Doppler effect
ICSE · Class 11 · Physics
Flashcards for Doppler effect — ICSE Class 11 Physics. Quick Q&A cards covering key concepts, definitions, and formulas.
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State the Doppler effect for sound.
Answer
The apparent change in the frequency of the source due to a relative motion between the source and the observer is known as Doppler effect.
Why does sound pitch change when a source and observer move relative to each other?
Answer
If the distance between source and observer decreases, wavefronts reach the observer more often, so frequency appears increased. If the distance increases, wavefronts reach less often, so frequency ap…
State the wavelength of sound from a stationary source.
Answer
Formula: λ = v/n Meaning: λ = wavelength, v = velocity of sound, n = true frequency. This is the wavelength when the source is stationary.
A source emits sound of frequency 500 Hz in air where v = 331 m/s. Find wavelength when the source is stationary.
Answer
Given: n = 500 Hz, v = 331 m/s Formula: λ = v/n Step 1: λ = 331/500 m Step 2: λ = 0.662 m Answer: 0.662 m…
State the apparent frequency when a source moves towards a stationary observer.
Answer
Formula: n' = n(v/(v-v_s)) Meaning: n = true frequency, v = sound velocity, v_s = source velocity. This gives a frequency greater than the true frequency.
A source of frequency 400 Hz moves towards a stationary observer with speed 20 m/s. Take v = 331 m/s. Find the apparent frequency.
Answer
Given: n = 400 Hz, v = 331 m/s, v_s = 20 m/s Formula: n' = n(v/(v-v_s)) Step 1: v - v_s = 331 - 20 = 311 m/s Step 2: n' = 400 × (331/311) Step 3: n' ≈ 425.7 Hz Answer: about 426 Hz…
State the apparent frequency when a source recedes from a stationary observer.
Answer
Formula: n' = n(v/(v+v_s)) Meaning: the apparent frequency is less than the true frequency when the source moves away.
A source of frequency 600 Hz moves away from a stationary observer with speed 31 m/s. Take v = 331 m/s. Find the apparent frequency.
Answer
Given: n = 600 Hz, v = 331 m/s, v_s = 31 m/s Formula: n' = n(v/(v+v_s)) Step 1: v + v_s = 331 + 31 = 362 m/s Step 2: n' = 600 × (331/362) Step 3: n' ≈ 548.6 Hz Answer: about 549 Hz…
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