Frequency, Time, Wavelength = Doppler Effect

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Homework Help Overview

The discussion revolves around a physics problem involving sound and light related to a firework's explosion, with a diver and a person on a boat as the primary subjects. The problem includes calculating the time it takes for the diver to see and hear the firework, determining the actual frequency of the sound heard by the person on the boat, and finding the wavelength of the light emitted by the firework as perceived by the diver.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore various aspects of the problem, including time calculations for sound and light travel, frequency adjustments due to motion, and wavelength determination. Some participants question the validity of initial calculations and conversions, particularly regarding the speed of the boat and the speed of light in water.

Discussion Status

The discussion has progressed with participants providing corrections and updates to initial attempts. Some have successfully recalculated parts of the problem, while others continue to seek clarification on specific aspects, particularly the time calculations for parts A and B.

Contextual Notes

Participants note the importance of unit conversions and the effects of medium on wave speed, particularly in relation to sound in air and water. There is an acknowledgment of misunderstandings regarding the refractive index of water and its impact on calculations.

UMDstudent
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Homework Statement



A small homemade firework goes off 3 meters about the surface of the water where a person is diving. The diver is 10 meters below the surface of the water.

(a). How much time passes from the instant the firework goes off until the diver sees it?

(b). How much time passes from the instant the firework goes off until the diver hears it?

When the firework goes off, another individual is traveling on a boat moving at a speed of
30 km per hour away from the firework. That person is standing on the upper deck of the
boat, also 3 m above the surface.

(c). If the person on the boat hears a sound with frequency 200 Hz, what is the actual
frequency of the firework’s sound?

(d). If the frequency of the light emitted by the fireworks is 5 × 10^14 Hz, what is the
wavelength of this light as seen by the diver?



Homework Equations



Formula A : Velocity = Wavelength (gamma) * Frequency ( we can change this around too)
Formula B : Frequency (receding sound) = (1- V(not)/V)*F(not)

The Attempt at a Solution



For beginners, I have no idea how to successfully start, or complete A. & B. (need help)

I however completed C & D and I'm hoping I completed them correctly :

For part C : We use Formula B and here's my setup for this:

(1 - (3.6*10^6(m/s))/(343(m/s)))*200hz = -2.1*10^6 hz

^^Seems very high but i had to convert the speed of the boat from 30 Kilometers per hour to meters per second.

For Part D : We use Formula A and here's my setup :

Wavelength (gamma) = Velocity / Frequency : 3 * 10^8 (speed of light) / 5*10^14 (hz) = 6 * 10^-7 .

^^Very small wavelength. I have no way of checking my answers so I ask please review this and give the best advice. As always, thank you for reviewing my work and hopefully i will get some much needed help with parts A & B.

Very Respectfully,

UMDstudent
 
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For part C : We use Formula B and here's my setup for this:

(1 - (3.6*10^6(m/s))/(343(m/s)))*200hz = -2.1*10^6 hz

First of all, the question says the person on the boat hears 200 Hz. That's F, not F-nought. Second, your answer is negative. How can a frequency by negative? Third, 30 km/h cannot be 3.6*10^6 m/s. Do you really think something going 30 km/h can travel 3.6 million meters--which is 3600 kilometers--every second? It's OK to make mistakes, but you should at least recognize when your answer is not reasonable.

For Part D : We use Formula A and here's my setup :

Wavelength (gamma) = Velocity / Frequency : 3 * 10^8 (speed of light) / 5*10^14 (hz) = 6 * 10^-7 .

The speed of light in water is not 3*10^8; it's 3*10^8 divided by the refractive index of water. Other than that, your work is correct. 6*10^-7 m, or 600 nm, is bluish-green light.
 
I understand I should check my work over, and I did; but sometimes it takes someone else to inform you of that error. When I did the unit conversion, I wasn't paying attention and believe me, I recognized that C was a little funny. This is why I posted because if i was 100% confident, why would I ask someone to review a prefect computation?

Updated C : 200 Hz = (1 - 8.33 (m/s) / 343 (m/s)) * f(not)

f(not) = 205 Hz

Updated D : Wavelength (gamma) = (2.25*10^8) / (5*10^14) = 4.5 *10^-7

**P.S : I wasn't informed that you needed to divide the speed of light by the refractive index of water because I failed to understand that section after reading the chapter. Thank You. I'm still looking for pointers for part A & B. They can't be that difficult but I just can't figure out how I will solve for T time?

Thanks,

UMDstudent
 
[SOLVED]

Here is the complete work through (in case others need help)

Part A : Time = Distance / Velocity T = (3 meters / 2.99 * 10^8) + (10 meters / 2.26 *10^8) = 5.43 * 10^-8 seconds

Part B : Time = Distance / Velocity

T = (3 meters / 343 ) + (10 meters / 1480 ) = .02 seconds.

343 = speed of sound through air. 1480 speed of sound through water

Part C : Frequency (moving) = (1 - Vsource/Vspeed of sound) * Frequency stationary

200 Hz = (1 - 8.33/343) * Fs | Fs = 205 Hz

Part D : Wavelength = velocity / Frequency = 2.99*10^8 / 5 * 10^14 = 5.99 * 10^-7 = 600 nanometers

**I do believe this is perfectly correct; however, if you see something wrong here, please identify it asap.

Thanks and Enjoy :)

-UMDstudent
 

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