Question

Use the worked example above to help you solve this problem. An ambulance travels down a highway at a speed of 74.0 mi/h, its siren emitting sound at a frequency of 3.90 x 10^2 Hz. Take the speed of sound in air to be v = 345 m/s. What frequency is heard by a passenger in a car traveling at 58.0 mi/h in the opposite direction as the car and ambulance? (a) approach each other Hz (b) pass and move away from each other? Hz Repeat this problem, but assume the ambulance and the car are going in the same direction, with the ambulance initially behind the car. The speeds and frequency of the siren are the same as in the example. (a) Find the frequency heard before the ambulance passes the car. fbefore = Hz (b) Find the frequency heard after the ambulance passes the car. [Note: The highway patrol subsequently gives the driver of the car a ticket for not pulling over for an emergency vehicle!] fafter = Hz

          Use the worked example above to help you solve this problem. An ambulance travels down a highway at a speed of 74.0 mi/h, its siren emitting sound at a frequency of 3.90 x 10^2 Hz. Take the speed of sound in air to be v = 345 m/s. What frequency is heard by a passenger in a car traveling at 58.0 mi/h in the opposite direction as the car and ambulance?

(a) approach each other
Hz

(b) pass and move away from each other?
Hz

Repeat this problem, but assume the ambulance and the car are going in the same direction, with the ambulance initially behind the car. The speeds and frequency of the siren are the same as in the example.

(a) Find the frequency heard before the ambulance passes the car.
fbefore = Hz

(b) Find the frequency heard after the ambulance passes the car.
[Note: The highway patrol subsequently gives the driver of the car a ticket for not pulling over for an emergency vehicle!]
fafter = Hz

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Added by Kayla J.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Shaiju T

04/11/2022

Step 1

Given: Velocity of ambulance (v1) = 74.0 mi/h = 33.081 m/s Velocity of car (v2) = 58.0 mi/h = 25.92 m/s Velocity of sound (v) = 345 m/s Frequency of siren (f) = 3.90 x 10^2 Hz Using the formula for approaching each other: \[ f_{\text{approach}} = \left( \frac{v + Show more…

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Use the worked example above to help you solve this problem. An ambulance travels down a highway at a speed of 74.0 mi/h, its siren emitting sound at a frequency of 3.90 x 10^2 Hz. Take the speed of sound in air to be v = 345 m/s. What frequency is heard by a passenger in a car traveling at 58.0 mi/h in the opposite direction as the car and ambulance? (a) approach each other Hz (b) pass and move away from each other? Hz Repeat this problem, but assume the ambulance and the car are going in the same direction, with the ambulance initially behind the car. The speeds and frequency of the siren are the same as in the example. (a) Find the frequency heard before the ambulance passes the car. fbefore = Hz (b) Find the frequency heard after the ambulance passes the car. [Note: The highway patrol subsequently gives the driver of the car a ticket for not pulling over for an emergency vehicle!] fafter = Hz

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Transcript

00:01 Hello student, in this question there are two parts.

00:02 So we are considering part first in which s denote what? it denotes the source.

00:07 What is? what is? it denotes the observer.

00:11 It is denoting what? it is denoting the observer.

00:15 What is the value of bs? bs is the velocity of ambulance.

00:19 Velocity of ambulance and this is given to us which is equals to 74 mile per hour.

00:26 So it can be written as 33 .081 meter per.

00:30 Second okay b not is the velocity of car b not is what velocity of car and this is given to us that is 58 mile per hour and it can be written as 25 .92 meter per sex okay and b is the velocity of sound velocity of sound and this is equals to what 345 meter per sex okay now in the first part we have to put the value when both are approaching okay when they are approaching when they are approaching each other then f -dash this will be equals to v plus v -nod divided by v -s v -s and multiplied by the f -s.

01:10 Where what is the f -dash? f -dash is the observed frequency, f -s is the source -v -dains.

01:15 So f -dash will be equals to 345 plus 25 .92 divided by 345 minus bs is 33 .08 multiplied by 3 .9 multiplied by 10 power 2.

01:29 So from here f dash this will be equals to 4 .63 multiplied by 10 power to hurt.

01:35 So this is the observed frequency.

01:38 Now in the b part we will calculate the f dash pass move away, okay, when the move away, move away frequency.

01:47 So this will be close to what? this is equals to v minus vs divided by b plus vs, okay? this is b minus b node.

01:55 This is b minus b0 and multiplied by n.

01:59 So we put the value that is 345 minus 25 .92 divided by 345 plus 33 .08 multiplied by 3 .90 multiplied by 10 power 2...

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