The Doppler Effect AP Physics 2 MCQ – Exam Style Questions etc.
The Doppler Effect AP Physics 2 MCQ
Unit 14: Waves , Sound , and Physical Optics
Weightage : 15–18%
Exam Style Practice Questions ,The Doppler Effect AP Physics 2 MCQ
Question
A radar speed gun is used to measure the speed of a car. The car is moving with speed v away from the gun.
The radar emits microwaves of frequency f and speed c. Which of the following is the frequency of the microwaves measured at the gun after reflection by the car?
A. \(f + \left( {\frac{{2v}}{c}f} \right)\)
B. \(f + \left( {\frac{v}{c}f} \right)\)
C. \(f – \left( {\frac{{2v}}{c}f} \right)\)
D. \(f – \left( {\frac{v}{c}f} \right)\)
Answer/Explanation
Markscheme
C
change in frequency in term of gun is
\(\frac{\Delta f}{f_o}=\beta=\frac{v}{c}\)
Frequency observed by moving car can be calculated as
\(\frac{\Delta f’}{f_o}=\frac{v}{c}\)
\(\Delta f’=-\frac{v}{c}{f_o}\)
\(f’-f_o=-\frac{v}{c}{f_o}\)
\(f’=f_o -\frac{v}{c}{f_o}\)
Frequency observed by Gun after reflection from car
\(\frac{\Delta f”}{f’}=-\frac{v}{c}\)
or
\(\Delta f” =-\frac{v}{c}{f’} =-\frac{v}{c} \times (f_o -\frac{v}{c}{f_o})\)
\(f”-f’ =f_o (-\frac{v}{c} + \frac{v^2}{c^2})\)
\(f” =f’+f_o (-\frac{v}{c} + \frac{v^2}{c^2})\)
\(=f_o -\frac{v}{c}{f_o} +f_o (-\frac{v}{c} + \frac{v^2}{c^2})\)
\(=f_o(1-2\frac{v}{c} +\frac{v^2}{c^2})\)
\(\Delta f’=-\frac{v}{c}{f_o}\)
\(f’-f_o=-\frac{v}{c}{f_o}\)
\(f’=f_o -\frac{v}{c}{f_o}\)
Frequency observed by Gun after reflection from car
\(\frac{\Delta f”}{f’}=-\frac{v}{c}\)
or
\(\Delta f” =-\frac{v}{c}{f’} =-\frac{v}{c} \times (f_o -\frac{v}{c}{f_o})\)
\(f”-f’ =f_o (-\frac{v}{c} + \frac{v^2}{c^2})\)
\(f” =f’+f_o (-\frac{v}{c} + \frac{v^2}{c^2})\)
\(=f_o -\frac{v}{c}{f_o} +f_o (-\frac{v}{c} + \frac{v^2}{c^2})\)
\(=f_o(1-2\frac{v}{c} +\frac{v^2}{c^2})\)
ignoring \(\frac{v^2}{c^2}\) as \(c^2 \) is very large and given \(f_o=f\)
we get
\(f” ==f(1-2\frac{v}{c}) = f -(2\frac{v}{c}f)\)
Question
A stationary source of sound emits sound of frequency f. A moving observer measures the sound as having the frequency f ′. The observer is moving directly away from the source at a speed that is 30% of the speed of sound in air.
Which of the following gives the correct value for \(\frac{{f’}}{f}\)?
A. \(\frac{7}{{10}}\)
B. \(\frac{10}{{13}}\)
C. \(\frac{13}{{10}}\)
D. \(\frac{10}{{7}}\)
Answer/Explanation
Markscheme
A
The equation for doppler effect may be generalised as
\(f’=f(\frac{v\pm v_o}{v\mp v_s})\)
In the given case \(v_s=0\) and \(v_o = 0.3v_s\)
Hence
\(\frac{f’}{f}=\frac{v-v_o}{v}=\frac{v-0.3v}{v}=0.7=\frac{7}{10}\)