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AP Physics 2 : Doppler Effect

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Example Questions

Example Question #11 : Doppler Effect

How fast would a car have to go to see a red light ( 685nm ) appear green ( 535nm )?

Possible Answers:

$6.57*10^7\frac{m}{s}$

$3.88*10^7\frac{m}{s}$

$3.88*10^5\frac{m}{s}$

$6.57*10^5\frac{m}{s}$

$1.11*10^7\frac{m}{s}$

Correct answer:

$6.57*10^7\frac{m}{s}$

Explanation:

Use the following formula:

$\lambda_{observed}=(1-\frac{v_s}{v})\lambda_{source}$

Where

v_s is the velocity of the source

is the velocity of light in the medium

$\lambda_{source}$ is the source wavelength

$\lambda_{observed}$ is the observed wavelength

Solve for v_s :

$(1-\frac{\lambda_{observed}}{\lambda_{source}})*v=v_s$

Plug in values:

$(1-\frac{535}{685})*3.0*10^8=v_s$

$6.57*10^7\frac{m}{s}=v_s$

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Example Question #12 : Doppler Effect

A train is approaching at $60\frac{m}{s}$ with it's horn on. Normally, the horn has frequency 350Hz . Determine the perceived frequency if the speed of sound is $340.9\frac{m}{s}$ .

Possible Answers:

$f_{observed}=425Hz$

$f_{observed}=454Hz$

$f_{observed}=850Hz$

$f_{observed}=399Hz$

$f_{observed}=476Hz$

Correct answer:

$f_{observed}=425Hz$

Explanation:

Use the Doppler effect equation for approaching sources:

$f_{observed}=\frac{v}{v-v_{source}}f_{source}$

Where is the speed of sound in the current medium

Plug in values:

$f_{observed}=\frac{340.9}{340.9-15}700$

$f_{observed}=425Hz$

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Example Question #11 : Doppler Effect

A motorcycle is receding at $25\ \frac{\textup{m}}{\textup{s}}$ . Normally, the exhaust note has frequency $105\textup{ Hz}$ . Determine the perceived frequency if the speed of sound is $340.9\ \frac{\textup{m}}{\textup{s}}$ .

Possible Answers:

$f_{observed}=93.9\textup{ Hz}$

$f_{observed}=97.8\textup{ Hz}$

$f_{observed}=95.5\textup{ Hz}$

None of these

$f_{observed}=100.3\textup{ Hz}$

Correct answer:

$f_{observed}=97.8\textup{ Hz}$

Explanation:

Using the doppler effect equation for receding sources:

$f_{observed}=\frac{v}{v+v_{source}}f_{source}$

Where is the speed of sound in the current medium

Plugging in values:

$f_{observed}=\frac{340.9}{340.9+25}105$

$f_{observed}=97.8\textup{ Hz}$

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Example Question #11 : Doppler Effect

How fast towards an observer would a red $(\lambda=700\textup{ nm})$ car have to be moving in order to appear yellow $(\lambda=580\textup{ nm})$ ?

Possible Answers:

$8.98*10^7\ \frac{\textup{m}}{\textup{s}}$

$1.35*10^7\ \frac{\textup{m}}{\textup{s}}$

$6.19*10^7\ \frac{\textup{m}}{\textup{s}}$

$5.14*10^7\ \frac{\textup{m}}{\textup{s}}$

None of these

Correct answer:

$5.14*10^7\ \frac{\textup{m}}{\textup{s}}$

Explanation:

Using the Doppler effect equation for approaching sources:

$f_{observed}=\frac{v}{v-v_{source}}f_{source}$

Where is the speed of the wave in the medium, which in this case is the speed of light,

$c=\lambda f$

Combining equations

$\frac{c}{\lambda_o}=\frac{c}{c-v_{source}}\frac{c}{\lambda_s}$

Solving for $v_{source}$ :

$c-\frac{\lambda_o}{\lambda_s}c=v_s$

$3.0*10^8-\frac{580}{700}3.0*10^8=v_s$

$v_s=5.14*10^7\frac{m}{s}$

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Example Question #12 : Doppler Effect

An ambulance is receding at $30\ \frac{\textup{m}}{\textup{s}}$ with it's siren on. Normally, the siren has frequency $700\textup{ Hz}$ . Determine the perceived frequency if the speed of sound is $340.9\ \frac{\textup{m}}{\textup{s}}$ .

Possible Answers:

$f_{observed}=681\textup{ Hz}$

$f_{observed}=643\textup{ Hz}$

$f_{observed}=602\textup{ Hz}$

$f_{observed}=631\textup{ Hz}$

$f_{observed}=615\textup{ Hz}$

Correct answer:

$f_{observed}=643\textup{ Hz}$

Explanation:

Using the Doppler effect equation for receding sources:

$f_{observed}=\frac{v}{v+v_{source}}f_{source}$

Where is the speed of sound in the current medium

Plugging in values:

$f_{observed}=\frac{340.9}{340.9+30}700$

$f_{observed}=643Hz$

Report an Error

Example Question #12 : Doppler Effect

A train is receding at $45\ \frac{\textup{m}}{\textup{s}}$ with it's horn on. Normally, the siren has frequency $350\textup{ Hz}$ . Determine the perceived frequency if the speed of sound is $340.9\ \frac{\textup{m}}{\textup{s}}$ .

Possible Answers:

$f_{observed}=309\textup{ Hz}$

$f_{observed}=322\textup{ Hz}$

$f_{observed}=349\textup{ Hz}$

$f_{observed}=301\textup{ Hz}$

$f_{observed}=290\textup{ Hz}$

Correct answer:

$f_{observed}=309\textup{ Hz}$

Explanation:

Using the Doppler effect equation for receding sources:

$f_{observed}=\frac{v}{v+v_{source}}f_{source}$

Where is the speed of sound in the current medium

Plugging in values:

$f_{observed}=\frac{340.9}{340.9+45}350$

$f_{observed}=309Hz$

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Example Question #11 : Doppler Effect

A rocket ship traveling towards an observer flashes a red light $\lambda=700nm$ . How fast would it have to be traveling for the wavelength to be cut in half?

Possible Answers:

None of these

$3.0*10^{8}\frac{m}{s}$

$6.0*10^8\frac{m}{s}$

$.75*10^8\frac{m}{s}$

$1.5*10^{8}\frac{m}{s}$

Correct answer:

$1.5*10^{8}\frac{m}{s}$

Explanation:

Doppler effect:

$\frac{\Delta \lambda}{\lambda}=\frac{v_{source}}{c}$

Where $\Delta \lambda$ is the change in wavelength

$\lambda$ is the original wavelength

$v_{source}$ is the velocity of the source

is the speed of light

Plugging in values:

$\frac{350}{700}=\frac{v}{3.0*10^8 \frac{m}{s}}$

Solving for

$v=1.5*10^{8}\frac{m}{s}$

Report an Error

Example Question #11 : Doppler Effect

A rocket ship traveling towards an observer flashes a light, $\lambda=600nm$ . How fast would it have to be traveling for the wavelength to be cut to one third it's initial value?

Possible Answers:

$1.0*10^{8}\frac{m}{s}$

$2.5*10^{8}\frac{m}{s}$

$1.5*10^{8}\frac{m}{s}$

$9*10^{8}\frac{m}{s}$

None of these

Correct answer:

$1.0*10^{8}\frac{m}{s}$

Explanation:

Doppler effect:

$\frac{\Delta \lambda}{\lambda}=\frac{v_{source}}{c}$

Where $\Delta \lambda$ is the change in wavelength

$\lambda$ is the original wavelength

$v_{source}$ is the velocity of the source

is the speed of light

Plugging in values:

$\frac{200}{600}=\frac{v}{3.0*10^8 \frac{m}{s}}$

Solving for

$v=1.0*10^{8}\frac{m}{s}$

Report an Error

Example Question #15 : Doppler Effect

A certain shade of blue light has a laboratory rest wavelength of 500nm . The same shade of blue light is emitted from a newly discovered galaxy at a wavelength of 535nm . Using this information, what can we tell about this newly discovered galaxy?

Possible Answers:

The galaxy is moving towards the Earth at speed $1.9*10^7 \frac{m}{s}$

The galaxy is moving away from Earth at speed $1.9*10^7 \frac{m}{s}$

None of these

The galaxy is moving away from Earth at speed $2.1*10^7 \frac{m}{s}$

The galaxy is moving towards the Earth at speed $2.1*10^7 \frac{m}{s}$

Correct answer:

The galaxy is moving away from Earth at speed $2.1*10^7 \frac{m}{s}$

Explanation:

Here, we need to use the Doppler effect equation:

$\frac{\Delta \lambda}{\lambda_{0}}=\frac{v}{c},$

Where $\Delta \lambda$ refers to the wavelength difference between the two sources, $\lambda_{0}$ is the laboratory wavelength, is the speed of the source, and is the speed of light.

Now, let's plug in all of the values:

$\frac{\Delta \lambda}{\lambda_{0}}=\frac{535-500}{500}=\frac{v}{3.0*10^8}$ .

$v=2.1*10^7\frac{m}{s}$ .

Because the wavelength has been shifted to longer wavelengths (the number is larger than the rest wavelength down on Earth), we say the object is redshifted. Therefore, the source (the galaxy) is moving away from Earth at this speed.

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AP Physics 2 : Doppler Effect

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #11 : Doppler Effect

Example Question #12 : Doppler Effect

Example Question #11 : Doppler Effect

Example Question #11 : Doppler Effect

Example Question #12 : Doppler Effect

Example Question #12 : Doppler Effect

Example Question #11 : Doppler Effect

Example Question #11 : Doppler Effect

Example Question #15 : Doppler Effect

All AP Physics 2 Resources

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