A stationary police car emits a sound of frequency 1240 Hz that bounces off of a car on the highway and returns with a frequency of 1265Hz . The police car is right next to the highway, so the moving car is traveling directly toward or away from it. And what frequency would the police car have received if it had been traveling toward the other car at 25.0 m/s?

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wagonbelleville wagonbelleville · Answer 1 · 2019-12-23T06:40:33+01:00

Answer:

1467.56 Hz

Explanation:

given,

f₀ = 1215 Hz,

f₂ = 1265 Hz

speed of sound u = 340 m/s,

speed of the other car = v

when the police car is stationary:

the frequency the other car receives is

[tex]f_1= f_0\times \dfrac{u+v}{u}[/tex]

the frequency the police car receives is

[tex]f_2 = f_1\times \dfrac{u}{u-v}[/tex]

now,

[tex]\dfrac{f_2}{f_0}=\dfrac{u+v}{u-v}[/tex]

[tex]\dfrac{1265}{1240}=\dfrac{u+v}{u-v}[/tex]

[tex]v =\dfrac{1265-1240}{1265+1240}\times u[/tex]

[tex]v =\dfrac{1265-1240}{1265+1240}\times 344[/tex]

v = 3.43 m/s

when the police car is moving with v_p = 25.0 m/s toward the other car:

the frequency the other car receives is

[tex]f_1= f_0\times \dfrac{u+v}{u-v_p}[/tex]

the frequency the police car receives is

[tex]f_2 = f_1\times \dfrac{u+v_p}{u - v}[/tex]

now,

[tex]= f_0\times \dfrac{(u+v)(u + v_p)}{(u-v)(u-v_p)}[/tex]

[tex]=1240\times \dfrac{(344 +3.43)(344 + 25)}{(344-3.43)(344 - 25)}[/tex]

= 1467.56 Hz