Questions & explanations
1. Two identical sources each emitting 500 Hz move in opposite directions at 20 m/s. A stationary observer is between them. Speed of sound is 340 m/s. Find the beat frequency heard.
- 118 Hz
- 29.5 Hz
- 59.0 Hz
- 0 Hz
Answer: 118 Hz
Doppler effect: For source approaching observer, f' = f * v/(v - vs) = 500*340/320 = 531.25 Hz. For source receding, f'' = f * v/(v + vs) = 500*340/360 ≈ 472.22 Hz. Beat frequency = |f' - f''| ≈ 59.03 Hz. However, the observer hears both sources directly, so beat frequency is 2 * 59.03 ≈ 118 Hz because each source produces a wave that interferes with the other. The correct beat frequency is the difference between the two frequencies heard, which is 118 Hz.
2. A source emitting 500 Hz moves toward a wall at 20 m/s. An observer at the source location hears the echo from the wall. Speed of sound is 340 m/s. Find the beat frequency between the direct and reflected sound.
- 62 Hz
- 31 Hz
- 56 Hz
- 28 Hz
Answer: 62 Hz
The wall acts as a moving observer (approaching source) and then as a moving source (approaching observer). Frequency received by wall: f1 = f (v/(v - v_s)) = 500 × 340/320 = 531.25 Hz. Wall reflects same frequency. Observer hears reflected frequency: f2 = f1 (v + v_o)/v = 531.25 × 360/340 = 562.5 Hz. Direct frequency is 500 Hz. Beat frequency = |f2 - 500| = 62.5 ≈ 62 Hz.
3. A source and an observer move toward each other with the same speed u. Which gives a higher apparent frequency: source moving toward observer, or observer moving toward source?
- Source moving gives higher frequency
- Observer moving gives higher frequency
- Both give the same frequency
- It depends on the value of u
Answer: Observer moving gives higher frequency
Doppler effect: when observer moves toward source, f' = f(v+u)/v; when source moves toward observer, f' = f v/(v-u). For u>0, (v+u)/v > v/(v-u) because (v+u)/v = 1+u/v and v/(v-u) = 1/(1-u/v) ≈ 1+u/v+... but the exact inequality shows observer motion gives larger shift. Thus observer moving yields higher apparent frequency.
4. A stationary source emits sound of frequency 600 Hz. The sound reflects off a reflector moving directly toward the source at 20 m/s. Speed of sound is 340 m/s. What frequency does the source detect in the echo?
- 600 Hz
- 638 Hz
- 675 Hz
- 720 Hz
Answer: 675 Hz
The reflector first acts as an observer approaching the source: f₁ = f (v + v_R)/v = 600 × (340+20)/340 = 600 × 360/340 = 635.29 Hz. Then the reflector acts as a source moving toward the stationary observer (source): f₂ = f₁ v/(v - v_R) = 635.29 × 340/(340-20) = 635.29 × 340/320 = 675 Hz. So the echo frequency is 675 Hz.
5. Two trains approach each other with speeds 30 m/s and 20 m/s. The first train blows a horn of frequency 400 Hz. Speed of sound is 340 m/s. What frequency is heard by a passenger in the second train?
- 459 Hz
- 400 Hz
- 439 Hz
- 465 Hz
Answer: 465 Hz
Using the signed Doppler formula f' = f(v - v_o)/(v - v_s). Take direction from source (first train) to observer (second train) as positive. Source moves towards observer: v_s = +30 m/s. Observer moves towards source: v_o = -20 m/s. So f' = 400 × (340 - (-20))/(340 - 30) = 400 × 360/310 ≈ 464.5 Hz, rounded to 465 Hz.
6. A stationary source emits sound of frequency 500 Hz. A reflector moves toward the source at 30 m/s. Speed of sound is 330 m/s. Which expression gives the frequency heard by the source in the echo?
- 500 × (330 + 30) / (2 × 330)
- 500 × (330 + 30) / 330
- 500 × 330 / (330 - 30)
- 500 × (330 + 30) / (330 - 30)
Answer: 500 × (330 + 30) / (330 - 30)
Step 1: reflector as observer approaching stationary source gives f₁ = f (v + v_R)/v = 500 × (330+30)/330. Step 2: reflector as source moving toward stationary observer gives f₂ = f₁ v/(v - v_R) = 500 × (330+30)/330 × 330/(330-30) = 500 × (330+30)/(330-30). So the composite expression is 500 × (330+30)/(330-30).
7. A source of frequency 600 Hz moves at 30 m/s towards an observer moving at 10 m/s away from the source. Speed of sound is 330 m/s. Find the apparent frequency.
- 680 Hz
- 660 Hz
- 600 Hz
- 640 Hz
Answer: 680 Hz
Using Doppler effect formula f' = f (v + v_o)/(v - v_s) with sign convention: direction from source to observer positive. Source moving towards observer: v_s = +30 m/s. Observer moving away from source: v_o = -10 m/s. Then f' = 600 * (330 - 10)/(330 - 30) = 600 * 340/300 = 680 Hz.
8. A source of sound moves towards a stationary observer. What happens to the wavelength and frequency of the sound heard by the observer?
- Wavelength increases, frequency decreases
- Both wavelength and frequency decrease
- Both wavelength and frequency increase
- Wavelength decreases, frequency increases
Answer: Wavelength decreases, frequency increases
When the source moves toward the observer, each successive wavefront is emitted from a point closer to the observer, so the wavefronts bunch up. This reduces the wavelength (λ' < λ). Since the speed of sound v is constant, v = fλ implies that frequency increases (f' > f).
9. A source of frequency 500 Hz moves at 20 m/s along a line at a perpendicular distance of 40 m from a stationary observer. Speed of sound is 340 m/s. What is the apparent frequency when the source is at the closest point to the observer?
- 531 Hz
- 500 Hz
- 472 Hz
- 515 Hz
Answer: 500 Hz
At the closest point, the source velocity is perpendicular to the line joining source and observer. The component of source velocity along the line of sight is zero. Hence, there is no Doppler shift, and the apparent frequency equals the source frequency of 500 Hz.
10. A star emits light of frequency 5×10^14 Hz and moves away from Earth at 0.1c. Which formula should be used to find the observed frequency?
- f' = f·(c - v)/c
- f' = f·c/(c + v)
- f' = f·√((c - v)/(c + v))
- f' = f·(c + v)/c
Answer: f' = f·√((c - v)/(c + v))
Light does not require a medium; the relativistic Doppler formula f' = f·√((c - v)/(c + v)) for source receding is correct. The sound formulas (options a, b, d) are not applicable to light because there is no medium and the speed of light is constant in all frames.
11. A source emits a sound of constant frequency 500 Hz. When the source moves towards a stationary observer, the observer hears a frequency of 530 Hz. Which statement is correct?
- The source frequency remains 500 Hz; the observed frequency changes due to relative motion.
- The source frequency has increased to 530 Hz.
- The source frequency decreases to 470 Hz when moving away.
- The source frequency becomes 530 Hz only when the observer moves.
Answer: The source frequency remains 500 Hz; the observed frequency changes due to relative motion.
The Doppler effect is an apparent change in frequency due to relative motion between source and observer. The source itself continues to emit its original frequency (500 Hz). The observed frequency changes because the wavefronts are compressed or stretched.
12. A source of frequency 1000 Hz moves on a circle of radius 2 m with speed 10 m/s. An observer is at the centre. What frequency does the observer hear? (Speed of sound = 340 m/s)
- 1030 Hz
- 1000 Hz
- 971 Hz
- 1062 Hz
Answer: 1000 Hz
Doppler effect depends on the component of source velocity along the line joining source and observer. Here, the source moves tangentially, so its radial velocity is zero. Hence, no frequency shift occurs; observed frequency equals source frequency 1000 Hz.