9.5 Doppler Effect

A star orbits another much heavier one in an elliptical orbit as shown. The light received from the star at Earth will be alternately shifted to the red and blue, which of the following statements is true.

• The blue shift will be more than the red shift

• The red shift will be more than the blue shift.

• The red shift will be the same as the blue shift

• The time red shifted is shorter than the time blue shifted

Answer explanation

Since the orbit is elliptical the speed of the star is not constant. Applying the cons of energy we can deduce that when the PE is least the KE is greatest so the orbiting star moves fastest when it is closest to the central star.The light will be blue shifted when the star is coming towards the Earth which is also where it travels fastest.

A police speed trap calculates the speed of a car by using the Doppler shift produced when an electromagnetic wave is reflected off a moving car. To beat the trap the car could send a different frequency wave back to the policeman, the wave should be

• the same frequency as received.

• a higher frequency than received.

• a lower frequency than received.

• it's not possible to beat the trap.

Answer explanation

The speed is related to the difference in frequency, bigger difference = faster speed

If same signal sent back then there would be no difference in measured speed. If frequency was higher then speed would be higher. if frequency lower the calculated speed would be lower, if it was too low then it might appear that you were stationary or even going backward which wouldn't be a good idea at all.

Its not so easy to do this in practice so D might be the correct answer

An atom travelling away from a light detector at 0.8c emits a photon of light of wavelength 600 nm. The wavelength received by the detector is

• 480 nm

• 1080 nm

• 750 nm

• 1350 nm

Answer explanation

Δλ = vλ_o/c

Δλ = 0.8c x 600/c = 480 This will add on to the original wavelength

λ₁ = 600 + 480 = 1080 nm

An observer moves towards a source of sound at a speed of 30 ms^-1

If the frequency of the sound to a stationary observer is 500 Hz the frequency to the moving observer is

• 456 Hz

• 544 Hz

• 460 Hz

• 548 Hz

Answer explanation

f₁ = (c+v) f_o/c

= (340 + 30) x 500/340 = 544 Hz

An observer passes a 500 Hz sound source and moves away with a speed of 30 ms^-1.

• 456 Hz

• 544 Hz

• 460 Hz

• 548 Hz

Answer explanation

f₁ = f_o (c - v ) /c

= 500 x (340 -30)/340 = 456 Hz

The observer throws the 500 Hz sound source back at a speed of 40 ms^-1 . The frequency of the source as observed by A is now

• 447 Hz

• 358 Hz

• 320 Hz

• 300 Hz

Answer explanation

f₁ = cf_o/c+v

= 340 x 500/(340 +40) = 447 Hz

Observer A listens to source of sound moving at 60 ms^-1 towards her. If the frequency of the sound to an observer stationary relative to the sound is 500 Hz what is the frequncy to A. The speed of sound is 340 ms^-1

• 607 Hz

• 425 Hz

• 2800 Hz

• 800 Hz

Answer explanation

f₁ = cf_o/c-v

f₁ = 340 x 500 /(340 - 60) = 607 Hz