AP Physics 2- 14.5 The Doppler Effect- Study Notes- New Syllabus
AP Physics 2- 14.5 The Doppler Effect – Study Notes
AP Physics 2- 14.5 The Doppler Effect – Study Notes – per latest Syllabus.
Key Concepts:
- The Doppler Effect
The Doppler Effect
The Doppler Effect is the change in the observed frequency (or wavelength) of a wave due to the relative motion between the source of the wave and the observer.
Key Properties:
- Source moving toward observer: Observed frequency increases (wavelength decreases).
- Source moving away from observer: Observed frequency decreases (wavelength increases).
- Observer moving toward source: Observed frequency increases.
- Observer moving away from source: Observed frequency decreases.
Doppler Effect for Sound:
For a moving source or observer in a medium where the wave speed is \(\mathrm{v}\):
$ \mathrm{f’ = f \left( \dfrac{v \pm v_o}{v \mp v_s} \right)} $
- \(\mathrm{f’}\) = observed frequency
- \(\mathrm{f}\) = source frequency
- \(\mathrm{v}\) = wave speed in the medium
- \(\mathrm{v_o}\) = speed of observer (positive if moving toward source)
- \(\mathrm{v_s}\) = speed of source (positive if moving toward observer)
Doppler Effect for Light:
Since light does not require a medium, relativistic formulas are used when speeds are a significant fraction of \(\mathrm{c}\).
- Redshift: Source moving away → wavelength increases, frequency decreases.
- Blueshift: Source moving toward → wavelength decreases, frequency increases.
Approximate formula for small speeds (\(\mathrm{v \ll c}\)):
$\mathrm{\dfrac{\Delta f}{f} \approx \dfrac{v}{c}}$
Applications:
- Police radar and speed detection.
- Medical imaging (Doppler ultrasound for blood flow).
- Astronomy (measuring redshift of galaxies to determine universe expansion).
Example :
An ambulance siren emits sound at \(\mathrm{f = 800 \, Hz}\). The speed of sound is \(\mathrm{340 \, m/s}\). If the ambulance moves toward a stationary observer at \(\mathrm{30 \, m/s}\), what frequency does the observer hear?
▶️ Answer/Explanation
Step 1: Formula: \(\mathrm{f’ = f \left( \dfrac{v}{v – v_s} \right)}\)
Step 2: Substitute: \(\mathrm{f’ = 800 \left( \dfrac{340}{340 – 30} \right)}\)
Step 3: \(\mathrm{f’ = 800 \left( \dfrac{340}{310} \right) \approx 877 \, Hz}\)
Final Answer: Observer hears \(\mathrm{877 \, Hz}\).
Example :
Light from a distant galaxy is observed to have a wavelength of \(\mathrm{600 \, nm}\), while its emitted (rest) wavelength is \(\mathrm{500 \, nm}\). Estimate the speed of the galaxy relative to Earth.
▶️ Answer/Explanation
Step 1: Use redshift relation (non-relativistic): \(\mathrm{\dfrac{\Delta \lambda}{\lambda} \approx \dfrac{v}{c}}\)
Step 2: \(\mathrm{\Delta \lambda = 600 – 500 = 100 \, nm}\)
Step 3: \(\mathrm{\dfrac{100}{500} = \dfrac{v}{3.0 \times 10^8}}\)
Step 4: \(\mathrm{v = 0.2 \times 3.0 \times 10^8 = 6.0 \times 10^7 \, m/s}\)
Final Answer: The galaxy is moving away at \(\mathrm{6.0 \times 10^7 \, m/s}\).