Waves - AP Physics 2

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Question

A bat is flying towards a stationary wall at a constant speed of $50\frac{m}{s}$ . The bat emits a sound of 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?

Answer

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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Question

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?

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

Answer

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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Question

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?

Answer

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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Question

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 , what is the wavelength perceived by the other car?

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

Answer

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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Question

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

Answer

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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Question

A train is receding at $15\frac{m}{s}$ with it's horn on. Normally, the horn has frequency . Determine the perceived frequency if the speed of sound is $340.9\frac{m}{s}$ .

Answer

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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Question

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

Answer

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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Question

A train is approaching at $35\frac{m}{s}$ with it's horn on. Normally, the horn has frequency . Determine the perceived frequency if the speed of sound is $340.9\frac{m}{s}$ .

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

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 $340.9\ \frac{\textup{m}}{\textup{s}}$ .

Answer

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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Question

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

Answer

Use the following formula:

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

Where

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 :

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

Plug in values:

$(1-\frac{535}{685})3.010^8=v_s$

$6.5710^7\frac{m}{s}=v_s$

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Question

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

Answer

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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Question

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 $340.9\ \frac{\textup{m}}{\textup{s}}$ .

Answer

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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Question

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

Answer

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 $v_{source}$ :

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

$3.010^8-\frac{580}{700}3.010^8=v_s$

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

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Question

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 $340.9\ \frac{\textup{m}}{\textup{s}}$ .

Answer

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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Question

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?

Answer

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:

$\frac{350}{700}=\frac{v}{3.010^8 \frac{m}{s}}$

Solving for

$v=1.510^{8}\frac{m}{s}$

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Question

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?

Answer

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:

$\frac{200}{600}=\frac{v}{3.010^8 \frac{m}{s}}$

Solving for

$v=1.010^{8}\frac{m}{s}$

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Question

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

Answer

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:

$\frac{\Delta \lambda}{\lambda_{0}}=\frac{535-500}{500}=\frac{v}{3.010^8}$ .

$v=2.110^7\frac{m}{s}$ .

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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Question

Consider an electromagnetic wave travelling through a vacuum. If the magnetic field's wave is $3.7 \cdot 10^{-5} T$ , and the speed of light is $3 \cdot 10^8\frac{m}{s}$ , what is the strength of the electric field?

Answer

In an electromagnetic wave, the electric and magnetic portions are proprtional to each other. The ratio of electric field to magnetic field is , the speed of light.

$\frac{E}{B}= c$

Using this information, we can determine the strength of one part given the other.

We have and , so now we multiply the two.

$\begin{align} E&=(3\cdot 10^8)(3.7\cdot 10^{-5}) \ &= 11100 \end{align}$

Therefore, the strength of the electric field is $11100 \frac{N}{C}$ .

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