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AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

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6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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:

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

$f_{observed}=643Hz$

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Example Question #11 : 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:

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

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

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

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

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

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

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

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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

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

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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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Example Question #1 : Slit Experiments

What did the double-slit experiment (also called Young's Experiment) demonstrate?

Possible Answers:

Light is composed purely of waves

The double-slit experiment did not tell scientists anything useful

Light is composed purely of particles

Light cannot be diffracted by slits

Light demonstrates properties of both particles and waves

Correct answer:

Light demonstrates properties of both particles and waves

Explanation:

The process for the double-slit experiment consisted of a coherent light source aimed at a screen with a plate with two parallel slits in between them. The light traveled through each of the slits and had wave-interference (destructive and constructive interference) to produce bands of alternating light and dark along the screen. This result would not be expected if light consisted solely of particles, as was classically thought. This showed that light is a wave, because waves interfere in that manner. The light also was found to be hitting the screen at discrete points as individual particles (photons), with the alternating bands indicating the density of the particles that hit the screen. In versions of the experiment that featured detectors at the slits, the photon passed through a single slit (as would a particle) and not both (as would a wave). These outcomes demonstrate the wave-particle duality of light.

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Example Question #1 : Slit Experiments

Suppose a light with a wavelength of $8.5*10^{-7}{ m}$ passes a pair of slits with a separation of $5\*10^{-5}{ m}$ . Determine the angle which corresponds to the first bright fringe.

Possible Answers:

0.4873^o

0.9741^o

1.0362^o

89.0352^o

89.5136^o

Correct answer:

0.9741^o

Explanation:

To determine the angle, use the formula for bright fringes in a two-slit experiment.

$dsin(\theta)=m\lambda$

Rewrite the equation so that the angle term is isolated.

$\theta=sin^{-1}(\frac{m\lambda}{d})$

m=+1 represents the first bright fringe.

Substitute all the givens and solve for the angle.

$\theta=sin^{-1}(\frac{m\lambda}{d})=sin^{-1}\left(\frac{(1)(8.5* 10^{-7}{ m})}{5* 10^{-5} { m}}\right)= 0.9741^o$

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Example Question #3 : Slit Experiments

Suppose light pass through a slit with a width $1.3 *10^{-4}{ m}$ . The diffraction pattern angle is $\theta=5^o$ for the first dark fringe. What is the light's wavelength?

Possible Answers:

1.133{ m}

$1.133* 10^{-5}{ m}$

1.295m}

$1.295* 10^{-4} m}$

$1.865* 10^{-6}{ m}$

Correct answer:

$1.133* 10^{-5}{ m}$

Explanation:

To solve for the wavelength of the light, use the formula for dark fringes of a single-slit interference.

$Wsin(\theta)=m\lambda$

Rewrite the equation in order to solve for lambda (wavelength). Since this is the first dark fringe, the value of m=1 .

$\lambda=\frac{Wsin(\theta)}{m}=\frac{1.3* 10^{-4}{ m}(sin(5))}{1}=1.133* 10^{-5}{ m}$

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Example Question #1 : Slit Experiments

Suppose that a diffraction experiment is performed in order to determine the wavelength of light. As light of an unknown wavelength is passed through a $5.0\cdot10^{-6}\:m$ slit, a diffraction pattern is observed. If the first dark fringe is found to occur at an angle of $30^{o}$ away from the slit, what is the wavelength of this light?

Possible Answers:

$5.0\cdot10^{-6}\:m$

$1.25\cdot10^{-6}\:m$

$2.5\cdot10^{-6}\:m$

$6.25\cdot10^{-6}\:m$

$7.5\cdot10^{-6}\:m$

Correct answer:

$2.5\cdot10^{-6}\:m$

Explanation:

For this question, we're presented with a scenario in which light is passing through a small slit. In doing so, the light is diffracted, meaning that it is spread out along a viewing screen. In such a case, there will be a central bright fringe, with alternating dark and bright fringes as one moves above and below the central fringe.

In this question, we're told that the size of the slit is $5.0\cdot10^{-6}\:m$ , and the first of these dark fringes occurs at an angle of $30^{o}$ away from the slit. We're then asked to determine the wavelength of the light based on this information.

First and foremost, we'll need to use the light diffraction equation:

$sin(\Theta )=n\frac{\lambda }{a}$

In the above equation, $\lambda$ represents the wavelength of light, represents the length of the slit, and represents the location of the dark fringes (it can equal $1,\:2,\:3,...$ ). In this case, is equal to one because we're considering the first dark fringe (on the top/bottom of the central fringe).

Now that we have the equation, we can isolate the wavelength term:

$\lambda =\frac{asin(\Theta )}{n}$

Next all we have to do is plug in the values that we know, and we can solve for the wavelength:

$\lambda=\frac{(5.0\cdot 10^{-6}\: m)(sin(30^{o}))}{1}$

$\lambda=2.5\cdot10^{-6}\:m$

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All AP Physics 2 Resources

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AP Physics 2 : AP Physics 2

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 #11 : Doppler Effect

Example Question #15 : Doppler Effect

Example Question #1 : Slit Experiments

Example Question #1 : Slit Experiments

Example Question #3 : Slit Experiments

Example Question #1 : Slit Experiments

All AP Physics 2 Resources

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