Doppler effect for sound waves
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State the Doppler effect for sound waves in the context of a moving source and a stationary observer.
When a source of sound waves moves relative to a stationary observer, the observed frequency of the sound is different from the frequency emitted by the source.
A sound source is moving towards a stationary observer. Will the observed frequency be higher or lower than the source frequency?
The observed frequency will be higher than the source frequency. As the source approaches, the sound waves are compressed, leading to a shorter wavelength and thus a higher frequency.
Write the formula for calculating the observed frequency (f₀) when a sound source moves relative to a stationary observer.
f₀ = fₛv / (v ± vₛ), where fₛ is the source frequency, v is the speed of sound, and vₛ is the speed of the source. The '+' sign is used when the source is moving away and the '-' sign is used when the source is moving towards.
A car horn emits a sound at 400 Hz. The car is moving towards you (a stationary observer) at 20 m/s. The speed of sound is 340 m/s. Calculate the frequency you observe.
Using f₀ = fₛv / (v - vₛ): f₀ = (400 Hz * 340 m/s) / (340 m/s - 20 m/s) = 425 Hz (approximately).
A train is moving away from a stationary observer at 30 m/s and its whistle emits a sound at 500 Hz. If the speed of sound is 340 m/s, what frequency will the observer hear?
Using f₀ = fₛv / (v + vₛ): f₀ = (500 Hz * 340 m/s) / (340 m/s + 30 m/s) = 459.5 Hz (approximately).
Explain why the observed frequency changes when the sound source is moving.
As the sound source moves, it 'catches up' with the sound waves it emits in the direction of its motion, effectively compressing the wavelength. In the opposite direction, the source moves away from previously emitted waves, stretching the wavelength. Frequency is inversely proportional to wavelength.
What happens to the observed wavelength as a sound source moves towards a stationary observer?
The observed wavelength decreases (gets shorter). This is because the source is 'catching up' with the sound waves it is emitting, effectively compressing them.
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