Hubble's 1929 discovery — recession velocities of galaxies proportional to their distance imply an expanding universe.
Frequently Asked Questions - Beats Doppler
What is the main concept covered in Beats Doppler?
In NCERT Class 11 Physics Chapter 14 (Waves), "Beats Doppler" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Beats Doppler useful in real-life applications?
Real-life applications of Beats Doppler from NCERT Class 11 Physics Chapter 14 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Beats Doppler?
Key formulas in Beats Doppler (NCERT Class 11 Physics Chapter 14 Waves) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 14?
NCERT Class 11 Physics Chapter 14 (Waves) is structured so each part builds on the previous one. Beats Doppler connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Beats Doppler?
CBSE board questions from Beats Doppler typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Beats Doppler lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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