Question

An opera singer in a convertible sings a note at 600 Hz while cruising down the highway at 90 km/h. What is the frequency heard by
a. A person standing beside the road in front of the car?
b. A person standing beside the road behind the car?

Answer 1

To solve this problem we will apply the concepts related to the Doppler effect. The general expression to apply the Doppler effect is given by the function:

$f = (\frac{f_0}{1\pm v_s/v})$

Here,

f_0 = Original Frequency

f = Observed Frequency

$v_s$ = Speed of the object

v = Speed of the sound wave

PART A)

Converting the velocity to SI units we have,

$v = 90km/h(\frac{3600s}{1h})(\frac{1000m}{1km})$

$v = 25m/s$

Replacing at the equation we have,

$f = (\frac{f_0}{1 - v_s/v})$

$f = (\frac{600Hz}{1-25/343})$

$f = 650Hz$

Therefore the frequency for a person standing beside the road in front of the car is 650Hz

PART B) The frequency for a person standing beside the road behind the car would be,

$f = (\frac{f_0}{1+v_s/v})$

$f = (\frac{600Hz}{1-25/343})$

$f = 560Hz$