Doppler Effect Source Moving Closer

In summary, the equation f'=\frac{f}{\left(1-\frac{v_s}{v}\right)} gives the correct wavelength for a lantern with a 650nm wavelength.
  • #1
JSGandora
95
0

Homework Statement


Superman is carrying a red lantern with wavelength 650nm. He flies toward you at a speed of 2.7x108 m/s.
What is the observed wavelength?

Homework Equations


[itex]f'=\frac{f}{\left(1-\frac{v_s}{v}\right)}[/itex]
and
[itex]f=\frac{c}{\lambda}[/itex]

The Attempt at a Solution


I used the equation [itex]f'=\frac{f}{\left(1-\frac{v_s}{v}\right)}[/itex] and got my answer to be [itex]65[/itex] nm. What did I do wrong?
 
Last edited:
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  • #2
JSGandora said:

Homework Statement


Superman is carrying a red lantern with wavelength 650nm. He flies toward you at a speed of 2.7x108 m/s.
What is the observed wavelength?

Homework Equations


[itex]f'=\frac{f}{\left(1+\frac{v_s}{v}\right)}[/itex]
and
[itex]f=\frac{c}{\lambda}[/itex]

The Attempt at a Solution


I used the equation [itex]f'=\frac{f}{\left(1+\frac{v_s}{v}\right)}[/itex] and solved for [itex]v_s[/itex] and got my answer to be [itex]75\times 10^8[/itex] m/s. What did I do wrong?

Well, for starters, the result you provided is a velocity (m/s), when in fact, what was asked for was a distance(m) (of a wavelength).
 
  • #3
OOPS, I accidentally copied the wrong thing. I actualy got 65 nm for the wavelength using the correct formula with the minus sign in the denominator, not plus sign (edited original post). Sorry about that.
 
  • #4
JSGandora said:
OOPS, I accidentally copied the wrong thing. I actualy got 65 nm for the wavelength using the correct formula with the minus sign in the denominator, not plus sign (edited original post). Sorry about that.

So can you now do your other question thread for the source moving away?

https://www.physicsforums.com/showthread.php?t=546027

.
 
  • #5
Bleh, I also forgot to mention that the answer key said 149nm.

Sorry for not giving all the information, I made this all a big mess. I don't understand why the answer is 149nm, I'm pretty sure all my calculations were correct. I also forgot to mention that my answer was not what the answer key said in my other thread. What answer are you getting?
 
  • #6
Someone told me that the equation [itex] \lambda =\lambda_{0}\sqrt{{1-\beta}\over{1+\beta}} [/itex] give the correct answer. Why didn't the usual doppler effect equation work here? I'm confused.
 

1. What is the Doppler Effect?

The Doppler Effect is a phenomenon in which the frequency and wavelength of a wave appear to change when the source of the wave is moving relative to the observer. This effect is most commonly observed in sound waves, but it also applies to other types of waves such as light waves.

2. How does the Doppler Effect change when the source is moving closer?

When the source of the wave is moving closer to the observer, the frequency of the wave increases and the wavelength decreases. This means that the pitch of the sound will appear higher and the color of the light will appear bluer.

3. What factors affect the magnitude of the Doppler Effect?

The magnitude of the Doppler Effect is affected by the relative speeds of the source and the observer, as well as the speed of the wave itself. The closer the source is to the observer and the faster it is moving, the greater the effect will be.

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

The Doppler Effect has many practical applications in various fields such as astronomy, meteorology, and medicine. For example, it is used in weather radar to detect the speed and direction of moving storms, and in medical ultrasound to measure blood flow and detect abnormalities.

5. Can the Doppler Effect be observed with stationary sources?

Yes, the Doppler Effect can still be observed even if the source of the wave is stationary. This is because the observer's motion can also cause a change in frequency and wavelength of the wave. For example, when a car with a loud siren passes by a stationary person, the pitch of the siren will appear to change as the car moves closer and then farther away.

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