General Principles and Properties - MCAT Chemical and Physical Foundations of Biological Systems

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Question

What is the relationship between frequency and period of a sine wave?

Answer

The period of a wave is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

Respectively, frequency is the reciprocal of period. By definition, the product of two reciprocals is one.

$fT=f\frac{1}{f}=1$

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Question

A stretched string of length L, mass M, and tension T is vibrating at its fundamental frequency. Which of the following changes takes place if the vibration frequency of the string increases, but tension and mass density remain constant?

Answer

We can use the equation $v = \sqrt{\frac{T}{\frac{m}{L}}}$ together with $v = f\lambda$ . If T is constant, v cannot change assuming the mass density, m/L, is constant. Thus, $f\lambda$ must be constant; if f increases, $\lambda$ must decrease.

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Question

Picture a transverse wave traveling through water. After the crest of one wave hits a stationary object in the water, an observer counts eight more crests hitting the same object in fifteen seconds. The frequency of the waves is .

Answer

Right away you can rule out the answers with units in seconds, as the unit of frequency is an inverse second, or Hz. Frequency is measured in cycles per second. If eight crests pass a given point in fifteen seconds, the frequency is given by the number of crests divided by the time period.

$\frac{8}{15s} = 0.53 s^{-1}$

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Question

A transverse wave has a velocity of 5.2m/s. If ten cycles pass a given point in 1.6s, what are the wave’s period and wavelength?

Answer

First calculate the frequency of the wave (cycles/sec). The problem tells us that there are ten cycles in 1.6s.

$f = \frac{10}{1.6 s} = 6.25 Hz$

Next find the period by taking the inverse of the frequency.

$\tau = \frac{1}{f} = \frac{1}{6.25 Hz} = 0.16 s$
Finally, find the wavelength by dividing the velocity by the frequency.

$v=f\lambda$
$\lambda = \frac{v}{f} = \frac{5.2 m/s}{6.25 Hz} = 0.83 m$

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Question

What are the frequency and wavelength of a sound wave with a period of 0.04s and a velocity of 575m/s?

Answer

Solve for frequency by taking the inverse of the period.

$f = \frac{1}{\tau } = \frac{1}{0.04s} = 25 Hz$

Next, solve for wavelength by dividing velocity by frequency.

$v=f\lambda$
$\lambda = \frac{v}{f} = \frac{575 m/s}{25 Hz} = 23 m$

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Question

What is the beat frequency if f1 = 200Hz and f2 = 150Hz?

Answer

Beat frequency is the difference between the two frequencies.

200Hz – 150Hz = 50Hz

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Question

What is the frequency of a typical soundwave traveling at 340m/s with a wavelength of 40mm?

Answer

Using the equation $v=\lambda f$ we can find the frequency of the soundwave.

$340\frac{m}{s}=(0.04m) f$

$f=\frac{340\frac{m}{s}}{0.04m} = 8500Hz$

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Question

What is the wavelength of a sound traveling at a frequency of 3000Hz?

Answer

The wavelength of a sound can be found by utilizing the equation, $v=\lambda f$ . where v is the velocity of sound, $\lambda$ is the wavelength, and f is the frequency. You should know that sound normally travels with a speed of 340m/s, unless otherwise stated. With the information given we can find the wavelength of the traveling sound to be 0.11m.

$v=\lambda f$

$\lambda=\frac{v}{f}=\frac{340\frac{m}{s}}{3000s^{-1}}=0.11m$

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Question

An incandescent light bulb is shown through a glass prism. The certain wavlength of the light is then directed into a glass cuvette containing an unknown concentration of protein. Commonly, this process is called spectroscopy and is used to determine the concentrations of DNA, RNA, and proteins in solutions. The indices of reflection of air, glass, and the solution are 1, 1.5, and 1.3, respectively.

What property of light does not change when it enters the prism?

Answer

The frequency of light does not change when it enters a medium with a different index of refraction; in this case, that new medium is the glass of the prism. From the velocity of light equation we know the relationship between velocity and frequency.

$v = \lambda*f$

v is the velocity of light, $\lambda$ is the wavelength, and f is the frequency. When light enters the prism, its velocity changes due to the new index of refraction, but its frequency remains constant.

Because the frequency does not change, we can see that velocity is directly proportional to wavelength; thus, the shorter the wavelength, the slower the velocity. So both wavelength and velocity change when frequency is constant.

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Question

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.

What is the wavelength of a 150Hz beat at 20ºC?

Answer

This question is asking us to determine the wavelength of a sound wave at a specific temperature. From the equation presented to us in the paragraph above, we can calculate the wave velocity.

$v = 331 \frac{m}{s} + 0.6\frac{m}{s}(T)$

$v = 331 \frac{m}{s} + 0.6\frac{m}{s}(20) = 343\frac{m}{s}$

Now, we can use this velocity to calculate the wavelength.

$v=\lambda f; \lambda=\frac{v}{f}=\frac{343\frac{m}{s}}{150 Hz}=2.3m$

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Question

Two tuning forks, with frequencies of 442Hz and 444Hz, are struck and a beat frequency is observed. What is this beat frequency?

Answer

The beat frequency is simply the difference between two frequencies.

$f_B=\left |f_1-f_2 \right |$

We are given the frequency of each tuning fork, so we can use the equation to solve for the beat frequency.

$f_B=\left |442Hz-444Hz \right |=2Hz$

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Question

An electron falls from an excited state to its ground state, emitting a photon at . What is the frequency of the emitted light?

Answer

The relationship between wavelength and frequency is given by the equation:

$v=f\lambda$

In this case, the velocity will be equal to the speed of light.

$v=c=310^8\frac{m}{s}$

Using this value and the given wavelength, we can find the frequency of the photon. Keep in mind that the wavelength must be given in meters.

$310^8\frac{m}{s}=f(7.310^{-7}m)$

$f=\frac{310^8\frac{m}{s}}{7.310^{-7}m}=4.1110^{14}Hz$

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Question

Waves hit a beach every three seconds. The horizontal distance between an adjacent maximum and minimum is one meter. What is the speed of the waves?

Answer

Wave velocity is given by the product of frequency and wavelength:

$v=f\lambda$

In the question, we are given the period (waves per second). To find the frequency, we will need to take the reciprocal of the period.

$v=(\frac{1}{T})\lambda$

Using the values given in the question, we can find the velocity of the waves. The wavelength is twice the distance between adjacent maxima and minima, making our wavelength two meters.

$v=(\frac{1}{3\frac{waves}{s}})(2\frac{m}{wave})=0.67\frac{m}{s}$

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Question

Sound traveling at a velocity, V1, through a certain medium will travel at what velocity through a medium of twice the density?

Answer

The speed of sound depends on both the medium’s density and resistance to compression. We do not have enough information to solve for V2 in terms of V1.

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Question

A wave produced on a string travels with a velocity of $500\frac{m}{s}$ . If the tension on the string is increased by a factor of four, at what speed does the wave travel?

Answer

The velocity of a wave can be obtained with the formula $v=\sqrt{\frac{T}{\mu }}$ , where is the tension in the string and $\mu$ is the mass per unit length of the string. Since the tension is quadrupled, the velocity will be doubled.

Let's assume that a string with tension and a mass per unit length $\mu$ produces a wave with velocity $500\frac{m}{s}$ .

$v_1=500 \frac{m}{s} =\sqrt{\frac{T}{\mu}}$

If we increase the tension by a factor of four, we will get the below expression.

$v_2=\sqrt{\frac{4T}{\mu}}= 2 * \sqrt{\frac{T}{\mu}}$

We can see that , and we know that $v_1=500\frac{m}{s}$ .

$v_1=\sqrt{\frac{T}{\mu}}=500\frac{m}{s}$

$v_2=2\sqrt{\frac{T}{\mu}}=2v_1=2*(500\frac{m}{s})$

$v_2=1000\frac{m}{s}$

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Question

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 does the speed of sound in the summer (30oC) compare to the speed of sound in the winter (9oC)?

Answer

This question asks us to use information provided in the paragraph about how the speed of sound varies with temperature. We can see from the relationship provided that in warmer temperatures the speed of sound is faster. This intuitively makes sense—hotter temperatures mean that air molecules are moving around more, and thus have less resistance to compression or rarefaction by a propagating sound wave. Now that we have a qualitative understanding, we need to compute the ratio of the velocities.

$v = 331 \frac{m}{s} + 0.6\frac{m}{s}(T)$

$v_{30^oc} = 331 \frac{m}{s} + 0.6\frac{m}{s}(30) = 349\frac{m}{s}$

$v_{9^oC} = 331 \frac{m}{s} + 0.6\frac{m}{s}(9) = 336.4\frac{m}{s}$

$ratio= \frac{v_{30^oC}}{v_{9^oC}}=\frac{349\frac{m}{s}}{336.4\frac{m}{s}}=1.04$

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Question

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 long would it take for a 30Hz beat to reach an audience member 100m away when the ambient temperature is 21ºC?

Answer

The question asks us to determine how long it will take for a wave beat to reach an audience member at 100m away; thus, we need to calculate the velocity of the wave to determine the time.

We know from kinematics that $v = \frac{d}{t}$ . This can be rearranged to solve for t: $t = \frac{d}{v}$ .

$v = 331 \frac{m}{s} + 0.6\frac{m}{s}(T)$

$v = 331 \frac{m}{s} + 0.6\frac{m}{s}(21) = 343.6\frac{m}{s}$

$t = \frac{d}{v}=\frac{100m}{343.6\frac{m}{s}}=0.29s$

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Question

Through which of the following would you expect a photon to travel fastest?

Answer

A photon will travel fastest through a vacuum. Photons are generally massless and can be thought of as a light wave, which travels fastest in a vacuum and slowest through a metal or solid. This can be visualized using the concept of the index of refraction, which describes the speed of light through air compared to the speed through other mediums. A vacuum will be the least dense and cause the least hindrance to a photon as it travels, thus giving it the lowest index of refraction and allowing the fastest speed of light.

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Question

You are out snorkling off the coast of an exotic island when a pod of whales comes swimming by. The pod is 100m away. If they emit sounds underwater with an average frequency of 2200Hz and there are 500 complete sound waves between you and the pod, how fast is the speed of sound in the water?

$v_{s_{air}}=340 \frac{m}{s}$

Answer

If there are 500 waves over a distance of 100 meters, we can say that the wavelength is:

$\lambda = \frac{100m}{500} = 0.2 m$

Now we can use the formula for the speed of waves:

$v = \lambda f = (0.2m)(2200s^{-1}) = 440 \frac{m}{s}$

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Question

What is the frequency of a typical soundwave traveling at 340m/s with a wavelength of 40mm?

Answer

Using the equation $v=\lambda f$ we can find the frequency of the soundwave.

$340\frac{m}{s}=(0.04m) f$

$f=\frac{340\frac{m}{s}}{0.04m} = 8500Hz$

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