{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Homework 21

# Homework 21 - Gozick Brandon Homework 21 Due Dec 1 2006...

This preview shows pages 1–3. Sign up to view the full content.

Gozick, Brandon – Homework 21 – Due: Dec 1 2006, 5:00 pm – Inst: D Weathers 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points An ambulance is traveling South at 45 . 8 m / s, away from a car that is traveling North at 28 . 5 m / s. The ambulance driver hears his siren at a frequency of 503 Hz. Velocity of sound v sound = 343 m / s . Ambulance 45 . 8 m / s 28 . 5 m / s Police What wavelength does a person who is standing between the car and the ambulance detect from the sound of the ambulance’s siren? Correct answer: 0 . 772962 m. Explanation: The wavelength of the sound emitted in the back of the ambulance is λ 0 = v sound + v amb f . The positive sign arises because the ambu- lance driver is traveling in the opposite direc- tion as these sound waves. λ 0 = v sound + v amb f = (343 m / s) + (45 . 8 m / s) (503 Hz) = 0 . 772962 m . Here the wave form is stretched out, which results in a longer wavelength. 002 (part 2 of 2) 10 points At what frequency does the driver of the car hear the ambulance’s siren? Correct answer: 406 . 876 Hz. Explanation: The observed Frequency is f 0 = v 0 λ 0 . The car driver sees the sound wave overtaking him with a velocity of v 0 = v sound - v car . Therefore the car driver hears a sound with a frequency f 0 = ( v sound - v car ) λ 0 . In terms of the original frequency, f 0 = v sound - v car v sound + v amb f = (343 m / s) - (28 . 5 m / s) (343 m / s) + (45 . 8 m / s) × (503 Hz) = 406 . 876 Hz . Comment/Special Note: The argument above works for sound waves and other me- chanical waves traveling through dense media, but it is NOT correct for light waves, which must be treated using the special theory of relativity. The special theory of relativity requires all observers measure the same velocity for light, regardless of their velocity. Therefore, in or- der for a Doppler Effect for light to exist, different observers can no longer observe the same wavelength for a light wave. This is re- lated to the relativistic phenomenon known as length contraction. keywords: 003 (part 1 of 1) 10 points The Doppler shift was first tested in 1845 by the French scientist B. Ballot. He had a trumpet player sound a(n) 432 Hz note while riding on a flat-car pulled by a locomotive. At the same time, a stationary trumpeter played the same note. Ballot heard 6 beats / s. The speed of sound in air is 343 m / s . How fast was the train moving toward him? Correct answer: 4 . 69863 m / s. Explanation: Let : f = 432 Hz , b = 6 beats / s , and v = 343 m / s . The velocity of sound in air is v = 343 m/s. The equation for the Doppler shift of a sound

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Gozick, Brandon – Homework 21 – Due: Dec 1 2006, 5:00 pm – Inst: D Weathers 2 wave of speed v reaching a detector moving at v d is f 0 = f v + v d v - v tr .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 8

Homework 21 - Gozick Brandon Homework 21 Due Dec 1 2006...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online