The Doppler Effect

Chapter 15-7Physics: An Interactive Approach

What is the doppler effect?• When hearing sounds from a source travelling towards the receiver, the receiver

depicts higher frequencies while the source is approaching, and lower frequencies after the source has passed, and travelling away from the receiver.

• This is known as the Doppler effect, named after Christian Doppler, a physicist from Austria.

• In more understandable terms, when a racecar passes a bystander on a racing track, the car’s sound waves are at higher frequencies approaching the receiver, and lower frequencies when it passes the receiver. Meaning that the sound produced by the car is louder after it passes the bystander.

How to determine frequencies:

• Now that we know receivers depict frequencies based on the position of the source, we can determine the frequency at the

receiver by this equation

ƒr = [(v + vr) ÷ (v – vs)](ƒs)

• ƒr = Frequency at receiver• V = Speed of sound in the medium (air)• Vr = Speed of receiver

• Vs = Speed of source

• ƒs = Frequency emitted from source

Other form:

• The previous equation works in a situation where the source is travelling towards the receiver.

• However, when the source is travelling away from the receiver (having passed the receiver), the relationship is

ƒr = [(v - vr) ÷ (v + vs)](ƒs)

Question 1

• A plane is taking off at YVR with a velocity of 280km/h . The frequency of sound the engines produce are 870 Hz.

• You are one of those employees in charge to taxi the plane, and you watch it take off at a stationary position about 300m away. What is the frequency of sound waves as the plane approaches?

• The speed of sound in air is about 343m/s.

Question 2

• The plane has now passed you, and is now moving further down the runway preparing to take off. What is the frequency of the sound waves?

• Keep in mind that a receiver gets higher frequency waves as the plane is approaching, and lower as the plane is moving further away. So your answer should be less then that of Question 1.

Solution Question 1

• 280km/h = 77.8m/s• Sound speed = 343m/sƑr = [(343m/s + 0m/s) ÷ (343m/s – 77.8m/s)](870Hz)

= (343/265.2)(870)= 1125.226244

= 1125Hz

Solution Question 2

• 280km/h = 77.8m/s• Sound speed = 343m/sƑr = [(343m/s - 0m/s) ÷ (343m/s + 77.8m/s)](870Hz)

= (343/420.8)(870)= 709.1492395

= 709Hz

One last note:

• Note that the position of the stationary receiver (300m out) was negligible in the equation. It was only meant to show that the receiver was far away from the plane, and that the plane was travelling towards the receiver.