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AP Physics 2 : Doppler Effect

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Example Question #1 : Doppler Effect

A bat is flying towards a stationary wall at a constant speed of $50\frac{m}{s}$ . The bat emits a sound of $125\:Hz$ towards the wall, which is then reflected back at the bat. If the speed of sound in air is $343\frac{m}{s}$ , what is the frequency of sound that the bat experiences?

Possible Answers:

$167.6\:Hz$

$146.3\:Hz$

$125\:Hz$

$143.2\:Hz$

$164.1\:Hz$

Correct answer:

$167.6\:Hz$

Explanation:

To answer this question, it's imperative to realize that we'll need to use the equation for the doppler effect. First, we'll need to calculate the frequency of the sound that reaches the wall. Then, we'll have to calculate the frequency of the reflected wave that reaches the bat.

The doppler effect equation is:

$f_{observed}=f\frac{\left ( v\pm v_{observer}\right )}{\left ( v\pm v_{source}\right )}$

In the first case, we'll consider the frequency received by the wall. The bat is the source in this scenario, which is moving, while the wall is the stationary observer. Therefore, the $v_{observer}$ term in the above equation is 0. Moreover, since the bat is moving towards the wall, we should expect the frequency received by the wall to be larger than the original frequency. Hence, we will need to subtract the speed of the source in the denominator, since that will result in the expected increase in observed frequency.

$f^{'}=f\frac{v}{\left ( v-v_{source}\right )}=125\:Hz\left ( \frac{343\: \frac{m}{s}}{343\: \frac{m}{s}-50\: \frac{m}{s}} \right )=146.3\:Hz$

Now that we have the frequency relfected from the wall, we can calculate the frequency that the bat will experience. In this scenario, the wall is now the source. But because it isn't moving, we can say that the $v_{source}$ term in the doppler equation is 0. Likewise, the bat is now the observer in this case and is still moving at a speed of $50\: \frac{m}{s}$ . Also, because the bat is moving towards the source, then conceptually we should expect the bat to observe a frequency that is greater than that reflected by the wall. To ensure this, we will need to add the $v_{observer}$ term in the numerator of the doppler equation.

$f^{''}=f^{_{'}}\left ( \frac{v+v_{observer}}{v}\right )=146.3\:Hz\left ( \frac{343\: \frac{m}{s}+50\: \frac{m}{s}}{343\: \frac{m}{s}} \right )=167.6\:Hz$

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Example Question #2 : Doppler Effect

If a music box produces a tone of $500 \textup{Hz}$ as a boy is running towards the music box at $2 \frac{m}{s}$ , what is the frequency the boy hears?

$v_{sound\:in\:air} = 343\frac{m}{s}$

Possible Answers:

$480.329 \textup{Hz}$

$498.258\textup{Hz}$

$525.374 \textup{Hz}$

$502.915 \textup{Hz}$

$497.085\textup{Hz}$

Correct answer:

$502.915 \textup{Hz}$

Explanation:

The formula for the Doppler effect of the moving observer is:

$f'=(1\pm \frac{u}{v})f$

Since the boy is approaching, the positive sign will be used. The velocity of sound is $343 \frac{\textup{m}}{\textup{s}}$ . Substitute the knowns into the formula.

$f'=(1+ \frac{2\frac{\textup{m}}{\textup{s}}}{343\frac{\textup{m}}{\textup{s}}})(500 \frac{1}{\textup{s}})=502.915 \textup{Hz}$

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Example Question #1 : Doppler Effect

Suppose a car moves at $15 \frac{\textup{m}}{\textup{s}}$ and produces a $500 \textup{Hz}$ honk. A runner running at $3 \frac{\textup{m}}{\textup{s}}$ approaches the car. About what frequency does the runner hear?

Possible Answers:

$475 \textup{Hz}$

$562\textup{Hz}$

$483\textup{Hz}$

$518 \textup{Hz}$

$527 \textup{Hz}$

Correct answer:

$527 \textup{Hz}$

Explanation:

This scenario deals with both a moving source and a moving observer.

Write the correct Doppler effect formula for this case.

$\large f'=(\frac{1\pm\frac{u_o}{v}}{1\pm\frac{u_s}{v}}) f$

Since the observer and the source are both approaching, the numerator will have a positive sign and the denominator will have a negative sign. The speed of sound is $343 \frac{\textup{m}}{\textup{s}}$ . Substitute all the knowns and find the frequency.

$\large f'=(\frac{1+\frac{3\frac{\textup{m}}{\textup{s}}}{343\frac{\textup{m}}{\textup{s}}}}{1-\frac{15\frac{\textup{m}}{\textup{s}}}{343\frac{\textup{m}}{\textup{s}}}})(500 \textup{ Hz})=(\frac{173}{164})(500)$

$f=527 \textup{ Hz}$

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Example Question #4 : Doppler Effect

Suppose that two cars are moving towards one another, and each is traveling at a speed of $30\: \frac{m}{s}$ . If one of the cars begins to beep its horn at a frequency of 300 Hz , what is the wavelength perceived by the other car?

$v_{sound}=343\frac{m}{s}$

Possible Answers:

105cm

The perceived wavelength will be identical to the source wavelength because the two cars are moving toward one another

95.9cm

357.5cm

Correct answer:

95.9cm

Explanation:

We are being told that two cars are moving towards one another, and one of the cars is emitting a sound at a certain frequency. The other car will, in turn, perceive this sound at a different frequency because both cars are moving relative to one another. Therefore, we can classify this problem as one involving the concept of the Doppler effect.

$f_{observed}=f_{source}\frac{v\pm v_{detector}}{v\pm v_{source}}$

Since the two cars are moving towards one another, we can conclude that the observed frequency should be greater than the source frequency. In order to make that true, we'll need to add in the numerator above, and subtract in the denominator.

$f_{observed}=300Hz\left(\frac{343+30}{343-30}\right)$

$f_{observed}=357.5 Hz$

But we're not done yet. The question is asking for the perceived wavelength, not the perceived frequency. Hence, we'll need to convert frequency into wavelength using the following formula:

$v=f\lambda$

$\lambda =\frac{v}{f}$

$\lambda =\frac{343 \frac{m}{s}}{357.5 s^{-1}}=0.959m=95.9cm$

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Example Question #1 : Doppler Effect

A motorcycle is receding at $15\frac{m}{s}$ . Normally, the exhaust note has frequency 105Hz . Determine the perceived frequency if the speed of sound is $340.9\frac{m}{s}$ .

Possible Answers:

$f_{observed}=100.6Hz$

$f_{observed}=120Hz$

None of these

$f_{observed}=93.1Hz$

$f_{observed}=95.5Hz$

Correct answer:

$f_{observed}=100.6Hz$

Explanation:

Use 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

Plug in values:

$f_{observed}=\frac{340.9}{340.9+15}105$

$f_{observed}=100.6Hz$

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Example Question #6 : Doppler Effect

A train is receding at $15\frac{m}{s}$ 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:

$f_{observed}=390Hz$

None of these

$f_{observed}=410Hz$

$f_{observed}=490Hz$

$f_{observed}=335Hz$

Correct answer:

$f_{observed}=335Hz$

Explanation:

Use 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

Plug in values:

$f_{observed}=\frac{340.9}{340.9+15}350$

$f_{observed}=335Hz$

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Example Question #7 : Doppler Effect

An ambulance is receding at $15\frac{m}{s}$ with it's siren on. Normally, the siren has frequency 700Hz . Determine the perceived frequency if the speed of sound is $340.9\frac{m}{s}$ .

Possible Answers:

$f_{observed}=670Hz$

$f_{observed}=650Hz$

$f_{observed}=470Hz$

$f_{observed}=730Hz$

None of these

Correct answer:

$f_{observed}=670Hz$

Explanation:

Use 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

Plug in values:

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

$f_{observed}=670Hz$

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Example Question #8 : Doppler Effect

A motorcycle is approaching at $55\frac{m}{s}$ . Normally, the exhaust note has frequency 105Hz . Determine the perceived frequency if the speed of sound is $340.9\frac{m}{s}$ .

Possible Answers:

None of these

$f_{observed}=125Hz$

$f_{observed}=131Hz$

$f_{observed}=98Hz$

$f_{observed}=115Hz$

Correct answer:

$f_{observed}=125Hz$

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:

$f_{observed}=\frac{340.9}{340.9-55}105$

$f_{observed}=125Hz$

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Example Question #9 : Doppler Effect

A train is approaching at $35\frac{m}{s}$ 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:

$f_{observed}=360Hz$

None of these

$f_{observed}=377Hz$

$f_{observed}=420Hz$

$f_{observed}=390Hz$

Correct answer:

$f_{observed}=390Hz$

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:

$f_{observed}=\frac{340.9}{340.9-35}350$

$f_{observed}=390Hz$

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Example Question #10 : Doppler Effect

An ambulance is approaching at $15\frac{m}{s}$ with it's siren on. Normally, the siren has frequency 700Hz . Determine the perceived frequency if the speed of sound is $340.9\frac{m}{s}$ .

Possible Answers:

$f_{observed}=711Hz$

None of these

$f_{observed}=668Hz$

$f_{observed}=732Hz$

$f_{observed}=700Hz$

Correct answer:

$f_{observed}=732Hz$

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:

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

$f_{observed}=732Hz$

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AP Physics 2 : Doppler Effect

Study concepts, example questions & explanations for AP Physics 2

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