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MCAT Physical › Sound

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At a local concert, a speaker is set up to produce low-pitched base sounds with a frequency range of 20Hz to 200Hz, which can be modeled as sine waves. In a simplified model, the sound waves the speaker produces can be modeled as a cylindrical pipe with one end closed that travel through the air at a velocity of $v = 331 \frac{m}{s} + 0.6\frac{m}{s}(T)$ , where T is the temperature in °C.

How are the first three harmonics of the base speaker designated?

1, 3, 5

1, 2, 3

1, 2, 5

1, 3, 4

Explanation

First, notice that the paragraph above tells us that the wave can be modeled as a pipe with one end closed. This is in contrast to the other possibility, where the wave is modeled as a pipe with two ends open. It is critical to recognize this difference, as the definition of sequential harmonics and the formula used to calculate them changes depending on whether both ends are open or not. In the situation where one end is closed, the harmonics are odd numbers, meaning that the first three harmonics are 1st, 3rd, and 5th. In the situation where both ends are open or both ends are closed, the harmonics are sequential, meaning that the first three harmonics are 1st, 2nd, and 3rd.

How are the first three harmonics of the base speaker designated?

1, 3, 5

1, 2, 3

1, 2, 5

1, 3, 4

Explanation

Two cars approach each other at $50 {\frac{m}{s}}$ when one car starts to beep its horn at a frequency of 475Hz. What is the wavelength of the horn as heard by the other driver?

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

Explanation

The Doppler equation is:

$f_{o} = f_{s} * \frac{v \pm v_{o}}{v \mp v_{s}}$ .

Because the cars are approaching each other, the frquency heard will be increased. This fundamental knowledge allows you to determine the signs of the equation. The top of the fraction will be addition and the bottom will be subtraction to make a coefficient greater than 1.

$f_{o} = f_{s} * \frac{v + v_{o}}{v - v_{s}}$

Use the given values to solve:

$f_{o} = 475 * \frac{343+50}{343 - 50} = 637 \textup{ Hz}$

Now that we know the frequency, we can solve for the wavelength:

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

$\lambda= \frac{(343 \frac{m}{s})}{637 \textup{ Hz}} = 0.54 \textup{ m}$

Two cars approach each other at $50 {\frac{m}{s}}$ when one car starts to beep its horn at a frequency of 475Hz. What is the wavelength of the horn as heard by the other driver?

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

Explanation

The Doppler equation is:

$f_{o} = f_{s} * \frac{v \pm v_{o}}{v \mp v_{s}}$ .

$f_{o} = f_{s} * \frac{v + v_{o}}{v - v_{s}}$

Use the given values to solve:

$f_{o} = 475 * \frac{343+50}{343 - 50} = 637 \textup{ Hz}$

Now that we know the frequency, we can solve for the wavelength:

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

$\lambda= \frac{(343 \frac{m}{s})}{637 \textup{ Hz}} = 0.54 \textup{ m}$

You are watching a launching space shuttle that produces 10MW of sound from a horizantal distance of 300 meters away. After the shuttle has risen 500 meters straight up from its launch point, what intensity of sound do you percieve?

$4.68 \frac{W}{m^2}$

$7.34 \frac{W}{m^2}$

$23.55 \frac{W}{m^2}$

$0.13 \frac{W}{m^2}$

$16.34\frac{W}{m^2}$

Explanation

As sound radiates from a point source, it does so in concentric circles. Therefore, we can calculate the intensity of sound using the following formula:

$I = \frac{P}{4\pi r^2}$

We simply need to calculate : your distance from the space shuttle.

The shuttle is 500 meters from the ground, and you are 300 meters from its launch point. We can calculate the distance from you to the current position of the shuttle by using the Pythagorean Theorem:

There is no need to take the square root, since it's already in the form we need to do the calculation. Plugging in this value to the original equation, we get:

$I=\frac{10\mathrm{x}10^6W}{2\pi(340,000m^2)} = 4.68 \frac{W}{m^2}$

$4.68 \frac{W}{m^2}$

$7.34 \frac{W}{m^2}$

$23.55 \frac{W}{m^2}$

$0.13 \frac{W}{m^2}$

$16.34\frac{W}{m^2}$

Explanation

As sound radiates from a point source, it does so in concentric circles. Therefore, we can calculate the intensity of sound using the following formula:

$I = \frac{P}{4\pi r^2}$

We simply need to calculate : your distance from the space shuttle.

There is no need to take the square root, since it's already in the form we need to do the calculation. Plugging in this value to the original equation, we get:

$I=\frac{10\mathrm{x}10^6W}{2\pi(340,000m^2)} = 4.68 \frac{W}{m^2}$

A fire truck emitting a siren at moves at $45\frac{m}{s}$ towards a jogger. The jogger is moving at $5\frac{m}{s}$ towards the fire truck. Take the speed of sound to be $345\frac{m}{s}$ .

At what frequency does the jogger perceive the siren?

Explanation

In order to solve this problem we must know how to utilize the Doppler formula.

$f_{o}=f_{s}\left ( \frac{v\pm v_{o}}{v\mp v_{s}} \right )$

$v_{o}$ is the velocity of the observer and $v_{s}$ is the velocity of the source. Notice that the frequency must increase as the observer and source move closer, and therefore the plus sign is used in the numerator and the minus sign is used in the denominator. Had the jogger been moving away from the fire truck, the subtraction function would be used in both the top and bottom.

In this case we see that the source is the fire truck, moving at $45\frac{m}{s}$ , and the observer is the jogger, moving at $5\frac{m}{s}$ . By plugging these numbers into the formula and for $f_{s}$ , we find the perceived frequency or $f_{o}$ to be .

$f_{o}=(1000 Hz)\left ( \frac{345\frac{m}{s}+5\frac{m}{s}}{345\frac{m}{s}-45\frac{m}{s}} \right )$

$f_{o}= (1000 Hz) \left ( \frac{350\frac{m}{s}}{300\frac{m}{s}} \right ) = 1000 Hz \left ( \frac{7\frac{m}{s}}{6\frac{m}{s}} \right )$

$f_{o} = 1,166 Hz$

A fire truck emitting a siren at moves at $45\frac{m}{s}$ towards a jogger. The jogger is moving at $5\frac{m}{s}$ towards the fire truck. Take the speed of sound to be $345\frac{m}{s}$ .

At what frequency does the jogger perceive the siren?

Explanation

In order to solve this problem we must know how to utilize the Doppler formula.

$f_{o}=f_{s}\left ( \frac{v\pm v_{o}}{v\mp v_{s}} \right )$

$f_{o}=(1000 Hz)\left ( \frac{345\frac{m}{s}+5\frac{m}{s}}{345\frac{m}{s}-45\frac{m}{s}} \right )$

$f_{o}= (1000 Hz) \left ( \frac{350\frac{m}{s}}{300\frac{m}{s}} \right ) = 1000 Hz \left ( \frac{7\frac{m}{s}}{6\frac{m}{s}} \right )$

$f_{o} = 1,166 Hz$

Which of the following best describes the effect of the Doppler shift on the appearance of stars moving towards Earth?

They appear more blue

They appear more red

They appear larger

They appear smaller

They appear brighter

Explanation

The Doppler shift equation for light is $\small \small f'=f \sqrt{\frac{1 + \frac{v}{c}}{1 - \frac{v}{c}}}$ , where f is the source frequency, f' is the observed frequency, v is the relative velocity between source and observer, and c is the speed of light.

When the source and observer are moving closer together, v is positive, so the observed frequency is greater than the source frequency. Greater frequency also implies shorter wavelength, so visible light is shifted towards the blue end of the spectrum.

A sound source with a frequency of 790Hz moves away from a stationary observer at a rate of 15m/s. What frequency does the observer hear?

The speed of sound is 340m/s.

757Hz

826Hz

655Hz

775Hz

Explanation

In this scenario the Doppler effect is described by the following equation.
$fo = fs \frac{\left ( v + vo\right )}{\left ( v + vs}\right )}$
Using the values from the problem, we know that vo is zero and vf is 15m/s. v is 340m/s and fs is 790Hz.

$fo = 790 Hz \frac{\left ( 340 m/s + 0 m/s\right )}{\left ( 340 m/s + 15 m/s}\right )} = 757 Hz$

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