Frequency, Time, Wavelength = Doppler Effect

In summary, a small homemade firework goes off 3 meters above the surface of the water where a person is diving at a depth of 10 meters. The time it takes for the diver to see the firework is 5.43 * 10^-8 seconds, while the time it takes for the diver to hear it is 0.02 seconds. When the firework goes off, another person on a boat moving at a speed of 30 km/h away from the firework hears a sound with a frequency of 200 Hz, resulting in an actual frequency of 205 Hz. The frequency of the light emitted by the firework is 5 * 10^14 Hz, resulting in a wavelength of 600 nan
  • #1
UMDstudent
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0

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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  • #2
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.
 
  • #3
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
 
  • #4
[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
 

1. What is the Doppler Effect?

The Doppler Effect is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the source of the wave. This phenomenon is commonly observed with sound waves, where the pitch of a sound appears to change as the source of the sound moves closer or farther away from the listener.

2. How does the Doppler Effect relate to frequency, time, and wavelength?

The Doppler Effect is directly related to frequency, time, and wavelength. As an observer moves closer to a wave source, the frequency increases and the wavelength decreases. Conversely, as an observer moves away from a wave source, the frequency decreases and the wavelength increases.

3. What factors affect the Doppler Effect?

The two main factors that affect the Doppler Effect are the relative motion between the wave source and observer, and the speed of the wave. The greater the relative motion and the higher the speed of the wave, the more pronounced the Doppler Effect will be.

4. Is the Doppler Effect only observed with sound waves?

No, the Doppler Effect can be observed with any type of wave, including light waves, water waves, and seismic waves. However, it is most commonly associated with sound waves because of its noticeable effect on the pitch of a sound.

5. How is the Doppler Effect used in real life?

The Doppler Effect has many practical applications in various fields. It is used in weather radar to determine the speed and direction of moving storms. It is also used in medical imaging techniques, such as ultrasound, to measure the velocity of blood flow. In astronomy, the Doppler Effect is used to measure the speed and direction of celestial objects. Additionally, the Doppler Effect is utilized in police radar and speed guns to measure the speed of moving vehicles.

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