How To Solve Doppler Effect Physics Problems
Doppler Effect Overview
The Doppler effect refers to the change in frequency detected by an observer due to the motion of the source or observer.
If the source moves toward the observer (or vice versa), the frequency increases.
If the source moves away from the observer (or vice versa), the frequency decreases.
Key Concepts
Observed Frequency (f_o) vs. Source Frequency (f_s):
f_o > f_s when moving towards each other.
f_o < f_s when moving away from each other.
Example:
Source frequency = 800 Hz.
If source moves toward observer, heard frequency could be 850 Hz (increased).
If source moves away, heard frequency would decrease to values like 900 or 800 Hz.
Visualization
Wave Patterns:
When a source approaches, crests are closer (wavelength decreases, frequency increases).
When a source retreats, crests are farther apart (wavelength increases, frequency decreases).
Formula for Calculating Observed Frequency
Formula:[ f_o = f_s \times \frac{v + v_o}{v - v_s} ]
( v ): speed of sound (around 343 m/s at 20°C).
( v_o ): speed of observer (positive if moving towards, negative if away).
( v_s ): speed of source (negative if moving towards, positive if away).
Speed of Sound Depending on Temperature
Speed can be calculated with:[ v = 331 + 0.6t ]
Where ( t ) is temperature in Celsius.
Example calculations:
At 0 °C, ( v \approx 331 ) m/s;
At 20 °C, ( v \approx 343 ) m/s;
At 25 °C, ( v \approx 346 ) m/s.
Sign Conventions for Using the Formula
Source Moving Towards Observer:
( v_s ): negative (decreases denominator, increases f_o).
Source Moving Away from Observer:
( v_s ): positive (increases denominator, decreases f_o).
Observer Moving Towards Source:
( v_o ): positive (increases numerator, increases f_o).
Observer Moving Away from Source:
( v_o ): negative (decreases numerator, decreases f_o).
Example Problems
Problem 1: Ambulance Truck
Frequency produced: 800 Hz.
If moving towards observer at 30 m/s: [ f_o = 800 \times \frac{343 + 0}{343 - 30} \approx 877 ext{ Hz} ]
If moving away from observer at 30 m/s: [ f_o = 800 \times \frac{343 + 0}{343 + 30} \approx 736 ext{ Hz} ]
Problem 2: Stationary Ambulance Truck
Frequency produced: 1200 Hz; observer moving towards at 25 m/s: [ f_o = 1200 \times \frac{343 + 25}{343 + 0} = 1287 ext{ Hz} ]
Observer moving away at 25 m/s: [ f_o = 1200 \times \frac{343 - 25}{343 + 0} \approx 1113 ext{ Hz} ]
Problem 3: Police Car and Driver
Police car emitting 900 Hz; moving at 20 m/s towards driver going 25 m/s: [ f_o = 900 \times \frac{343 + 25}{343 - 20} \approx 1258 ext{ Hz} ]
Conclusion
Understand how to manipulate the observed frequency based on the movement of the source and observer.
Practice with various scenarios to become comfortable with the formula and sign conventions.