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A police car is traveling east at 40.0 m/s along a straight road, overtaking a car ahead of it moving east at 30.0 m/s. The police car has a malfunctioning siren that is stuck at 1 000 Hz. (a) What would be the wavelength in air of the siren sound if the police car were at rest? (b) What is the wavelength in front of the police car? (c) What is it behind the police car? (d) What is the frequency heard by the driver being chased?

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MrRoyal

Answer:

a. The wavelength in air of the siren sound if the police car were at rest is 0.343m

b. The wavelength in front of the police car is 0.303m

c. The wavelength behind the police car is 0.383m

d. 1001.29Hz

Explanation:

Given

Speed 1 = 40m/s

Speed 2 = 30m/s

F = Frequency = 1000Hz

S = Speed of sound = 343 m/s

a.

What would be the wavelength in air of the siren sound if the police car were at rest

Stationary Wavelength λ = S/F= 343/1000= 0.343m

b.

What is the wavelength in front of the police car

Approaching Velocity = V = 40m/s

Approaching Wavelength = λ * (1 - V/S)

Where

λ = 0.343

S = 343m/s

Approaching Velocity = V = 40m/s

Approaching Wavelength = 0.343 * (1-40/343)

= 0.343 *(1 - 0.117)

= 0.343 * 0.883

= = 0.303m

c. .

What is the wavelength behind the police car

Approaching Velocity = V = 40m/s

Approaching Wavelength = λ * (1 + V/S)

Where

λ = 0.343

S = 343m/s

Approaching Velocity = V = 40m/s

Wavelength = 0.343 * (1 + 40/343)

= 0.343 *(1 + 0.117)

= 0.343 * 1.117

= 0.383m

d.

What is the frequency heard by the driver being chased?

Speed Of Approaching Police Car = 40 - 30 m/s = 10 m/s

Frequency heard, F = 1000 * (1 + 10/343) = 1029 Hz = 1.03 kHz

F = 1000 + 1+ 0.29

F = 1001.29Hz

fichoh

Using the appropriate wave equation, the solution the question posed are as follows :

  • 0.343 meters
  • 0.303 metres
  • 0.383 meters
  • 1001.29 Hz

A.)

Wavelength of siren of car were at rest :

Wavelength at rest ; V0 = 343

λ = [tex] \frac{V_{0}}{freqency}[/tex]

Wavelength = 343/1000= 0.343m

B.)

Wavelength in front of the police car

  • Approaching Velocity = V = 40m/s

Wavelength in front of ploca car = [tex] λ \times (1 - \frac{V}{V_{0}})[/tex]

Where

  • λ = stationary wavelength

Approaching Wavelength = [tex] 0.343 \times (1 - \frac{(1 - 40}{343})[/tex]

[tex] 0.343 - 0.883 = 0.303 metres [/tex]

C.)

Wavelength behind the police car :

  • Approaching Velocity = V = 40m/s

Wavelength behind = [tex] λ \times (1 + \frac{V}{V_{0}})[/tex]

Where

  • λ = stationary wavelength

Approaching Wavelength = [tex] 0.343 \times (1 + \frac{(1 - 40}{343})[/tex]

[tex] 0.343 + 0.883 = 0.383 metres [/tex]

d.

Frequency heard by the driver being chased:

Speed Of Approaching vehicle = (40 - 30) = 10 m/s

Frequency heard :

[tex] F = 1000 + (1 + \frac{10}{343})[/tex]

[tex] F = 1000 + (1.029) = 1001.29 Hz[/tex]

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