Cramster.com - Homework Solutions, Lecture Notes, Exams, and Free Online Homework Help
Sign Up Now! Login Customer Support Cramster Blog
McAfee Secure sites help keep you safe from identity theft, credit card fraud, spyware, spam, viruses and online scams
Problem Solved.
    Home    
    Homework Help    
   Answer Board   
    Resources (Beta)    
   
Member's Topic Headline:

Please Help!

Know the answer? Have a better solution? Share it.
Sign Up Now for FREE!
Join the thousands of students
getting ahead in their classes.
Member Testimonials

Question:

Advertisement:

Answer | Ask New Question | Customize Profile | Leaderboards | 
FAQ

Member's Avatar

Mentor
Karma Points: 453
(Houghton College)
Respect (91%):
Date Posted: 11/30/2007 4:55:59 PM  Status: Live
Please Help!
Course Textbook Chapter Problem
Calculus Based Physics N/A 17 33
Question Details:
A driver travels northbound on a highway at a speed of 25.0 m/s. A police car, traveling southbound at a speed of 40.0 m/s, approaches with its siren producing sount at a frequency of 2500 Hz. (a) What frequency does the driver observe as the police car approaches? (b) What frequency does the driver detect after the police car passes him? (c) Repeat parts a and b for the case when the police car is traveling northbound.
Bonus Point Alert! Earn +2 additional karma points for helping this annual member.

Answers:

Member's Avatar

Expert
Karma Points: 1,031
Date Posted: 11/30/2007 8:54:35 PM  Status: Live
Asker's Rating: Helpful   
Response:
The apparent frequency or the frequency whic you hear is  f1 = ()f

where f is the actual frequency, v is the velocity of sound and vs is the velocity of source. - sign is used when car is approaching you and + sign is used when car is moving away from you. This is called Doppler effect. When approaching, the frequency increases and while moving away, frequency decreases. Since you have not specified velocity of sound, i will consider it 340m/s
 
a) here relative to the dirver the velocity of police car is 25 + 40 (since they are approaching each other) = 65m/s = vs
so freq heard = [340 / (340 - 65) ]*2500 =  3090 Hz
 
b) here the relative to the driver, the velocity of police car is 40 - 25 =15m/s (as police car is moving away) = Vs
 
so freq heard = [340/(340+15)]*2500 = 2394 Hz
 
I think now you can go ahead with (c). If you are having difficulty ask me.
 
TIP: To check whether your answer is correct or not, remember when approaching, higher freq is heard and while moving away lower freq is heard compared to the actual freq
 
ktqu's Comment:
How does it change if they are both going northbound?





By reading or posting messages on these forums, you are agreeing to the Answer Board's Terms of Service and Conduct (TSC).


About Cramster | Terms of Use | Privacy Policy | Contact Us | Press Room | Site Map | Support | Anti-Cheating Policy

Cramster.com is not affiliated with any publisher. Book covers, title and author names appear for reference only.
Copyright © 2008 Cramster, Inc.