Physics clicker question: Doppler effect

Physics 5A, Spring Semester 2008
Cuesta College, San Luis Obispo, CA

Cf. Giambattista/Richardson/Richardson, Physics, 1/e, Multiple-Choice Question 12.8

Students were asked the following clicker question (Classroom Performance System, einstruction.com) near the end of their learning cycle:

[0.6 participation points.] A source emits a 440.0 Hz sound. In which situation would an observer hear the highest frequency? (Take the speed of sound to be 340.0 m/s.)
(A) Source is stationary, observer moves at 5.0 m/s towards source.
(B) Source moves at 5.0 m/s towards observer, observer is stationary.
(C) Source and observer both move at 5.0 m/s towards each other.
(D) (More than one of the above choices.)
(E) (I'm lost, and don't know how to answer this.)

Sections 4987, 4988
(A) : 3 students
(B) : 5 students
(C) : 17 students
(D) : 4 students
(E) : 1 student

Correct answer: (C)

The frequency detected by the observer is given by:

f_observer = ((1 - (v_observer/v_sound))/(1 - (v_source/v_sound))*f_source.

In each of cases (A)-(C), the numerator in the parenthesis above is less than the numerator, so f_observer is always higher than f_source (i.e., in each case the motions of the source and observer relative to each other are towards each other). However, the Doppler shift is the greatest when both the source and observer are moving towards each other with respect to the lab frame. Thus for (C), f_observer = (345.0/335.0)*f_source = 453.1 Hz, although most students had determined this without having done any calculations.

[Follow-up question]

[0.6 participation points.] In which situation would the observer hear the second highest frequency? (Use the same (A)-(E) choices from question (10).)

Sections 4987, 4988
(A) : 4 students
(B) : 6 students
(C) : 1 student
(D) : 18 students
(E) : 0 students

Correct answer: (B)

Inexplicably, one student chose (C) as the second highest frequency heard by the observer.

The modal response (D) means that most students said that (A) and (B) would both give the same higher f_observer. However, for (A) v_observer = -5 m/s, v_source = 0, then f_observer = (345.0/340.0)*f_source = 446.5 Hz; and for (B) v_observer = 0, v_source = +5 m/s, then f_observer = (340.0/335.0)*f_source = 446.6 Hz.