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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?

Respuesta :

cjmejiab

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:

[tex]f = (\frac{f_0}{1\pm v_s/v})[/tex]

Here,

f_0 = Original Frequency

f = Observed Frequency

[tex]v_s[/tex] = Speed of the object

v = Speed of the sound wave

PART A)

Converting the velocity to SI units we have,

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

[tex]v = 25m/s[/tex]

Replacing at the equation we have,

[tex]f = (\frac{f_0}{1 - v_s/v})[/tex]

[tex]f = (\frac{600Hz}{1-25/343})[/tex]

[tex]f = 650Hz[/tex]

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,

[tex]f = (\frac{f_0}{1+v_s/v})[/tex]

[tex]f = (\frac{600Hz}{1-25/343})[/tex]

[tex]f = 560Hz[/tex]

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