The waves are getting kind of 'squished' together, changing their wavelength and frequency, but not their speed.
CloseSince they are moving away the wavelengths must get longer, hence the lines in the spectrum will shift to the right, also known as red-shift.
CloseThat is Hubble's law, essentially saying, the further away a galaxy is from us, the faster it moves away from us.
CloseWith light we have the same case as with sounds, since we are observing a larger wavelength, it must be moving away.
CloseFor a source that is moving away, this is the equation: \( f = f_0 \frac{v}{v+o} \), plugging these values in gives B.
CloseThe observed frequency \(f'\) for sound waves is given by the Doppler effect formula: \[ f' = f \left(\frac{v }{v - v_s}\right) \]
Where:
Substituting the values: \[ f' = 500 \left(\frac{343}{343 - 20}\right) = 500 \left(\frac{343}{323}\right) \approx 500 \times 1.062 = 531 \, \mathrm{Hz}. \] Thus, the frequency observed is approximately \(531 \, \mathrm{Hz}\).
CloseThe observed frequency \(f'\) for a moving observer and stationary source is given by: \[ f' = f \left(\frac{v + v_o}{v}\right) \]
Where:
Substituting the values: \[ f' = 600 \left(\frac{340 + 10}{340}\right) = 600 \left(\frac{350}{340}\right) \approx 600 \times 1.029 = 617.4 \, \mathrm{Hz}. \] Thus, the observed frequency is approximately \(615 \, \mathrm{Hz}\).
CloseFor a moving source and stationary observer, the Doppler effect formula is: \[ f' = f \left(\frac{v}{v + v_s}\right) \]
Where:
Substituting the values: \[ f' = 600 \left(\frac{340}{340 + 25}\right) = 600 \left(\frac{340}{365}\right) \approx 600 \times 0.930 \approx 558 \, \mathrm{Hz}. \] Thus, the observed frequency is approximately \(558 \, \mathrm{Hz}\).
CloseThis question involves two Doppler shifts: one when the radar wave is emitted from the police officer, and one when it reflects off the moving car.
Step 1: Doppler effect when the radar wave is emitted towards the car (moving source).
The first Doppler shift happens as the radar wave, emitted by the police officer, reaches the car, which is moving away from the police. The frequency of the wave received by the car will be shifted downwards due to the motion of the car away from the radar. The Doppler shift formula for a moving observer is: \[ f_{\text{car}} = f_{\text{radar}} \left( \frac{v}{v + v_s} \right) \] Where:
Substituting the values: \[ f_{\text{car}} = 10.0 \times 10^9 \left( \frac{3.0 \times 10^8}{3.0 \times 10^8 + 25} \right) = 10.0 \times 10^9 \left( \frac{3.0 \times 10^8}{3.000025 \times 10^8} \right) \] \[ \approx 10.0 \times 10^9 \times 0.99991 = 9.9991 \, \text{GHz}. \]
Step 2: Doppler effect when the wave reflects off the car and returns to the police officer.
Now, the car acts as a moving source of the reflected wave, and the police officer acts as a stationary observer. The frequency of the reflected wave will be shifted again as it travels back to the officer, according to the Doppler shift formula for a moving source:
\[ f_{\text{officer}} = f_{\text{car}} \left( \frac{v + v_o}{v} \right) \]
Substituting the values:
\[ f_{\text{officer}} = 9.9991 \times 10^9 \left( \frac{3.0 \times 10^8 + 25}{3.0 \times 10^8} \right) \] \[ f_{\text{officer}} = 9.9991 \times 10^9 \left( 1 + 8.33 \times 10^{-5} \right) \approx 9.9991 \times 10^9 \times 1.0000833 = 9.9980 \, \text{GHz}. \]
Thus, the frequency of the reflected wave received by the police officer is approximately \(9.9980 \, \text{GHz}\).
Close